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使用Python进行单词频率排序统计

pythonsortingword-frequency
45

我需要使用Python计算文本中每个单词的出现频率。我的想法是将单词存入字典中，并为每个单词计数。

现在如果我需要按出现次数对单词进行排序，是否可以使用同一个字典而不是使用新字典，其中键为计数，值为单词数组？

可能是重复的问题：Python：按值对字典进行排序 - S.Lott
有趣的事实：一个单词的频率总是大约等于1除以它的排名。因此，第二个最常见的单词出现的次数大约是第一位的一半。第三个出现了1/3次，依此类推。第333个出现的次数是第一位的1/333。这适用于所有语言，没有人真正知道为什么。（齐普夫定律） - ashleedawg
12个回答
63

警告：此示例要求使用Python 2.7或更高版本。

Python内置的Counter对象正是你所需要的。数单词甚至是文档中的第一个示例：

>>> # Tally occurrences of words in a list
>>> from collections import Counter
>>> cnt = Counter()
>>> for word in ['red', 'blue', 'red', 'green', 'blue', 'blue']:
...     cnt[word] += 1
>>> cnt
Counter({'blue': 3, 'red': 2, 'green': 1})

如评论中所指定，Counter 接受一个可迭代对象作为参数，因此上面的示例仅用于说明，并等同于以下形式：

>>> mywords = ['red', 'blue', 'red', 'green', 'blue', 'blue']
>>> cnt = Counter(mywords)
>>> cnt
Counter({'blue': 3, 'red': 2, 'green': 1})
5
请注意，Counter 只支持 Python 2.7 及以上版本。 - Frédéric Hamidi
7
在这之前可能需要添加 from collections import Counter :) - Tim Pietzcker
1
这更多是为了说明...而且它来自Python文档！（我已经更新了示例。） - jathanism
6
按照单词出现的频率进行排序怎么样？ - martineau
10
你可以调用cnt.most_common()来按照频率（从高到低）进行排序。 - jfs
显示剩余2条评论
33

你可以使用相同的字典：

>>> d = { "foo": 4, "bar": 2, "quux": 3 }
>>> sorted(d.items(), key=lambda item: item[1])
第二行输出的内容是：
[('bar', 2), ('quux', 3), ('foo', 4)]
如果您只想要一个排序后的单词列表，请执行以下操作：
>>> [pair[0] for pair in sorted(d.items(), key=lambda item: item[1])]

那行代码会打印：

['bar', 'quux', 'foo']
如果这些值相等怎么办？如果这些值相等，是否有一种方法可以根据它们的长度进行排序呢？ - RoadRunner
如果“bar”和“quux”的计数相等，有没有办法使“quux”排在“bar”之前？ - RoadRunner
对于第二个选项，您可以使用sorted(d.keys(), key=lambda item: d[item])。 - Peter
6

我刚刚写了一个类似的程序，借助 Stack Overflow 的帮助：

from string import punctuation
from operator import itemgetter

N = 100
words = {}

words_gen = (word.strip(punctuation).lower() for line in open("poi_run.txt")
                                             for word in line.split())

for word in words_gen:
    words[word] = words.get(word, 0) + 1

top_words = sorted(words.items(), key=itemgetter(1), reverse=True)[:N]

for word, frequency in top_words:
    print ("%s %d" % (word, frequency))
5
>>> d = {'a': 3, 'b': 1, 'c': 2, 'd': 5, 'e': 0}
>>> l = d.items()
>>> l.sort(key = lambda item: item[1])
>>> l
[('e', 0), ('b', 1), ('c', 2), ('a', 3), ('d', 5)]
这将有一个额外的元组列表。那么这是一种额外的数据结构，对吗？ - AlgoMan
是的，列表是额外的，但如果您只想要单词或计数，则可以消除元组。此外，在构建第二个字典之后，您可能需要使用列表，因此它会创建一个不必要的结构。 - user470379
5
你可以在Python 2.7的collections模块中使用Counter和defaultdict来完成一个两步骤的过程。首先，使用Counter创建一个字典，其中每个单词都是一个键，并对应其频率计数。这个步骤比较简单。
其次，可以使用defaultdict创建一个反向字典，其中键是出现频率，关联的值是遇到那么多次的单词或单词列表。以下是我的意思：
from collections import Counter, defaultdict

wordlist = ['red', 'yellow', 'blue', 'red', 'green', 'blue', 'blue', 'yellow']

