Question

我有一长串以下表格列表---

a = [[1.2,'abc',3],[1.2,'werew',4],........,[1.4,'qew',2]]

即。列表中的值是不同的类型 - 浮点数，整数，字符串。如何将其写入csv文件，以便我的输出csv文件看起来像

1.2,abc,3
1.2,werew,4
.
.
.
1.4,qew,2

Answer 1

Python内置的CSV module可以轻松处理这个问题：

import csv

with open("output.csv", "wb") as f:
    writer = csv.writer(f)
    writer.writerows(a)

这假设您的列表定义为a，就像您的问题一样。您可以通过各种可选参数将输出CSV的确切格式调整为csv.writer()，如上面链接的库参考页中所述。

Answer 2

import csv
with open(file_path, 'a') as outcsv:   
    #configure writer to write standard csv file
    writer = csv.writer(outcsv, delimiter=',', quotechar='|', quoting=csv.QUOTE_MINIMAL, lineterminator='\n')
    writer.writerow(['number', 'text', 'number'])
    for item in list:
        #Write item to outcsv
        writer.writerow([item[0], item[1], item[2]])

官方文档：http://docs.python.org/2/library/csv.html

Answer 3

您可以使用pandas：

In [1]: import pandas as pd

In [2]: a = [[1.2,'abc',3],[1.2,'werew',4],[1.4,'qew',2]]

In [3]: my_df = pd.DataFrame(a)

In [4]: my_df.to_csv('my_csv.csv', index=False, header=False)

Answer 4

在我非常大的列表中使用csv.writer花了很长时间。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")

你可以改变一些事情以制作更好的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")

Answer 5

如果出于某种原因您想手动执行此操作（不使用csv，pandas，numpy等模块）：

with open('myfile.csv','w') as f:
    for sublist in mylist:
        for item in sublist:
            f.write(item + ',')
        f.write('\n')

当然，滚动自己的版本可能容易出错并且效率低下......这通常就是为什么有一个模块。但有时候写自己的东西可以帮助你理解它们是如何工作的，有时它会更容易。

Answer 6

Ambers的解决方案也适用于numpy数组：

from pylab import *
import csv

array_=arange(0,10,1)
list_=[array_,array_*2,array_*3]
with open("output.csv", "wb") as f:
    writer = csv.writer(f)
    writer.writerows(list_)

Answer 7

确保在创建作者时指出lineterinator='\n';否则，当数据源来自其他csv文件时，可能会在每个数据行之后将额外的空行写入文件...

这是我的解决方案：

with open('csvfile', 'a') as csvfile:
    spamwriter = csv.writer(csvfile, delimiter='    ',quotechar='|', quoting=csv.QUOTE_MINIMAL, lineterminator='\n')
for i in range(0, len(data)):
    spamwriter.writerow(data[i])

Answer 8

如何将列表列表转储到pickle中并用pickle module进行恢复？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]]]
>>>

Answer 9

如果您不想为此导入var log = console.log; var console = {}; console.log = log; log('foo');模块，则可以仅使用Python内置插件将列表列表写入csv文件中。

<Alter ObjectExpansion="ExpandFull" xmlns="http://schemas.microsoft.com/analysisservices/2003/engine">
    <Object>
        <DatabaseID>YourOlapDbNameHere</DatabaseID>
    </Object>
    <ObjectDefinition>
        <Database xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:ddl2="http://schemas.microsoft.com/analysisservices/2003/engine/2" xmlns:ddl2_2="http://schemas.microsoft.com/analysisservices/2003/engine/2/2" xmlns:ddl100_100="http://schemas.microsoft.com/analysisservices/2008/engine/100/100" xmlns:ddl200="http://schemas.microsoft.com/analysisservices/2010/engine/200" xmlns:ddl200_200="http://schemas.microsoft.com/analysisservices/2010/engine/200/200" xmlns:ddl300="http://schemas.microsoft.com/analysisservices/2011/engine/300" xmlns:ddl300_300="http://schemas.microsoft.com/analysisservices/2011/engine/300/300" xmlns:ddl400="http://schemas.microsoft.com/analysisservices/2012/engine/400" xmlns:ddl400_400="http://schemas.microsoft.com/analysisservices/2012/engine/400/400" xmlns:ddl500="http://schemas.microsoft.com/analysisservices/2013/engine/500" xmlns:ddl500_500="http://schemas.microsoft.com/analysisservices/2013/engine/500/500">
            <ID>YourOlapDbNameHere</ID>
            <Name>YourOlapDbNameHere</Name>
            <ddl200:CompatibilityLevel>1100</ddl200:CompatibilityLevel>
            <Language>1033</Language>
            <Collation>Latin1_General_CI_AS</Collation>
            <DataSourceImpersonationInfo>
                <ImpersonationMode>Default</ImpersonationMode>

Answer 10

在csv.writer函数中使用 newline 参数跟踪示例时，出现错误消息。以下代码对我有用。

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

Answer 11

如果将 lists 的 lists 的 lll list 导出到 .csv，这将适用于 Python3： >

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

将Python列表列表写入csv文件

11 个答案: