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

另一种迭代和添加换行符的方法:

for item in items:
    filewriter.write(f"{item}" + "\n")

其他回答

with open ("test.txt","w")as fp:
   for line in list12:
       fp.write(line+"\n")

你为什么不试试

file.write(str(list))

你还可以通过以下步骤:

例子:

my_list=[1,2,3,4,5,"abc","def"]
with open('your_file.txt', 'w') as file:
    for item in my_list:
        file.write("%s\n" % item)

输出:

在your_file.txt中,项目的保存方式如下:

1

2

3

4

5

abc

def

您的脚本也按上述方式保存。

否则,你可以用泡菜

import pickle
my_list=[1,2,3,4,5,"abc","def"]
#to write
with open('your_file.txt', 'wb') as file:
    pickle.dump(my_list, file)
#to read
with open ('your_file.txt', 'rb') as file:
    Outlist = pickle.load(file)
print(Outlist)

输出: [1,2,3,4,5, 'abc', 'def']

它保存转储列表,就像一个列表,当我们加载它时,我们能够读取。

同样由simplejson可能输出与上面相同的结果

import simplejson as sj
my_list=[1,2,3,4,5,"abc","def"]
#To write
with open('your_file.txt', 'w') as file:
    sj.dump(my_list, file)

#To save
with open('your_file.txt', 'r') as file:
    mlist=sj.load(file)
print(mlist)

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 

简单:

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