Exercises

Try to solve exercises in every chapter using only the features discussed until that chapter. Some of the exercises will be easier to solve with techniques presented in later chapters, but the aim of these exercises is to explore the features presented so far.

For solutions, see Exercise_solutions.md.

re introduction

a) Check whether the given strings contain 0xB0. Display a boolean result as shown below.

>>> line1 = 'start address: 0xA0, func1 address: 0xC0'
>>> line2 = 'end address: 0xFF, func2 address: 0xB0'

>>> bool(re.search(r'', line1))     ##### add your solution here
False
>>> bool(re.search(r'', line2))     ##### add your solution here
True

b) Replace all occurrences of 5 with five for the given string.

>>> ip = 'They ate 5 apples and 5 oranges'

>>> re.sub()        ##### add your solution here
'They ate five apples and five oranges'

c) Replace first occurrence of 5 with five for the given string.

>>> ip = 'They ate 5 apples and 5 oranges'

>>> re.sub()       ##### add your solution here
'They ate five apples and 5 oranges'

d) For the given list, filter all elements that do not contain e.

>>> items = ['goal', 'new', 'user', 'sit', 'eat', 'dinner']

>>> [w for w in items if not re.search()]        ##### add your solution here
['goal', 'sit']

e) Replace all occurrences of note irrespective of case with X.

>>> ip = 'This note should not be NoTeD'

>>> re.sub()        ##### add your solution here
'This X should not be XD'

f) Check if at is present in the given byte input data.

>>> ip = b'tiger imp goat'

>>> bool(re.search())     ##### add your solution here
True

g) For the given input string, display all lines not containing start irrespective of case.

>>> para = '''good start
... Start working on that
... project you always wanted
... stars are shining brightly
... hi there
... start and try to
... finish the book
... bye'''

>>> pat = re.compile()      ##### add your solution here
>>> for line in para.split('\n'):
...     if not pat.search(line):
...         print(line)
... 
project you always wanted
stars are shining brightly
hi there
finish the book
bye

h) For the given list, filter all elements that contains either a or w.

>>> items = ['goal', 'new', 'user', 'sit', 'eat', 'dinner']

##### add your solution here
>>> [w for w in items if re.search() or re.search()]
['goal', 'new', 'eat']

i) For the given list, filter all elements that contains both e and n.

>>> items = ['goal', 'new', 'user', 'sit', 'eat', 'dinner']

##### add your solution here
>>> [w for w in items if re.search() and re.search()]
['new', 'dinner']

j) For the given string, replace 0xA0 with 0x7F and 0xC0 with 0x1F.

>>> ip = 'start address: 0xA0, func1 address: 0xC0'

##### add your solution here
'start address: 0x7F, func1 address: 0x1F'

Anchors

a) Check if the given strings start with be.

>>> line1 = 'be nice'
>>> line2 = '"best!"'
>>> line3 = 'better?'
>>> line4 = 'oh no\nbear spotted'

>>> pat = re.compile()       ##### add your solution here

>>> bool(pat.search(line1))
True
>>> bool(pat.search(line2))
False
>>> bool(pat.search(line3))
True
>>> bool(pat.search(line4))
False

b) For the given input string, change only whole word red to brown

>>> words = 'bred red spread credible'

>>> re.sub()     ##### add your solution here
'bred brown spread credible'

c) For the given input list, filter all elements that contains 42 surrounded by word characters.

>>> words = ['hi42bye', 'nice1423', 'bad42', 'cool_42a', 'fake4b']

>>> [w for w in words if re.search()]   ##### add your solution here
['hi42bye', 'nice1423', 'cool_42a']

d) For the given input list, filter all elements that start with den or end with ly.

>>> items = ['lovely', '1\ndentist', '2 lonely', 'eden', 'fly\n', 'dent']

>>> [e for e in items if ]        ##### add your solution here
['lovely', '2 lonely', 'dent']

e) For the given input string, change whole word mall to 1234 only if it is at the start of a line.

>>> para = '''\
... ball fall wall tall
... mall call ball pall
... wall mall ball fall
... mallet wallet malls'''

>>> print(re.sub())    ##### add your solution here
ball fall wall tall
1234 call ball pall
wall mall ball fall
mallet wallet malls

f) For the given list, filter all elements having a line starting with den or ending with ly.

>>> items = ['lovely', '1\ndentist', '2 lonely', 'eden', 'fly\nfar', 'dent']

##### add your solution here
['lovely', '1\ndentist', '2 lonely', 'fly\nfar', 'dent']

g) For the given input list, filter all whole elements 12\nthree irrespective of case.

>>> items = ['12\nthree\n', '12\nThree', '12\nthree\n4', '12\nthree']
##### add your solution here
['12\nThree', '12\nthree']

h) For the given input list, replace hand with X for all elements that start with hand followed by at least one word character.

>>> items = ['handed', 'hand', 'handy', 'unhanded', 'handle', 'hand-2']

##### add your solution here
['Xed', 'hand', 'Xy', 'unhanded', 'Xle', 'hand-2']

i) For the given input list, filter all elements starting with h. Additionally, replace e with X for these filtered elements.

>>> items = ['handed', 'hand', 'handy', 'unhanded', 'handle', 'hand-2']

##### add your solution here
['handXd', 'hand', 'handy', 'handlX', 'hand-2']

Alternation and Grouping

a) For the given input list, filter all elements that start with den or end with ly

>>> items = ['lovely', '1\ndentist', '2 lonely', 'eden', 'fly\n', 'dent']

##### add your solution here
['lovely', '2 lonely', 'dent']

b) For the given list, filter all elements having a line starting with den or ending with ly.

>>> items = ['lovely', '1\ndentist', '2 lonely', 'eden', 'fly\nfar', 'dent']

##### add your solution here
['lovely', '1\ndentist', '2 lonely', 'fly\nfar', 'dent']

c) For the given input strings, replace all occurrences of removed or reed or received or refused with X.

>>> s1 = 'creed refuse removed read'
>>> s2 = 'refused reed redo received'

>>> pat = re.compile()        ##### add your solution here

>>> pat.sub('X', s1)
'cX refuse X read'
>>> pat.sub('X', s2)
'X X redo X'

d) For the given input strings, replace all matches from the list words with A.

>>> s1 = 'plate full of slate'
>>> s2 = "slated for later, don't be late"
>>> words = ['late', 'later', 'slated']

>>> pat = re.compile()        ##### add your solution here

>>> pat.sub('A', s1)
'pA full of sA'
>>> pat.sub('A', s2)
"A for A, don't be A"

e) Filter all whole elements from the input list items based on elements listed in words.

>>> items = ['slate', 'later', 'plate', 'late', 'slates', 'slated ']
>>> words = ['late', 'later', 'slated']

>>> pat = re.compile()       ##### add your solution here

##### add your solution here
['later', 'late']

Escaping metacharacters

a) Transform the given input strings to the expected output using same logic on both strings.

>>> str1 = '(9-2)*5+qty/3'
>>> str2 = '(qty+4)/2-(9-2)*5+pq/4'

##### add your solution here for str1
'35+qty/3'
##### add your solution here for str2
'(qty+4)/2-35+pq/4'

b) Replace (4)\| with 2 only at the start or end of given input strings.

>>> s1 = r'2.3/(4)\|6 foo 5.3-(4)\|'
>>> s2 = r'(4)\|42 - (4)\|3'
>>> s3 = 'two - (4)\\|\n'

>>> pat = re.compile()        ##### add your solution here

>>> pat.sub('2', s1)
'2.3/(4)\\|6 foo 5.3-2'
>>> pat.sub('2', s2)
'242 - (4)\\|3'
>>> pat.sub('2', s3)
'two - (4)\\|\n'

c) Replace any matching element from the list items with X for given the input strings. Match the elements from items literally. Assume no two elements of items will result in any matching conflict.

>>> items = ['a.b', '3+n', r'x\y\z', 'qty||price', '{n}']
>>> pat = re.compile()      ##### add your solution here

>>> pat.sub('X', '0a.bcd')
'0Xcd'
>>> pat.sub('X', 'E{n}AMPLE')
'EXAMPLE'
>>> pat.sub('X', r'43+n2 ax\y\ze')
'4X2 aXe'

d) Replace backspace character \b with a single space character for the given input string.

>>> ip = '123\b456'
>>> ip
'123\x08456'
>>> print(ip)
12456

>>> re.sub()        ##### add your solution here
'123 456'

e) Replace all occurrences of \e with e.

>>> ip = r'th\er\e ar\e common asp\ects among th\e alt\ernations'

>>> re.sub()     ##### add your solution here
'there are common aspects among the alternations'

f) Replace any matching item from the list eqns with X for given the string ip. Match the items from eqns literally.

>>> ip = '3-(a^b)+2*(a^b)-(a/b)+3'
>>> eqns = ['(a^b)', '(a/b)', '(a^b)+2']

##### add your solution here

>>> pat.sub('X', ip)
'3-X*X-X+3'

Dot metacharacter and Quantifiers

Since . metacharacter doesn't match newline character by default, assume that the input strings in the following exercises will not contain newline characters.

a) Replace 42//5 or 42/5 with 8 for the given input.

>>> ip = 'a+42//5-c pressure*3+42/5-14256'

>>> re.sub()      ##### add your solution here
'a+8-c pressure*3+8-14256'

b) For the list items, filter all elements starting with hand and ending with at most one more character or le.

>>> items = ['handed', 'hand', 'handled', 'handy', 'unhand', 'hands', 'handle']

##### add your solution here
['hand', 'handy', 'hands', 'handle']

c) Use re.split to get the output as shown for the given input strings.

>>> eqn1 = 'a+42//5-c'
>>> eqn2 = 'pressure*3+42/5-14256'
>>> eqn3 = 'r*42-5/3+42///5-42/53+a'

##### add your solution here for eqn1
['a+', '-c']
##### add your solution here for eqn2
['pressure*3+', '-14256']
##### add your solution here for eqn3
['r*42-5/3+42///5-', '3+a']

d) For the given input strings, remove everything from the first occurrence of i till end of the string.

>>> s1 = 'remove the special meaning of such constructs'
>>> s2 = 'characters while constructing'

>>> pat = re.compile()        ##### add your solution here

>>> pat.sub('', s1)
'remove the spec'
>>> pat.sub('', s2)
'characters wh'

e) For the given strings, construct a RE to get output as shown.

>>> str1 = 'a+b(addition)'
>>> str2 = 'a/b(division) + c%d(#modulo)'
>>> str3 = 'Hi there(greeting). Nice day(a(b)'

>>> remove_parentheses = re.compile()     ##### add your solution here

>>> remove_parentheses.sub('', str1)
'a+b'
>>> remove_parentheses.sub('', str2)
'a/b + c%d'
>>> remove_parentheses.sub('', str3)
'Hi there. Nice day'

f) Correct the given RE to get the expected output.

