需要为每个字符串创建一个数组数组答案

【问题标题】：Need to create an array of arrays for each string需要为每个字符串创建一个数组数组
【发布时间】：2018-12-12 08:11:47
【问题描述】：

所以我的数据看起来像这样：

>> print data
"12345","string1","string2","string3","string4","string5"
"67890","string6","string7","string8","string9","string10"

我希望能够将每一行作为一个数组放入一个数组中。所以最后，我希望它看起来像这样：

>> print array_data
[["12345","string1","string2","string3","string4","string5"],
["67890","string6","string7","string8","string9","string10"]]

我已经尝试了一些东西，这是我得到的最接近的：

>> temp_list = []
>> for line in data.splitlines(): temp_list.append([line])
>> print temp_list
[['"12345","string1","string2","string3","string4","string5"'],
['"67890","string6","string7","string8","string9","string10"']]

我在每个数组周围都得到了那个单引号。我应该如何继续获得我正在寻找的结果？

提前致谢！

【问题讨论】：

如果您仍然感到困惑，请查看 Python 中引号如何表示字符串：docs.python.org/2.0/ref/strings.html

标签： python arrays string python-2.7 split

【解决方案1】：

您可以使用ast 和split

设置

import ast
d = '''
"12345","string1","string2","string3","string4","string5"
"67890","string6","string7","string8","string9","string10"
'''
templist=[]

那么就

for line in d.strip().splitlines(): 
    templist.append([ast.literal_eval(st) for st in line.split(',')])

请注意，您也可以使用列表推导式制作单行器

templist = [[ast.literal_eval(st) for st in line.split(',')] for line in d.strip().splitlines()]

【讨论】：

【解决方案2】：

使用熊猫：

import pandas as pd

d = '''\
"12345","string1","string2","string3","string4","string5"
"67890","string6","string7","string8","string9","string10"'''

l = pd.read_csv(pd.compat.StringIO(d), header=None, dtype='object').values

你有 l:

[['12345', 'string1', 'string2', 'string3', 'string4', 'string5'],
 ['67890', 'string6', 'string7', 'string8', 'string9', 'string10']]

【讨论】：

【解决方案3】：

使用标准库中的csv 和io 模块：

import csv
from io import StringIO

d = '''"12345","string1","string2","string3","string4","string5"
"67890","string6","string7","string8","string9","string10"'''

res = list(csv.reader(StringIO(d)))

print(res)

[['12345', 'string1', 'string2', 'string3', 'string4', 'string5'],
 ['67890', 'string6', 'string7', 'string8', 'string9', 'string10']]

【讨论】：

【解决方案4】：

给定：

>>> data
'"12345","string1","string2","string3","string4","string5"\n"67890","string6","string7","string8","string9","string10"'

您可以这样做（如果' 没有另外指明可能包含, 的csv 字段！）：

>>> [line.replace('"','').split(',') for line in data.splitlines()]
[['12345', 'string1', 'string2', 'string3', 'string4', 'string5'], ['67890', 'string6', 'string7', 'string8', 'string9', 'string10']]

或者您可以使用csv 正确处理字段中的逗号：

>>> import csv
>>> [list(e) for e in csv.reader(data.splitlines())]
[['12345', 'string1', 'string2', 'string3', 'string4', 'string5'], ['67890', 'string6', 'string7', 'string8', 'string9', 'string10']]

【讨论】：

【解决方案5】：

你快到了。

>> temp_list = []
>> temp_list = [list(eval(line)) for line in data.splitlines()]
>> print temp_list
[['12345', 'string1', 'string2', 'string3', 'string4', 'string5'], ['67890', 'string6', 'string7', 'string8', 'string9', 'string10']]

【讨论】：

如果您对某个答案投了反对票，请同时说明原因。