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Update perceptron_dichotomy.py #25

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Update perceptron_dichotomy.py #25

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8 changes: 4 additions & 4 deletions perceptron/perceptron_dichotomy.py
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Original file line number Diff line number Diff line change
Expand Up @@ -29,16 +29,16 @@ def loadData(fileName):
fr = open(fileName, 'r')
# 将文件按行读取
for line in fr.readlines():
# 对每一行数据按切割福','进行切割，返回字段列表
# 对每一行数据按切割符','进行切割，返回字段列表
curLine = line.strip().split(',')

# Mnsit有0-9是个标记，由于是二分类任务，所以将>=5的作为1，<5为-1
# Mnist有0-9是个标记，由于是二分类任务，所以将>=5的作为1，<5为-1
if int(curLine[0]) >= 5:
labelArr.append(1)
else:
labelArr.append(-1)
#存放标记
#[int(num) for num in curLine[1:]] -> 遍历每一行中除了以第一哥元素（标记）外将所有元素转换成int类型
#[int(num) for num in curLine[1:]] -> 遍历每一行中除了以第一个元素（标记）外将所有元素转换成int类型
#[int(num)/255 for num in curLine[1:]] -> 将所有数据除255归一化(非必须步骤，可以不归一化)
dataArr.append([int(num)/255 for num in curLine[1:]])

Expand All @@ -58,7 +58,7 @@ def perceptron(dataArr, labelArr, iter=50):
#转换后的数据中每一个样本的向量都是横向的
dataMat = np.mat(dataArr)
#将标签转换成矩阵，之后转置(.T为转置)。
#转置是因为在运算中需要单独取label中的某一个元素，如果是1xN的矩阵的话，无法用label[i]的方式读取
#转置是因为在运算中需要单独取label中的某一个元素，如果不是1xN的矩阵的话，无法用label[i]的方式读取
#对于只有1xN的label可以不转换成矩阵，直接label[i]即可，这里转换是为了格式上的统一
labelMat = np.mat(labelArr).T
#获取数据矩阵的大小，为m*n
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