将文本文件(逐行)拆分为不同的文件

Question

使用 python 逐行查找数据拆分器

• 正则表达式？
• 包含？

作为示例文件"文件"包含：

X
X
Y
Z
Z
Z

我需要一种干净的方式来根据字母将此文件拆分为 3 个不同的文件

作为示例：

def split_by_platform(FILE_NAME):
with open(FILE_NAME, "r+") as infile:
Data = infile.read()
If the file contains "X"
write to x.txt
If the file contains "Y"
write to y.txt
If the file contains "Z"
write to z.txt

x.txt文件将如下所示：

X
X

y.txt文件将如下所示：

Y

z.txt文件将如下所示：

Z
Z
Z

Answer 1

EDIT感谢@bruno desthuilliers，他提醒我正确的方式：

遍历文件对象(不是"readlines"(：

def split_by_platform(FILE_NAME, out1, out2, out3):
with open(FILE_NAME, "r") as infile, open(out1, 'a') as of1, open(out2, 'a') as of2, open(out3, 'a') as of3:
for line in infile:
if "X" in line:
of1.write(line)
elif "Y" in line:
of2.write(line)
elif "Z" in line:
of3.write(line)

编辑提示@dim： 以下是任意长度的标志字符列表的更通用方法：

def loop(infilename, flag_chars):
with open(infilename, 'r') as infile:
for line in infile:
for c in flag_chars:
if c in line:
with open(c+'.txt', 'a') as outfile:
outfile.write(line)            

Answer 2

这应该可以做到：

with open('my_text_file.txt') as infile, open('x.txt', 'w') as x, open('y.txt', 'w') as y, open('z.txt', 'w') as z:
for line in infile:
if line.startswith('X'):
x.write(line)
elif line.startswith('Y'):
y.write(line)
elif line.startswith('Z'):
z.write(line)

Answer 3

以下是执行相同工作的更通用方法：

from collections import Counter
with open("file.txt", "r+") as file:
data = file.read().splitlines()
counter = Counter(data)
array2d = [[key, ] * value for key, value in counter.items()]
print array2d # [['Y'], ['X', 'X'], ['Z', 'Z', 'Z']]
for el in array2d:
with open(str(el[0]) + ".txt", "w") as f:
[f.write(e + "n") for e in el]

上面的代码将生成具有相应值的X.txt、Y.txt和Z.txt。例如，如果您有多个C字母，则代码将生成一个文件C.txt。