How can I use dict comprehension to build a dictionary of punctuation counts for a list of strings? I was able to do it for a single string like this:
import string
test_string = "1990; and 1989', \ '1975/97', '618-907 CE"
counts = {p:test_string.count(p) for p in string.punctuation}
Edit: For anyone who may want this in the future, here's Patrick Artner's answer copied from below, with a very small modification to keep only punctuation counts:
# 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
Why count yourself?
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)
Output:
! 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})
Counter is dictionary-like: see https://docs.python.org/2/library/collections.html#collections.Counter
Edit: multiple strings in a list:
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)
Output: (removed 0-Entries)
! 2
' 5
, 2
- 1
. 3
/ 1
; 1
? 1
\ 1
Counter({' ': 15, 'o': 8, '9': 7, 'n': 6, "'": 5, 'e': 5, 't': 5, '1': 4, 'a': 3, '7': 3, 's': 3, '.': 3, '0': 2, '8': 2, ',': 2, 'm': 2, 'h': 2, '!': 2, ';': 1, 'd': 1, '\\': 1, '5': 1, '/': 1, '6': 1, '-': 1, 'C': 1, 'E': 1, 'r': 1, '?': 1, 'q': 1, 'u': 1, 'i': 1, 'N': 1, 'I': 1})