>>> words = 'plink incoming tint winter in caution sentient'
>>> change = re.compile(r'int|in|ion|ing|inco|inter|ink')

# wrong output
>>> change.sub('X', words)
'plXk XcomXg tX wXer X cautX sentient'

# expected output
>>> change = re.compile()       ##### add your solution here
>>> change.sub('X', words)
'plX XmX tX wX X cautX sentient'

g) For the given greedy quantifiers, what would be the equivalent form using {m,n} representation?

? is same as
* is same as
+ is same as

h) (a*|b*) is same as (a|b)* — True or False?

i) For the given input strings, remove everything from the first occurrence of test (irrespective of case) till end of the string, provided test isn't at the end of the string.

>>> s1 = 'this is a Test'
>>> s2 = 'always test your RE for corner cases'
>>> s3 = 'a TEST of skill tests?'

>>> pat = re.compile()      ##### add your solution here

>>> pat.sub('', s1)
'this is a Test'
>>> pat.sub('', s2)
'always '
>>> pat.sub('', s3)
'a '

j) For the input list words, filter all elements starting with s and containing e and t in any order.

>>> words = ['sequoia', 'subtle', 'exhibit', 'asset', 'sets', 'tests', 'site']

##### add your solution here
['subtle', 'sets', 'site']

k) For the input list words, remove all elements having less than 6 characters.

>>> words = ['sequoia', 'subtle', 'exhibit', 'asset', 'sets', 'tests', 'site']

##### add your solution here
['sequoia', 'subtle', 'exhibit']

l) For the input list words, filter all elements starting with s or t and having a maximum of 6 characters.

>>> words = ['sequoia', 'subtle', 'exhibit', 'asset', 'sets', 'tests', 'site']

##### add your solution here
['subtle', 'sets', 'tests', 'site']

m) Can you reason out why this code results in the output shown? The aim was to remove all <characters> patterns but not the <> ones. The expected result was 'a 1<> b 2<> c'.

>>> ip = 'a<apple> 1<> b<bye> 2<> c<cat>'

>>> re.sub(r'<.+?>', '', ip)
'a 1 2'

n) Use re.split to get the output as shown below for given input strings.

>>> s1 = 'go there  //   "this // that"'
>>> s2 = 'a//b // c//d e//f // 4//5'
>>> s3 = '42// hi//bye//see // carefully'

>>> pat = re.compile()     ##### add your solution here

>>> pat.split()     ##### add your solution here for s1
['go there', '"this // that"']
>>> pat.split()     ##### add your solution here for s2
['a//b', 'c//d e//f // 4//5']
>>> pat.split()     ##### add your solution here for s3
['42// hi//bye//see', 'carefully']

Working with matched portions

a) For the given strings, extract the matching portion from first is to last t.

>>> str1 = 'This the biggest fruit you have seen?'
>>> str2 = 'Your mission is to read and practice consistently'

>>> pat = re.compile()     ##### add your solution here

##### add your solution here for str1
'is the biggest fruit'
##### add your solution here for str2
'ission is to read and practice consistent'

b) Find the starting index of first occurrence of is or the or was or to for the given input strings.

>>> s1 = 'match after the last newline character'
>>> s2 = 'and then you want to test'
>>> s3 = 'this is good bye then'
>>> s4 = 'who was there to see?'

>>> pat = re.compile()      ##### add your solution here

##### add your solution here for s1
12
##### add your solution here for s2
4
##### add your solution here for s3
2
##### add your solution here for s4
4

c) Find the starting index of last occurrence of is or the or was or to for the given input strings.

>>> s1 = 'match after the last newline character'
>>> s2 = 'and then you want to test'
>>> s3 = 'this is good bye then'
>>> s4 = 'who was there to see?'

>>> pat = re.compile()      ##### add your solution here

##### add your solution here for s1
12
##### add your solution here for s2
18
##### add your solution here for s3
17
##### add your solution here for s4
14

d) The given input string contains : exactly once. Extract all characters after the : as output.

>>> ip = 'fruits:apple, mango, guava, blueberry'

##### add your solution here
'apple, mango, guava, blueberry'

e) The given input strings contains some text followed by - followed by a number. Replace that number with its log value using math.log().

>>> s1 = 'first-3.14'
>>> s2 = 'next-123'

>>> pat = re.compile()      ##### add your solution here

>>> import math
>>> pat.sub()     ##### add your solution here for s1
'first-1.144222799920162'
>>> pat.sub()     ##### add your solution here for s2
'next-4.812184355372417'

f) Replace all occurrences of par with spar, spare with extra and park with garden for the given input strings.

>>> str1 = 'apartment has a park'
>>> str2 = 'do you have a spare cable'
>>> str3 = 'write a parser'

##### add your solution here

>>> pat.sub()        ##### add your solution here for str1
'aspartment has a garden'
>>> pat.sub()        ##### add your solution here for str2
'do you have a extra cable'
>>> pat.sub()        ##### add your solution here for str3
'write a sparser'

g) Extract all words between ( and ) from the given input string as a list. Assume that the input will not contain any broken parentheses.

>>> ip = 'another (way) to reuse (portion) matched (by) capture groups'

>>> re.findall()        ##### add your solution here
['way', 'portion', 'by']

h) Extract all occurrences of < up to next occurrence of >, provided there is at least one character in between < and >.

>>> ip = 'a<apple> 1<> b<bye> 2<> c<cat>'

>>> re.findall()        ##### add your solution here
['<apple>', '<> b<bye>', '<> c<cat>']

i) Use re.findall to get the output as shown below for the given input strings. Note the characters used in the input strings carefully.