# invert a temporary Counter(wordlist) dictionary so keys are
# frequency of occurrence and values are lists the words encountered
freqword = defaultdict(list)
for word, freq in Counter(wordlist).items():
    freqword[freq].append(word)

# print in order of occurrence (with sorted list of words)
for freq in sorted(freqword):
    print('count {}: {}'.format(freq, sorted(freqword[freq])))

输出：

count 1: ['green']
count 2: ['red', 'yellow']
count 3: ['blue']
3

要计算这些项的频率比你们想象的要容易。只需要将所有单词放入列表中（使用字符串分割函数很容易实现）。然后：

#(Pseudo Python Code) 

listOfWords = inputString.split() # splits the words up from whitespace
setOfWords = Set(listOfWords) #  Gives you all the unique words (no duplicates)

for each word in setOfWords  #Count how many words are in the list
   print word + " appears: " + listOfWords.Count(word) + "times"
3

如果您需要进行额外的文本处理，可能值得将 nltk（自然语言工具包）导入到您的项目中。以下是一个示例，使用肯尼迪总统就职演讲：

import nltk

speech_text = "Vice President Johnson, Mr. Speaker, Mr. Chief Justice, President Eisenhower, Vice President Nixon, President Truman, reverend clergy, fellow citizens: We observe today not a victory of party, but a celebration of freedom — symbolizing an end, as well as a beginning — signifying renewal, as well as change. For I have sworn before you and Almighty God the same solemn oath our forebears prescribed nearly a century and three-quarters ago. The world is very different now. For man holds in his mortal hands the power to abolish all forms of human poverty and all forms of human life. And yet the same revolutionary beliefs for which our forebears fought are still at issue around the globe — the belief that the rights of man come not from the generosity of the state, but from the hand of God. We dare not forget today that we are the heirs of that first revolution. Let the word go forth from this time and place, to friend and foe alike, that the torch has been passed to a new generation of Americans — born in this century, tempered by war, disciplined by a hard and bitter peace, proud of our ancient heritage, and unwilling to witness or permit the slow undoing of those human rights to which this nation has always been committed, and to which we are committed today at home and around the world. Let every nation know, whether it wishes us well or ill, that we shall pay any price, bear any burden, meet any hardship, support any friend, oppose any foe, to assure the survival and the success of liberty. This much we pledge — and more. To those old allies whose cultural and spiritual origins we share, we pledge the loyalty of faithful friends. United there is little we cannot do in a host of cooperative ventures. Divided there is little we can do — for we dare not meet a powerful challenge at odds and split asunder. To those new states whom we welcome to the ranks of the free, we pledge our word that one form of colonial control shall not have passed away merely to be replaced by a far more iron tyranny. We shall not always expect to find them supporting our view. But we shall always hope to find them strongly supporting their own freedom — and to remember that, in the past, those who foolishly sought power by riding the back of the tiger ended up inside. To those people in the huts and villages of half the globe struggling to break the bonds of mass misery, we pledge our best efforts to help them help themselves, for whatever period is required — not because the Communists may be doing it, not because we seek their votes, but because it is right. If a free society cannot help the many who are poor, it cannot save the few who are rich. To our sister republics south of our border, we offer a special pledge: to convert our good words into good deeds, in a new alliance for progress, to assist free men and free governments in casting off the chains of poverty. But this peaceful revolution of hope cannot become the prey of hostile powers. Let all our neighbors know that we shall join with them to oppose aggression or subversion anywhere in the Americas. And let every other power know that this hemisphere intends to remain the master of its own house. To that world assembly of sovereign states, the United Nations, our last best hope in an age where the instruments of war have far outpaced the instruments of peace, we renew our pledge of support — to prevent it from becoming merely a forum for invective, to strengthen its shield of the new and the weak, and to enlarge the area in which its writ may run. Finally, to those nations who would make themselves our adversary, we offer not a pledge but a request: that both sides begin anew the quest for peace, before the dark powers of destruction unleashed by science engulf all humanity in planned or