问如何计算Python列表中出现的事件？

EN

"""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))

$ 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

$ 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

# 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))

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

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)]