列表中的regex筛选器后输出中的分隔符

Question

我对蟒蛇很陌生，所以请容忍我。

我有一份名单-

print(listoffiles)
["{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n"]

我只需要名单上的文件名，

0_HONOR_1524204351030, 0_HONOR_1524204351030,0_HONOR_1524204351030

我先把它分了-

listoffiles=[i.split(":")[0] for i in listoffiles]
print(listoffiles)
["{'0_HONOR_1524204351030'", "{'0_HONOR_1524204351030'", "{'0_HONOR_1524204351030'"]

然后加入它-

listoffiles=(','.join(str(f) for f in listoffiles))
print(listoffiles)
{'0_HONOR_1524204351030',{'0_HONOR_1524204351030',{'0_HONOR_1524204351030'

在这里，我试着脱光它-

listoffiles=(str(listoffiles).strip("'{'"))
print(listoffiles)
0_HONOR_1524204351030',{'0_HONOR_1524204351030',{'0_HONOR_1524204351030'

这似乎只适用于第一个元素，所以我也试过了-

listoffiles=(str(f).strip("'{'") for f in listoffiles)
print(listoffiles)
<generator object <genexpr> at 0x7fda38837460>

因为这也不起作用，所以我试着用正则表达式-

regex = re.compile(r'^[a-zA-Z0-9_]*$')
newlistoffiles=filter(regex.match,listoffiles)

这个看起来很管用，但我无法把输出分开-

0_HONOR_15242043510300_HONOR_15242043510300_HONOR_1524204351030

我需要帮助把它们分开。我张贴了整个问题，以便我可以得到任何其他更好和聪明的方式来完成这项任务的建议。

Answer 1

使用ast模块将string转换为python，并将keys()转换为所需的键。

Ex:

import ast
listoffiles = ["{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n"]
listoffiles = map(ast.literal_eval, listoffiles)
print([i.keys()[0] for i in listoffiles])

输出：

['0_HONOR_1524204351030', '0_HONOR_1524204351030', '0_HONOR_1524204351030', '0_HONOR_1524204351030', '0_HONOR_1524204351030']

Answer 2

您可以使用拆分来分离数据：

#!/usr/bin/env python

Data = ["{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n", "{'0_HONOR_1524204351030': 'NEW'}\n"]

for i in Data:
    i = i.strip()
    i = i[2:]
    i = i[0:i.index("'")]
    print(i)

将印刷：

0_HONOR_1524204351030
0_HONOR_1524204351030
0_HONOR_1524204351030
0_HONOR_1524204351030
0_HONOR_1524204351030

Answer 3

下面是另一种将其分成几个列表的方法：

all_strs = [j for i, j in enumerate(lst)]

#assumes that this string format will remain constant
lsts = [list(all_strs[i])[2:23] for i, j in enumerate(all_strs)]

keys = [''.join(j) for i, j in enumerate(lsts)]
print(keys)

['0_HONOR_1524204351030',
'0_HONOR_1524204351030',
'0_HONOR_1524204351030',
'0_HONOR_1524204351030',
'0_HONOR_1524204351030']