高效计算字符串中的词频

Question

我正在解析一长串文本并计算每个单词在 Python 中出现的次数。我有一个有效的函数，但我正在寻找关于是否有方法可以使它更有效率（在速度方面）以及是否有 python 库函数可以为我做这件事的建议，所以我不会重新发明轮子？

你能建议一种更有效的方法来计算长字符串（通常超过 1000 个单词）中出现的最常见单词吗？

另外，将字典排序到列表中的最佳方法是什么，其中第一个元素是最常见的单词，第二个元素是第二个最常见的单词等等？

 test = """abc def-ghi jkl abc
abc"""

def calculate_word_frequency(s):
    # Post: return a list of words ordered from the most
    # frequent to the least frequent

    words = s.split()
    freq  = {}
    for word in words:
        if freq.has_key(word):
            freq[word] += 1
        else:
            freq[word] = 1
    return sort(freq)

def sort(d):
    # Post: sort dictionary d into list of words ordered
    # from highest freq to lowest freq
    # eg: For {"the": 3, "a": 9, "abc": 2} should be
    # sorted into the following list ["a","the","abc"]

    #I have never used lambda's so I'm not sure this is correct
    return d.sort(cmp = lambda x,y: cmp(d[x],d[y]))

print calculate_word_frequency(test)

原文由 sazr 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

使用 collections.Counter ：

 >>> from collections import Counter
>>> test = 'abc def abc def zzz zzz'
>>> Counter(test.split()).most_common()
[('abc', 2), ('zzz', 2), ('def', 2)]

原文由 Burhan Khalid 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 2

>>>> test = """abc def-ghi jkl abc
abc"""
>>> from collections import Counter
>>> words = Counter()
>>> words.update(test.split()) # Update counter with words
>>> words.most_common()        # Print list with most common to least common
[('abc', 3), ('jkl', 1), ('def-ghi', 1)]

原文由 jamylak 发布，翻译遵循 CC BY-SA 3.0 许可协议