Question

如何使用dict理解为字符串列表构建标点符号字典？我能够为这样的单个字符串做到这一点：

import string

test_string = "1990; and 1989', \ '1975/97', '618-907 CE"
counts = {p:test_string.count(p) for p in string.punctuation}

编辑：对于将来可能会想要此内容的任何人，请在下面复制Patrick Artner的答案，并进行非常小的修改以仅保留标点符号：< / p>

# return punctuation Counter dict for string/list/pd.Series

import string
from collections import Counter
from itertools import chain

def count_punctuation(str_series_or_list):
    c = Counter(chain(*str_series_or_list))
    unwanted = set(c) - set(string.punctuation)
    for unwanted_key in unwanted: del c[unwanted_key]
    return c

Answer 1

为何选择自己？

GONE

输出：

import string
from collections import Counter


test_string = "1990; and 1989', \ '1975/97', '618-907 CE"

c = Counter(test_string)  # counts all occurences

for p in string.punctuation:   # prints the one in string.punctuation
    print(p , c[p])            # access like dictionary (its a subclass of dict)
print(c)

计数器类似字典：请参阅https://docs.python.org/2/library/collections.html#collections.Counter

编辑：列表中的多个字符串：

! 0
" 0
# 0
$ 0
% 0
& 0
' 4
( 0
) 0
* 0
+ 0
, 2
- 1
. 0
/ 1
: 0
; 1
< 0
= 0
> 0
? 0
@ 0
[ 0
\ 1
] 0
^ 0
_ 0
` 0
{ 0
| 0
} 0
~ 0
Counter({'9': 7, ' ': 6, '1': 4, "'": 4, '7': 3, '0': 2, '8': 2, ',': 2, ';': 1, 'a': 1, 'n': 1, 'd': 1, '\\': 1, '5': 1, '/': 1, '6': 1, '-': 1, 'C': 1, 'E': 1})

输出:(删除0条目）

import string
from collections import Counter
from itertools import chain

test_strings = [ "1990; and 1989', \ '1975/97', '618-907 CE" , "someone... or no one? that's the question!", "No I am not!"]

c = Counter(chain(*test_strings))

for p in string.punctuation:
    print(p , c[p])

print(c)

标点符号字典用于字符串列表

1 个答案: