如何在Python中将此字符串转换为多维列表？

Question

我有下一个string ：

string = 'tuned     1372                root    6u      REG                8,3      4096  102029349 /tmp/ffiabNswC (deleted)\ngmain     1372 2614           root    6u      REG                8,3      4096  102029349 /tmp/ffiabNswC (deleted)\n'

我需要将string每个元素放入list1[0][..] ，但是当我看到新行'\\ n'时，我必须将下一个元素放入list1[1][..]

多维列表，如下所示：

list1 = [["tuned", "1372", "root", "6u", "REG", "8,3", "4096", "102029349", "/tmp/ffiabNswC", "(deleted)"], 
         ["gmain", "1372", "2614", "root", "6u", "REG", "8,3", "4096", "102029349", "/tmp/ffiabNswC", "(deleted)"]]

我使用split ，但是这使我所有人处于同一维度。

Answer 1

首先用新行拆分（以获取行），然后按空格拆分每个元素（以获取每一列）：

data = "tuned 1372 root 6u REG 8,3 4096 102029349 /tmp/ffiabNswC (deleted)\ngmain 1372 2614 root 6u REG 8,3 4096 102029349 /tmp/ffiabNswC (deleted)\n"

parsed = [elements.split() for elements in data.strip().split("\n")]  # `strip()` removes the last whitespace so we don't get blank elements

print(parsed)

# [['tuned', '1372', 'root', '6u', 'REG', '8,3', '4096', '102029349', '/tmp/ffiabNswC', '(deleted)'], ['gmain', '1372', '2614', 'root', '6u', 'REG', '8,3', '4096', '102029349', '/tmp/ffiabNswC', '(deleted)']]

Answer 2

以下功能应为您执行此操作：

f = lambda list: [sublist.split(' ') for sublist in list.split('\n')]

只需通过f(string)调用它即可。

另外，如果您不希望子列表中有任何空白条目，则可以这样做

f = lambda list: [sublist.split(' ') for sublist in list.split('\n') if sublist]

Answer 3

输入：

string = 'tuned 1372 root 6u REG 8,3 4096 102029349 /tmp/ffiabNswC 
(deleted)\ngmain 1372 2614 root 6u REG 8,3 4096 102029349 /tmp/ffiabNswC 
(deleted)\n'

代码：-只需写

 mylist=string.split()

输出：-

[tuned
 1372
 root
6u
REG
8,3
4096
102029349
/tmp/ffiabNswC
(deleted)
gmain
1372
2614
root
6u
REG
8,3
4096
102029349
/tmp/ffiabNswC
(deleted)]

Answer 4

由于您要将字符串拆分为几行，然后将这些行拆分为多个单词，因此可以使用理解

result = [line.split() for line in string.split('\n')]

输出：

[['tuned', '1372', 'root', '6u', 'REG', '8,3', '4096', '102029349', '/tmp/ffiabNswC', '(deleted)'],
 ['gmain', '1372', '2614', 'root', '6u', 'REG', '8,3', '4096', '102029349', '/tmp/ffiabNswC', '(deleted)'],
 []]

如果您不希望最后一个空格（由于最后一个\\n ），则可以在split之前strip它。

Answer 5

可以使用列表推导如下：

string = 'tuned     1372                root    6u      REG                8,3      4096  102029349 /tmp/ffiabNswC (deleted)\ngmain     1372 2614           root    6u      REG                8,3      4096  102029349 /tmp/ffiabNswC (deleted)\n'
print [x.split() for x in string.strip().split('\n')]

输出：

[['tuned', '1372', 'root', '6u', 'REG', '8,3', '4096', '102029349', '/tmp/ffiabNswC', '(deleted)'], ['gmain', '1372', '2614', 'root', '6u', 'REG', '8,3', '4096', '102029349', '/tmp/ffiabNswC', '(deleted)']]

Answer 6

在您的帮助下，我尝试用列表中的值制作字典，这很棒。

在这里，它起作用。 我开始爱上列表理解了！

dict1 = { i.split()[0]: (i.split()[1:]) for i in data.strip().split('\n') }

输出：

dict1
{'gmain': ['1372',
 '2614',
 'root',
 '6u',
 'REG',
 '8,3',
 '4096',
 '102029349',
 '/tmp/ffiabNswC',
 '(deleted)'],
'tuned': ['1372',
 'root',
 '6u',
 'REG',
 '8,3',
 '4096',
 '102029349',
 '/tmp/ffiabNswC',
 '(deleted)']}