在 Python 中如何计算一个列表中的出现次数？

#在 Python 中如何计算一个列表中的出现次数？| 来源: 网络整理| 查看: 265

正如上面所建议的，来自collections模块的Counter类绝对是计数应用的途径。

这一解决方案还解决了在multiple文件，使用fileinput.input()的方法来迭代所有 the filenames specified on the command line (or if no filenames specified on the command line then will read from STDIN, typic所有y the keyboard)

Fin所有y it uses a little more sophisticated approach for breaking the line into 'words' with a regular expression as a delimiter. As noted in the code it will handle contractions more gracefully (however it will be confused by apostrophes being used a single quotes)

"""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))

输出示例。

$ 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

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