如何计算 Python 列表中出现的次数？

Question

我是 python 新手，我想计算每个单词在所有文件中出现的次数。显示每个单词，它出现的次数和出现的时间百分比。对列表进行排序，使出现频率最高的单词首先出现，最不频繁出现的单词最后出现。 我正在处理小样本，只知道一个文件，但我无法正常工作，

 from collections import defaultdict

words = "apple banana apple strawberry banana lemon"

d = defaultdict(int)
for word in words.split():
    d[word] += 1

Answer 1

如上所述，collections 模块中的Counter 类绝对是计数应用程序的最佳选择。

此解决方案还解决了使用fileinput.input() 方法迭代所有在命令行上指定的文件名的内容的多个文件中的单词计数请求（或者如果在命令行上没有指定文件名，那么将从STDIN 读取，通常是键盘）

最后，它使用了一种更复杂的方法，以正则表达式作为分隔符将行分成“单词”。如代码中所述，它将更优雅地处理收缩（但是它会被使用单引号的撇号混淆）

"""countwords.py
   count all words across all files
"""

import fileinput
import re
import collections

# create a regex delimiter that is any character that is  not 1 or
# more word character or an apostrophe, this allows contractions
# to be treated as a word (eg can't  won't  didn't )
# Caution: this WILL get confused by a line that uses apostrophe
# as a single quote: eg 'hello' would be treated as a 7 letter word

word_delimiter = re.compile(r"[^\w']+")

# create an empty Counter

counter = collections.Counter()

# use fileinput.input() to open and read ALL lines from ALL files
# specified on the command line, or if no files specified on the
# command line then read from STDIN (ie the keyboard or redirect)

for line in fileinput.input():
    for word in word_delimiter.split(line):
        counter[word.lower()] += 1   # count case insensitively

del counter['']   # handle corner case of the occasional 'empty' word

# compute the total number of words using .values() to get an
# generator of all the Counter values (ie the individual word counts)        
# then pass that generator to the sum function which is able to 
# work with a list or a generator

total = sum(counter.values())

# iterate through the key/value pairs (ie word/word_count) in sorted
# order - the lambda function says sort based on position 1 of each
# word/word_count tuple (ie the word_count) and reverse=True does
# exactly what it says = reverse the normal order so it now goes
# from highest word_count to lowest word_count

print("{:>10s}  {:>8s} {:s}".format("occurs", "percent", "word"))

for word, count in sorted(counter.items(),
                          key=lambda t: t[1],
                          reverse=True):
    print ("{:10d} {:8.2f}% {:s}".format(count, count/total*100, word))

示例输出：

$ python3 countwords.py
I have a dog, he is a good dog, but he can't fly
^D

occurs   percent word
     2    15.38% a
     2    15.38% dog
     2    15.38% he
     1     7.69% i
     1     7.69% have
     1     7.69% is
     1     7.69% good
     1     7.69% but
     1     7.69% can't
     1     7.69% fly

还有：

$ python3 countwords.py text1 text2
    occurs   percent word
         2    11.11% hello
         2    11.11% i
         1     5.56% there
         1     5.56% how
         1     5.56% are
         1     5.56% you
         1     5.56% am
         1     5.56% fine
         1     5.56% mark
         1     5.56% where
         1     5.56% is
         1     5.56% the
         1     5.56% dog
         1     5.56% haven't
         1     5.56% seen
         1     5.56% him

Answer 2

正如cmets中提到的，这正是collections.Counter

words = 'a b c a'.split()
print(Counter(words).most_common())

来自文档：https://docs.python.org/2/library/collections.html

most_common([n])
Return a list of the n most common elements and their counts
from the most common to the least. If n is omitted or None,
most_common() returns all elements in the counter.
Elements with equal counts are ordered arbitrarily:

>>> Counter('abracadabra').most_common(3)
[('a', 5), ('r', 2), ('b', 2)]

Answer 3

使用您的代码，这是一种更简洁的方法：

# Initializing Dictionary
d = {}
with open(sys.argv[1], 'r') as f:

    # counting number of times each word comes up in list of words (in dictionary)
    for line in f: 
        words = line.lower().split() 
        # Iterate over each word in line 
        for word in words: 
            if word not in d.keys():
                d[word] = 1
            else:
                d[word]+=1

n_all_words = sum([k.values])

# Print percentage occurance
for k, v in d.items():
    print(f'{k} occurs {v} times and is {(100*v/n_all_words):,.2f}% total of words.')


# Sort a dictionary using this useful solution
# https://stackoverflow.com/a/613218/10521959
import operator
sorted_d = sorted(d.items(), key=operator.itemgetter(1))

Answer 4

最直接的方法就是使用 Counter 函数：

from collections import Counter
c = Counter(words.split())

输出：

Counter({'apple': 2, 'banana': 2, 'strawberry': 1, 'lemon': 1})

按顺序排列单词或计数：

list(c.keys())
list(c.values())

或将其放入普通字典中：

dict(c.items())

或元组列表：

c.most_common()