我如何写一个列表文件?writelines()不插入换行符,所以我需要这样做:

f.writelines([f"{line}\n" for line in lines])

当前回答

使用循环:

with open('your_file.txt', 'w') as f:
    for line in lines:
        f.write(f"{line}\n")

对于Python <3.6:

with open('your_file.txt', 'w') as f:
    for line in lines:
        f.write("%s\n" % line)

对于Python 2,还可以使用:

with open('your_file.txt', 'w') as f:
    for line in lines:
        print >> f, line

如果您热衷于单个函数调用,至少要删除方括号[],以便每次生成一个要打印的字符串(genexp而不是listcomp)——没有理由占用物化整个字符串列表所需的所有内存。

其他回答

如果在python3上,还可以使用print函数,如下所示。

f = open("myfile.txt","wb")
print(mylist, file=f)

简单:

with open("text.txt", 'w') as file:
    file.write('\n'.join(yourList))

序列化列表到文本文件与逗号分隔值

mylist = dir()
with open('filename.txt','w') as f:
    f.write( ','.join( mylist ) )

我最近发现Path很有用。帮助我绕过不得不打开('file')作为f,然后写入文件。希望这对某些人有用:)。

from pathlib import Path
import json
a = [[1,2,3],[4,5,6]]
# write
Path("file.json").write_text(json.dumps(a))
# read
json.loads(Path("file.json").read_text())

将stdout重定向到一个文件也可以用于此目的:

from contextlib import redirect_stdout
with open('test.txt', 'w') as f:
  with redirect_stdout(f):
     for i in range(mylst.size):
        print(mylst[i])