>>> row1 = '-2,5 4,+3 +42,-53 4356246,-357532354 '
>>> row2 = '1.32,-3.14 634,5.63 63.3e3,9907809345343.235 '

>>> pat = re.compile()       ##### add your solution here

>>> pat.findall(row1)
[('-2', '5'), ('4', '+3'), ('+42', '-53'), ('4356246', '-357532354')]
>>> pat.findall(row2)
[('1.32', '-3.14'), ('634', '5.63'), ('63.3e3', '9907809345343.235')]

j) This is an extension to previous question.

For row1, find the sum of integers of each tuple element. For example, sum of -2 and 5 is 3.
For row2, find the sum of floating-point numbers of each tuple element. For example, sum of 1.32 and -3.14 is -1.82.

>>> row1 = '-2,5 4,+3 +42,-53 4356246,-357532354 '
>>> row2 = '1.32,-3.14 634,5.63 63.3e3,9907809345343.235 '

# should be same as previous question
>>> pat = re.compile()       ##### add your solution here

##### add your solution here for row1
[3, 7, -11, -353176108]

##### add your solution here for row2
[-1.82, 639.63, 9907809408643.234]

k) Use re.split to get the output as shown below.

>>> ip = '42:no-output;1000:car-truck;SQEX49801'

>>> re.split()        ##### add your solution here
['42', 'output', '1000', 'truck', 'SQEX49801']

l) For the given list of strings, change the elements into a tuple of original element and number of times t occurs in that element.

>>> words = ['sequoia', 'attest', 'tattletale', 'asset']

##### add your solution here
[('sequoia', 0), ('attest', 3), ('tattletale', 4), ('asset', 1)]

m) The given input string has fields separated by :. Each field contains four uppercase alphabets followed optionally by two digits. Ignore the last field, which is empty. See docs.python: Match.groups and use re.finditer to get the output as shown below. If the optional digits aren't present, show 'NA' instead of None.

>>> ip = 'TWXA42:JWPA:NTED01:'

##### add your solution here
[('TWXA', '42'), ('JWPA', 'NA'), ('NTED', '01')]

Note that this is different from re.findall which will just give empty string instead of None when a capture group doesn't participate.

n) Convert the comma separated strings to corresponding dict objects as shown below.

>>> row1 = 'name:rohan,maths:75,phy:89,'
>>> row2 = 'name:rose,maths:88,phy:92,'

>>> pat = re.compile()      ##### add your solution here

##### add your solution here for row1
{'name': 'rohan', 'maths': '75', 'phy': '89'}
##### add your solution here for row2
{'name': 'rose', 'maths': '88', 'phy': '92'}

Character class

a) For the list items, filter all elements starting with hand and ending with s or y or le.

>>> items = ['-handy', 'hand', 'handy', 'unhand', 'hands', 'handle']

##### add your solution here
['handy', 'hands', 'handle']

b) Replace all whole words reed or read or red with X.

>>> ip = 'redo red credible :read: rod reed'

##### add your solution here
'redo X credible :X: rod X'

c) For the list words, filter all elements containing e or i followed by l or n. Note that the order mentioned should be followed.

>>> words = ['surrender', 'unicorn', 'newer', 'door', 'empty', 'eel', 'pest']

##### add your solution here
['surrender', 'unicorn', 'eel']

d) For the list words, filter all elements containing e or i and l or n in any order.

>>> words = ['surrender', 'unicorn', 'newer', 'door', 'empty', 'eel', 'pest']

##### add your solution here
['surrender', 'unicorn', 'newer', 'eel']

e) Extract all hex character sequences, with 0x optional prefix. Match the characters case insensitively, and the sequences shouldn't be surrounded by other word characters.

>>> str1 = '128A foo 0xfe32 34 0xbar'
>>> str2 = '0XDEADBEEF place 0x0ff1ce bad'

>>> hex_seq = re.compile()        ##### add your solution here

##### add your solution here for str1
['128A', '0xfe32', '34']

##### add your solution here for str2
['0XDEADBEEF', '0x0ff1ce', 'bad']

f) Delete from ( to the next occurrence of ) unless they contain parentheses characters in between.

>>> str1 = 'def factorial()'
>>> str2 = 'a/b(division) + c%d(#modulo) - (e+(j/k-3)*4)'
>>> str3 = 'Hi there(greeting). Nice day(a(b)'

>>> remove_parentheses = re.compile()      ##### add your solution here

>>> remove_parentheses.sub('', str1)
'def factorial'
>>> remove_parentheses.sub('', str2)
'a/b + c%d - (e+*4)'
>>> remove_parentheses.sub('', str3)
'Hi there. Nice day(a'

g) For the list words, filter all elements not starting with e or p or u.

>>> words = ['surrender', 'unicorn', 'newer', 'door', 'empty', 'eel', 'pest']

##### add your solution here
['surrender', 'newer', 'door']

h) For the list words, filter all elements not containing u or w or ee or -.

>>> words = ['p-t', 'you', 'tea', 'heel', 'owe', 'new', 'reed', 'ear']

##### add your solution here
['tea', 'ear']

i) The given input strings contain fields separated by , and fields can be empty too. Replace last three fields with WHTSZ323.

>>> row1 = '(2),kite,12,,D,C,,'
>>> row2 = 'hi,bye,sun,moon'

>>> pat = re.compile()      ##### add your solution here

>>> pat.sub()       ##### add your solution here for row1
'(2),kite,12,,D,WHTSZ323'
>>> pat.sub()       ##### add your solution here for row2
'hi,WHTSZ323'

j) Split the given strings based on consecutive sequence of digit or whitespace characters.

>>> str1 = 'lion \t Ink32onion Nice'
>>> str2 = '**1\f2\n3star\t7 77\r**'

>>> pat = re.compile()       ##### add your solution here

>>> pat.split(str1)
['lion', 'Ink', 'onion', 'Nice']
>>> pat.split(str2)
['**', 'star', '**']

k) Delete all occurrences of the sequence <characters> where characters is one or more non > characters and cannot be empty.

>>> ip = 'a<apple> 1<> b<bye> 2<> c<cat>'

##### add your solution here
'a 1<> b 2<> c'

l) \b[a-z](on|no)[a-z]\b is same as \b[a-z][on]{2}[a-z]\b. True or False? Sample input lines shown below might help to understand the differences, if any.

>>> print('known\nmood\nknow\npony\ninns')
known
mood
know
pony
inns

m) For the given list, filter all elements containing any number sequence greater than 624.

>>> items = ['hi0000432abcd', 'car00625', '42_624 0512', '3.14 96 2 foo1234baz']

##### add your solution here
['car00625', '3.14 96 2 foo1234baz']

n) Count the maximum depth of nested braces for the given strings. Unbalanced or wrongly ordered braces should return -1. Note that this will require a mix of regular expressions and Python code.

>>> def max_nested_braces(ip):
##### add your solution here

>>> max_nested_braces('a*b')
0
>>> max_nested_braces('}a+b{')
-1
>>> max_nested_braces('a*b+{}')
1
>>> max_nested_braces('{{a+2}*{b+c}+e}')
2
>>> max_nested_braces('{{a+2}*{b+{c*d}}+e}')
3
>>> max_nested_braces('{{a+2}*{\n{b+{c*d}}+e*d}}')
4
>>> max_nested_braces('a*{b+c*{e*3.14}}}')
-1

o) By default, str.split method will split on whitespace and remove empty strings from the result. Which re module function would you use to replicate this functionality?

>>> ip = ' \t\r  so  pole\t\t\t\n\nlit in to \r\n\v\f  '

>>> ip.split()
['so', 'pole', 'lit', 'in', 'to']
##### add your solution here
['so', 'pole', 'lit', 'in', 'to']

p) Convert the given input string to two different lists as shown below.

>>> ip = 'price_42 roast^\t\n^-ice==cat\neast'

##### add your solution here
['price_42', 'roast', 'ice', 'cat', 'east']

##### add your solution here
['price_42', ' ', 'roast', '^\t\n^-', 'ice', '==', 'cat', '\n', 'east']

q) Filter all elements whose first non-whitespace character is not a # character. Any element made up of only whitespace characters should be ignored as well.

>>> items = ['    #comment', '\t\napple #42', '#oops', 'sure', 'no#1', '\t\r\f']

##### add your solution here
['\t\napple #42', 'sure', 'no#1']

Groupings and backreferences

a) Replace the space character that occurs after a word ending with a or r with a newline character.

>>> ip = 'area not a _a2_ roar took 22'

>>> print(re.sub())      ##### add your solution here
area
not a
_a2_ roar
took 22

b) Add [] around words starting with s and containing e and t in any order.

>>> ip = 'sequoia subtle exhibit asset sets tests site'

##### add your solution here
'sequoia [subtle] exhibit asset [sets] tests [site]'

c) Replace all whole words with X that start and end with the same word character. Single character word should get replaced with X too, as it satisfies the stated condition.

>>> ip = 'oreo not a _a2_ roar took 22'

##### add your solution here
'X not X X X took X'

d) Convert the given markdown headers to corresponding anchor tag. Consider the input to start with one or more # characters followed by space and word characters. The name attribute is constructed by converting the header to lowercase and replacing spaces with hyphens. Can you do it without using a capture group?

>>> header1 = '# Regular Expressions'
>>> header2 = '## Compiling regular expressions'

##### add your solution here for header1
'# <a name="regular-expressions"></a>Regular Expressions'
##### add your solution here for header2
'## <a name="compiling-regular-expressions"></a>Compiling regular expressions'

e) Convert the given markdown anchors to corresponding hyperlinks.

>>> anchor1 = '# <a name="regular-expressions"></a>Regular Expressions'
>>> anchor2 = '## <a name="subexpression-calls"></a>Subexpression calls'

##### add your solution here for anchor1
'[Regular Expressions](#regular-expressions)'
##### add your solution here for anchor2
'[Subexpression calls](#subexpression-calls)'

f) Count the number of whole words that have at least two occurrences of consecutive repeated alphabets. For example, words like stillness and Committee should be counted but not words like root or readable or rotational.

>>> ip = '''oppressed abandon accommodation bloodless
... carelessness committed apparition innkeeper
... occasionally afforded embarrassment foolishness
... depended successfully succeeded
... possession cleanliness suppress'''

##### add your solution here
13

g) For the given input string, replace all occurrences of digit sequences with only the unique non-repeating sequence. For example, 232323 should be changed to 23 and 897897 should be changed to 897. If there no repeats (for example 1234) or if the repeats end prematurely (for example 12121), it should not be changed.

>>> ip = '1234 2323 453545354535 9339 11 60260260'

##### add your solution here
'1234 23 4535 9339 1 60260260'

h) Replace sequences made up of words separated by : or . by the first word of the sequence. Such sequences will end when : or . is not followed by a word character.

>>> ip = 'wow:Good:2_two:five: hi-2 bye kite.777.water.'

##### add your solution here
'wow hi-2 bye kite'

i) Replace sequences made up of words separated by : or . by the last word of the sequence. Such sequences will end when : or . is not followed by a word character.

>>> ip = 'wow:Good:2_two:five: hi-2 bye kite.777.water.'

##### add your solution here
'five hi-2 bye water'

j) Split the given input string on one or more repeated sequence of cat.

>>> ip = 'firecatlioncatcatcatbearcatcatparrot'

##### add your solution here
['fire', 'lion', 'bear', 'parrot']

k) For the given input string, find all occurrences of digit sequences with at least one repeating sequence. For example, 232323 and 897897. If the repeats end prematurely, for example 12121, it should not be matched.

>>> ip = '1234 2323 453545354535 9339 11 60260260'

>>> pat = re.compile()      ##### add your solution here

# entire sequences in the output
##### add your solution here
['2323', '453545354535', '11']

# only the unique sequence in the output
##### add your solution here
['23', '4535', '1']

l) Convert the comma separated strings to corresponding dict objects as shown below. The keys are name, maths and phy for the three fields in the input strings.

>>> row1 = 'rohan,75,89'
>>> row2 = 'rose,88,92'

>>> pat = re.compile()      ##### add your solution here

##### add your solution here for row1
{'name': 'rohan', 'maths': '75', 'phy': '89'}
##### add your solution here for row2
{'name': 'rose', 'maths': '88', 'phy': '92'}

m) Surround all whole words with (). Additionally, if the whole word is imp or ant, delete them. Can you do it with single substitution?

>>> ip = 'tiger imp goat eagle ant important'

##### add your solution here
'(tiger) () (goat) (eagle) () (important)'

n) Filter all elements that contains a sequence of lowercase alphabets followed by - followed by digits. They can be optionally surrounded by {{ and }}. Any partial match shouldn't be part of the output.

>>> ip = ['{{apple-150}}', '{{mango2-100}}', '{{cherry-200', 'grape-87']

##### add your solution here
['{{apple-150}}', 'grape-87']

o) The given input string has sequences made up of words separated by : or . and such sequences will end when : or . is not followed by a word character. For all such sequences, display only the last word followed by - followed by first word.

>>> ip = 'wow:Good:2_two:five: hi-2 bye kite.777.water.'

##### add your solution here
['five-wow', 'water-kite']

Lookarounds

Please use lookarounds for solving the following exercises even if you can do it without lookarounds. Unless you cannot use lookarounds for cases like variable length lookbehinds.

a) Replace all whole words with X unless it is preceded by ( character.

>>> ip = '(apple) guava berry) apple (mango) (grape'

##### add your solution here
'(apple) X X) X (mango) (grape'

b) Replace all whole words with X unless it is followed by ) character.

>>> ip = '(apple) guava berry) apple (mango) (grape'

##### add your solution here
'(apple) X berry) X (mango) (X'

c) Replace all whole words with X unless it is preceded by ( or followed by ) characters.

>>> ip = '(apple) guava berry) apple (mango) (grape'

##### add your solution here
'(apple) X berry) X (mango) (grape'

d) Extract all whole words that do not end with e or n.

>>> ip = 'at row on urn e note dust n'

##### add your solution here
['at', 'row', 'dust']

e) Extract all whole words that do not start with a or d or n.

>>> ip = 'at row on urn e note dust n'

##### add your solution here
['row', 'on', 'urn', 'e']

f) Extract all whole words only if they are followed by : or , or -.

>>> ip = 'poke,on=-=so:ink.to/is(vast)ever-sit'

##### add your solution here
['poke', 'so', 'ever']

g) Extract all whole words only if they are preceded by = or / or -.

>>> ip = 'poke,on=-=so:ink.to/is(vast)ever-sit'

##### add your solution here
['so', 'is', 'sit']

h) Extract all whole words only if they are preceded by = or : and followed by : or ..

>>> ip = 'poke,on=-=so:ink.to/is(vast)ever-sit'

##### add your solution here
['so', 'ink']

i) Extract all whole words only if they are preceded by = or : or . or ( or - and not followed by . or /.

>>> ip = 'poke,on=-=so:ink.to/is(vast)ever-sit'

##### add your solution here
['so', 'vast', 'sit']

j) Remove leading and trailing whitespaces from all the individual fields where , is the field separator.

>>> csv1 = ' comma  ,separated ,values \t\r '
>>> csv2 = 'good bad,nice  ice  , 42 , ,   stall   small'

>>> remove_whitespace = re.compile()     ##### add your solution here

>>> remove_whitespace.sub('', csv1)
'comma,separated,values'
>>> remove_whitespace.sub('', csv2)
'good bad,nice  ice,42,,stall   small'

k) Filter all elements that satisfy all of these rules:

should have at least two alphabets
should have at least 3 digits
should have at least one special character among % or * or # or $
should not end with a whitespace character

>>> pwds = ['hunter2', 'F2H3u%9', '*X3Yz3.14\t', 'r2_d2_42', 'A $B C1234']

##### add your solution here
['F2H3u%9', 'A $B C1234']

l) For the given string, surround all whole words with {} except for whole words par and cat and apple.

>>> ip = 'part; cat {super} rest_42 par scatter apple spar'

##### add your solution here
'{part}; cat {{super}} {rest_42} par {scatter} apple {spar}'

m) Extract integer portion of floating-point numbers for the given string. A number ending with . and no further digits should not be considered.

>>> ip = '12 ab32.4 go 5 2. 46.42 5'

##### add your solution here
['32', '46']

n) For the given input strings, extract all overlapping two character sequences.

>>> s1 = 'apple'
>>> s2 = '1.2-3:4'

>>> pat = re.compile()       ##### add your solution here

##### add your solution here for s1
['ap', 'pp', 'pl', 'le']
##### add your solution here for s2
['1.', '.2', '2-', '-3', '3:', ':4']

o) The given input strings contain fields separated by : character. Delete : and the last field if there is a digit character anywhere before the last field.

