Question

说我有一个空列表myNames = []

如何在每行上打开一个带有名称的文件，并将每个名称读入列表？

像：

>     names.txt
>     dave
>     jeff
>     ted
>     myNames = [dave,jeff,ted]

Answer 1

阅读documentation：

with open('names.txt', 'r') as f:
    myNames = f.readlines()

其他人已经提供了如何摆脱换行符的答案。

<强>更新：

Fred Larson在他的评论中提供了一个很好的解决方案：

with open('names.txt', 'r') as f:
    myNames = [line.strip() for line in f]

Answer 2

f = open('file.txt','r')

for line in f:
    myNames.append(line.strip()) # We don't want newlines in our list, do we?

Answer 3

names=[line.strip() for line in open('names.txt')]

Answer 4

#function call
read_names(names.txt)
#function def
def read_names(filename): 
with open(filename, 'r') as fileopen:
    name_list = [line.strip() for line in fileopen]
    print (name_list)

Answer 5

这应该是map和lambda的好例子

with open ('names.txt','r') as f :
   Names = map (lambda x : x.strip(),f_in.readlines())

我的立场得到了纠正（或至少得到了改善）。列表理解更加优雅

with open ('names.txt','r') as f :
    Names = [name.rstrip() for name in f]

Answer 6

Names = []
for line in open('names.txt','r').readlines():
    Names.append(line.strip())

strip（）在字符串之前和之后剪切空格......

Answer 7

读取文件并将每行放入列表的pythonic方法：

from __future__ import with_statement #for python 2.5
Names = []
with open('C:/path/txtfile.txt', 'r') as f:
    lines = f.readlines()
    Names.append(lines.strip())

Python，将txt文件中的项添加到列表中

7 个答案: