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

file.write('\n'.join(list))

其他回答

在一般情况下

下面是writelines()方法的语法

fileObject.writelines( sequence )

例子

#!/usr/bin/python

# Open a file
fo = open("foo.txt", "rw+")
seq = ["This is 6th line\n", "This is 7th line"]

# Write sequence of lines at the end of the file.
line = fo.writelines( seq )

# Close opend file
fo.close()

参考

http://www.tutorialspoint.com/python/file_writelines.htm

更简单的是:

with open("outfile", "w") as outfile:
    outfile.write("\n".join(itemlist))

要确保项目列表中的所有项目都是字符串,请使用生成器表达式:

with open("outfile", "w") as outfile:
    outfile.write("\n".join(str(item) for item in itemlist))

记住,项目列表占用内存,所以要注意内存消耗。

使用numpy。Savetxt也是一个选项:

import numpy as np

np.savetxt('list.txt', list, delimiter="\n", fmt="%s")

这个逻辑首先将list中的项转换为字符串(str)。有时列表包含一个元组,如

alist = [(i12,tiger), 
(113,lion)]

这个逻辑将把每个元组写入一个新行。我们可以在读取文件时加载每个元组时使用eval:

outfile = open('outfile.txt', 'w') # open a file in write mode
for item in list_to_persistence:    # iterate over the list items
   outfile.write(str(item) + '\n') # write to the file
outfile.close()   # close the file 

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。