在这个页面上甚至没有看到一个回答，包括如何包括头以及创建文件。这里有一种方法也包含了这一点。Method在python 3中工作得很好

csv_filename = 'abc.csv'
fieldnames = ['Col_1','Col_2','Col_3','Col_4']
with open(csv_filename, mode='w') as csv_file:
    writer = csv.DictWriter(csv_file, fieldnames=fieldnames)
    writer.writeheader()
    for i in my_list:
        writer.writerow({fieldnames[0]: i[0], fieldnames[1]: i[1], fieldnames[2]: i[2],fieldnames[3]: i[3]}) 

在将列表的列表的列表的列表导出为.csv的情况下，这将在Python3中工作:

import csv
with open("output.csv", "w") as f:
    writer = csv.writer(f)
    for element in lll:
        writer.writerows(element)

我在csv中使用换行参数跟随示例时得到了一个错误消息。作家的功能。下面的代码适用于我。

 with open(strFileName, "w") as f:
    writer = csv.writer(f, delimiter=',',  quoting=csv.QUOTE_MINIMAL)
    writer.writerows(result)

如果您不想为此导入csv模块，您可以只使用Python内置程序将列表的列表写入csv文件

with open("output.csv", "w") as f:
    for row in a:
        f.write("%s\n" % ','.join(str(col) for col in row))

如何转储列表的列表到pickle和恢复它与pickle模块?这很方便。

>>> import pickle
>>> 
>>> mylist = [1, 'foo', 'bar', {1, 2, 3}, [ [1,4,2,6], [3,6,0,10]]]
>>> with open('mylist', 'wb') as f:
...     pickle.dump(mylist, f) 


>>> with open('mylist', 'rb') as f:
...      mylist = pickle.load(f)
>>> mylist
[1, 'foo', 'bar', {1, 2, 3}, [[1, 4, 2, 6], [3, 6, 0, 10]]]
>>> 

使用csv。在我庞大的列表中，作者花了相当长的时间。我决定使用熊猫，它更快，更容易控制和理解:

 import pandas

 yourlist = [[...],...,[...]]
 pd = pandas.DataFrame(yourlist)
 pd.to_csv("mylist.csv")

好的部分是你可以改变一些东西来创建一个更好的csv文件:

 yourlist = [[...],...,[...]]
 columns = ["abcd","bcde","cdef"] #a csv with 3 columns
 index = [i[0] for i in yourlist] #first element of every list in yourlist
 not_index_list = [i[1:] for i in yourlist]
 pd = pandas.DataFrame(not_index_list, columns = columns, index = index)

 #Now you have a csv with columns and index:
 pd.to_csv("mylist.csv")