Python类似uniq -c命令？

Question

python是否有类似于linux命令的命令：

cat file.txt | sort -n | uniq -c

它在每个新行上对整数文本文件的频率进行排序和计算，并以下列形式输出：

76539  1 
100441 2 
108637 3 
108874 4 
103580 5 
 91869 6 
 78458 7 
 61955 8 
 46100 9 
 32701 10 
 21111 11 
 13577 12 
  7747 13 
  4455 14 
  2309 15 
  1192 16 
   554 17 
   264 18 
   134 19 
    63 20 
    28 21 
    15 22 
    12 23 
     7 24 
     5 25

如果没有，我可以简单地os.system(cat file.txt | sort -n | uniq -c)吗？

Answer 1

import collections

c = collections.Counter()

with open('file.txt') as f:
    for text in f:
        c.update( [int(text.strip())] )

c_sorted = sorted(c.most_common())

for key, val in c_sorted:
    print val, key

Answer 2

尝试 collections.Counter

>>> import collections
>>> collections.Counter(['asdf', 'sdfg', 'asdf', 'qwer', 'sdfg', 'asdf'])
Counter({'asdf': 3, 'sdfg': 2, 'qwer': 1})
>>> collections.Counter(map(str.strip, open('file.txt').readlines()))
Counter({'spam': 5, 'hello': 3, 'world': 2, 'eggs': 2})

Answer 3

您可以使用itertools.groupby

from itertools import groupby

words = ['blah', 'blah2']
my_result = dict((key, len(list(word_group))) for key, word_group in groupby(sorted(words)))

Answer 4

https://docs.scipy.org/doc/numpy/reference/generated/numpy.unique.html

Might be worth considering, but the return_counts option was not available in older versions of the library, so depends on what's available to you.