在计数器中对键和值进行排序

时间:2018-02-21 21:49:41

标签: python python-3.x sorting word-count

这是我正在处理的代码,我希望输出作为递减的计数,如果计数相同,则按名称排序。

from collections import Counter
import re
from nltk.corpus import stopwords
import operator
text = "The quick brown fox jumped over the lazy dogs bowl. The dog was angry with the fox considering him lazy."
def tokenize(text):
    tokens = re.findall(r"\w+|\S", text.lower())
    #print(tokens)
    tokens1 = []
    for i in tokens:
        x = re.findall(r"\w+|\S", i, re.ASCII)
        for j in x:
            tokens1.append(j)

    return tokens
tok = tokenize(text)

punctuations = ['(',')',';',':','[',']',',', '...', '.', '&']

keywords = [word for word in tok if not word in punctuations]

cnt = Counter()
d= {}
for word in keywords:
    cnt[word] += 1 
print(cnt)
freq = operator.itemgetter(1)

for k, v in sorted(cnt.items(), reverse=True, key=freq):
    print("%3d  %s" % (v, k))

当前输出:

  4  the
  2  fox
  2  lazy
  1  quick
  1  brown
  1  jumped
  1  over
  1  dogs
  1  bowl
  1  dog
  1  was
  1  angry
  1  with
  1  considering
  1  him

必需的输出:

  4  the
  2  fox
  2  lazy
  1  angry
  1  bowl
  1  brown
  1  considering
  1  dog
  1  dogs

1 个答案:

答案 0 :(得分:0)

使用返回元组的排序函数。元组中的第一项是计数的倒数(字典中的值),第二项是字符串(字典中的键)。您可以通过删除变量freq,删除对sorted的调用中的关键字reverse,并为字典中的每个项提供一个返回(-value,key)的lambda函数来完成此操作。该计划的最后几行是:

print(cnt)
for k, v in sorted(cnt.items(), key=lambda item: (-item[1], item[0])):
    print("%3d  %s" % (v, k))

签署lambda函数的原因是获取正确的排序顺序,因为默认排序顺序从最低到最高。