>>> s1 = '42:cat'
>>> s2 = 'twelve:a2b'
>>> s3 = 'we:be:he:0:a:b:bother'

>>> pat = re.compile()      ##### add your solution here

>>> pat.sub()       ##### add your solution here for s1
'42'
>>> pat.sub()       ##### add your solution here for s2
'twelve:a2b'
>>> pat.sub()       ##### add your solution here for s3
'we:be:he:0:a:b'

p) Extract all whole words unless they are preceded by : or <=> or ---- or #.

>>> ip = '::very--at<=>row|in.a_b#b2c=>lion----east'

##### add your solution here
['at', 'in', 'a_b', 'lion']

q) Match strings if it contains qty followed by price but not if there is whitespace or the string error between them.

>>> str1 = '23,qty,price,42'
>>> str2 = 'qty price,oh'
>>> str3 = '3.14,qty,6,errors,9,price,3'
>>> str4 = '42\nqty-6,apple-56,price-234,error'
>>> str5 = '4,price,3.14,qty,4'

>>> neg = re.compile()       ##### add your solution here

>>> bool(neg.search(str1))
True
>>> bool(neg.search(str2))
False
>>> bool(neg.search(str3))
False
>>> bool(neg.search(str4))
True
>>> bool(neg.search(str5))
False

r) Can you reason out why the output shown is different for these two regular expressions?

>>> ip = 'I have 12, he has 2!'

>>> re.sub(r'\b..\b', '{\g<0>}', ip)
'{I }have {12}{, }{he} has{ 2}!'

>>> re.sub(r'(?<!\w)..(?!\w)', '{\g<0>}', ip)
'I have {12}, {he} has {2!}'

Flags

a) Remove from first occurrence of hat to last occurrence of it for the given input strings. Match these markers case insensitively.

>>> s1 = 'But Cool THAT\nsee What okay\nwow quite'
>>> s2 = 'it this hat is sliced HIT.'

>>> pat = re.compile()       ##### add your solution here

>>> pat.sub('', s1)
'But Cool Te'
>>> pat.sub('', s2)
'it this .'

b) Delete from start if it is at the beginning of a line up to the next occurrence of the end at the end of a line. Match these markers case insensitively.

>>> para = '''\
... good start
... start working on that
... project you always wanted
... to, do not let it end
... hi there
... start and end the end
... 42
... Start and try to
... finish the End
... bye'''

>>> pat = re.compile()        ##### add your solution here

>>> print(pat.sub('', para))
good start

hi there

42

bye

c) For the given input strings, match all of these three patterns:

This case sensitively
nice and cool case insensitively

>>> s1 = 'This is nice and Cool'
>>> s2 = 'Nice and cool this is'
>>> s3 = 'What is so nice and cool about This?'

>>> pat = re.compile()       ##### add your solution here

>>> bool(pat.search(s1))
True
>>> bool(pat.search(s2))
False
>>> bool(pat.search(s3))
True

d) For the given input strings, match if the string begins with Th and also contains a line that starts with There.

>>> s1 = 'There there\nHave a cookie'
>>> s2 = 'This is a mess\nYeah?\nThereeeee'
>>> s3 = 'Oh\nThere goes the fun'

>>> pat = re.compile()     ##### add your solution here

>>> bool(pat.search(s1))
True
>>> bool(pat.search(s2))
True
>>> bool(pat.search(s3))
False

e) Explore what the re.DEBUG flag does. Here's some example patterns to check out.

re.compile(r'\Aden|ly\Z', flags=re.DEBUG)
re.compile(r'\b(0x)?[\da-f]+\b', flags=re.DEBUG)
re.compile(r'\b(?:0x)?[\da-f]+\b', flags=re.I|re.DEBUG)

Unicode

a) Output True or False depending on input string made up of ASCII characters or not. Consider the input to be non-empty strings and any character that isn't part of 7-bit ASCII set should give False. Do you need regular expressions for this?

>>> str1 = '123—456'
>>> str2 = 'good fοοd'
>>> str3 = 'happy learning!'
>>> str4 = 'İıſK'

##### add your solution here for str1
False
##### add your solution here for str2
False
##### add your solution here for str3
True
##### add your solution here for str4
False

b) Does . quantifier with re.ASCII flag enabled match non-ASCII characters?

c) Explore the following Q&A threads.

stackoverflow: remove powered number from string
stackoverflow: regular expression for French characters

regex module

a) Filter all elements whose first non-whitespace character is not a # character. Any element made up of only whitespace characters should be ignored as well.

>>> items = ['    #comment', '\t\napple #42', '#oops', 'sure', 'no#1', '\t\r\f']

##### add your solution here
['\t\napple #42', 'sure', 'no#1']

b) Replace sequences made up of words separated by : or . by the first word of the sequence and the separator. Such sequences will end when : or . is not followed by a word character.

>>> ip = 'wow:Good:2_two:five: hi bye kite.777.water.'

##### add your solution here
'wow: hi bye kite.'

c) The given list of strings has fields separated by : character. Delete : and the last field if there is a digit character anywhere before the last field.

>>> items = ['42:cat', 'twelve:a2b', 'we:be:he:0:a:b:bother']

##### add your solution here
['42', 'twelve:a2b', 'we:be:he:0:a:b']

d) Extract all whole words unless they are preceded by : or <=> or ---- or #.

>>> ip = '::very--at<=>row|in.a_b#b2c=>lion----east'

##### add your solution here
['at', 'in', 'a_b', 'lion']

e) The given input string has fields separated by : character. Extract all fields if the previous field contains a digit character.

>>> ip = 'vast:a2b2:ride:in:awe:b2b:3list:end'

##### add your solution here
['ride', '3list', 'end']

f) The given input string has fields separated by : character. Delete all fields, including the separator, unless the field contains a digit character. Stop deleting once a field with digit character is found.

>>> row1 = 'vast:a2b2:ride:in:awe:b2b:3list:end'
>>> row2 = 'um:no:low:3e:s4w:seer'

>>> pat = regex.compile()      ##### add your solution here

>>> pat.sub('', row1)
'a2b2:ride:in:awe:b2b:3list:end'
>>> pat.sub('', row2)
'3e:s4w:seer'

g) For the given input strings, extract if followed by any number of nested parentheses. Assume that there will be only one such pattern per input string.

>>> ip1 = 'for (((i*3)+2)/6) if(3-(k*3+4)/12-(r+2/3)) while()'
>>> ip2 = 'if+while if(a(b)c(d(e(f)1)2)3) for(i=1)'

>>> pat = regex.compile()       ##### add your solution here

>>> pat.search(ip1)[0]
'if(3-(k*3+4)/12-(r+2/3))'
>>> pat.search(ip2)[0]
'if(a(b)c(d(e(f)1)2)3)'

h) Read about POSIX flag from https://pypi.org/project/regex/. Is the following code snippet showing the correct output?

>>> words = 'plink incoming tint winter in caution sentient'

>>> change = regex.compile(r'int|in|ion|ing|inco|inter|ink', flags=regex.POSIX)

>>> change.sub('X', words)
'plX XmX tX wX X cautX sentient'

i) Extract all whole words for the given input strings. However, based on user input ignore, do not match words if they contain any character present in the ignore variable.

>>> s1 = 'match after the last newline character'
>>> s2 = 'and then you want to test'

>>> ignore = 'aty'
>>> regex.findall()     ##### add your solution here for s1
['newline']
>>> regex.findall()     ##### add your solution here for s2
[]

>>> ignore = 'esw'
>>> regex.findall()     ##### add your solution here for s1
['match']
>>> regex.findall()     ##### add your solution here for s2
['and', 'you', 'to']

j) Retain only punctuation characters for the given strings (generated from codepoints). Use Unicode character set definition for punctuation for solving this exercise.

>>> s1 = ''.join(chr(c) for c in range(0, 0x80))
>>> s2 = ''.join(chr(c) for c in range(0x80, 0x100))
>>> s3 = ''.join(chr(c) for c in range(0x2600, 0x27ec))

>>> pat = regex.compile()       ##### add your solution here

>>> pat.sub('', s1)
'!"#%&\'()*,-./:;?@[\\]_{}'
>>> pat.sub('', s2)
'¡§«¶·»¿'
>>> pat.sub('', s3)
'❨❩❪❫❬❭❮❯❰❱❲❳❴❵⟅⟆⟦⟧⟨⟩⟪⟫'

k) For the given markdown file, replace all occurrences of the string python (irrespective of case) with the string Python. However, any match within code blocks that start with whole line ```python and end with whole line ``` shouldn't be replaced. Consider the input file to be small enough to fit memory requirements.

Refer to github: exercises folder for files sample.md and expected.md required to solve this exercise.

>>> ip_str = open('sample.md', 'r').read()
>>> pat = regex.compile()      ##### add your solution here
>>> with open('sample_mod.md', 'w') as op_file:
...     ##### add your solution here
... 
305
>>> assert open('sample_mod.md').read() == open('expected.md').read()

l) For the given input strings, construct a word that is made up of last characters of all the words in the input. Use last character of last word as first character, last character of last but one word as second character and so on.

>>> s1 = 'knack tic pi roar what'
>>> s2 = '42;rod;t2t2;car'

>>> pat = regex.compile()       ##### add your solution here

##### add your solution here for s1
'trick'
##### add your solution here for s2
'r2d2'

m) Replicate str.rpartition functionality with regular expressions. Split into three parts based on last match of sequences of digits, which is 777 and 12 for the given input strings.

>>> s1 = 'Sample123string42with777numbers'
>>> s2 = '12apples'

##### add your solution here for s1
['Sample123string42with', '777', 'numbers']
##### add your solution here for s2
['', '12', 'apples']

n) Read about fuzzy matching on https://pypi.org/project/regex/. For the given input strings, return True if they are exactly same as cat or there is exactly one character difference. Ignore case when comparing differences. For example, Ca2 should give True. act will be False even though the characters are same because position should be maintained.

>>> pat = regex.compile()       ##### add your solution here

>>> bool(pat.fullmatch('CaT'))
True
>>> bool(pat.fullmatch('scat'))
False
>>> bool(pat.fullmatch('ca.'))
True
>>> bool(pat.fullmatch('ca#'))
True
>>> bool(pat.fullmatch('c#t'))
True
>>> bool(pat.fullmatch('at'))
False
>>> bool(pat.fullmatch('act'))
False
>>> bool(pat.fullmatch('2a1'))
False

Provide feedback

Saved searches

Use saved searches to filter your results more quickly

Exercises.md

Exercises.md

Exercises

re introduction

Anchors

Alternation and Grouping

Escaping metacharacters

Dot metacharacter and Quantifiers

Working with matched portions

Character class

Groupings and backreferences

Lookarounds

Flags

Unicode

regex module

Files

Exercises.md

Latest commit

History

Exercises.md

File metadata and controls

Exercises

re introduction

Anchors

Alternation and Grouping

Escaping metacharacters

Dot metacharacter and Quantifiers

Working with matched portions

Character class

Groupings and backreferences

Lookarounds

Flags

Unicode

regex module