accidental self-destruction. We dare not tempt them with weakness. For only when our arms are sufficient beyond doubt can we be certain beyond doubt that they will never be employed. But neither can two great and powerful groups of nations take comfort from our present course — both sides overburdened by the cost of modern weapons, both rightly alarmed by the steady spread of the deadly atom, yet both racing to alter that uncertain balance of terror that stays the hand of mankind’s final war. So let us begin anew — remembering on both sides that civility is not a sign of weakness, and sincerity is always subject to proof. Let us never negotiate out of fear, but let us never fear to negotiate. Let both sides explore what problems unite us instead of belaboring those problems which divide us. Let both sides, for the first time, formulate serious and precise proposals for the inspection and control of arms, and bring the absolute power to destroy other nations under the absolute control of all nations. Let both sides seek to invoke the wonders of science instead of its terrors. Together let us explore the stars, conquer the deserts, eradicate disease, tap the ocean depths, and encourage the arts and commerce. Let both sides unite to heed, in all corners of the earth, the command of Isaiah — to “undo the heavy burdens, and [to] let the oppressed go free.”¹ And, if a beachhead of cooperation may push back the jungle of suspicion, let both sides join in creating a new endeavor — not a new balance of power, but a new world of law — where the strong are just, and the weak secure, and the peace preserved. All this will not be finished in the first one hundred days. Nor will it be finished in the first one thousand days; nor in the life of this Administration; nor even perhaps in our lifetime on this planet. But let us begin. In your hands, my fellow citizens, more than mine, will rest the final success or failure of our course. Since this country was founded, each generation of Americans has been summoned to give testimony to its national loyalty. The graves of young Americans who answered the call to service surround the globe. Now the trumpet summons us again — not as a call to bear arms, though arms we need — not as a call to battle, though embattled we are — but a call to bear the burden of a long twilight struggle, year in and year out, “rejoicing in hope; patient in tribulation,”² a struggle against the common enemies of man: tyranny, poverty, disease, and war itself. Can we forge against these enemies a grand and global alliance, North and South, East and West, that can assure a more fruitful life for all mankind? Will you join in that historic effort? In the long history of the world, only a few generations have been granted the role of defending freedom in its hour of maximum danger. I do not shrink from this responsibility — I welcome it. I do not believe that any of us would exchange places with any other people or any other generation. The energy, the faith, the devotion which we bring to this endeavor will light our country and all who serve it. And the glow from that fire can truly light the world. And so, my fellow Americans, ask not what your country can do for you; ask what you can do for your country. My fellow citizens of the world, ask not what America will do for you, but what together we can do for the freedom of man. Finally, whether you are citizens of America or citizens of the world, ask of us here the same high standards of strength and sacrifice which we ask of you. With a good conscience our only sure reward, with history the final judge of our deeds, let us go forth to lead the land we love, asking His blessing and His help, but knowing that here on earth God’s work must truly be our own."

# Tokenize the words
all_words = speech_text.lower().split()

# Create a frequency distribution
freq = nltk.FreqDist(all_words)

# Show the top 10 words in the list, with counts
freq.items()[:10]

Out[5]: 
[('the', 86),
 ('of', 66),
 ('to', 42),
 ('and', 40),
 ('we', 30),
 ('a', 29),
 ('in', 24),
 ('our', 21),
 ('not', 19),
 ('that', 19)]

# Show the top 10 keys in the frequency dictionary
freq.keys()[:10]

Out[6]: ['the', 'of', 'to', 'and', 'we', 'a', 'in', 'our', 'not', 'that']

# Those frequent words aren't very interesting... let's strip common words
from nltk.corpus import stopwords

stop_words = set(stopwords.words('english'))
clean_words = [w for w in all_words if not w in stop_words]
freq_clean  = nltk.FreqDist(clean_words)

# This is a little more interesting
freq_clean.items()[1:10]
[('let', 16),
 ('us', 11),
 ('new', 7),
 ('sides', 7),
 ('pledge', 6),
 ('ask', 5),
 ('shall', 5),
 ('always', 4),
 ('call', 4)]

NLTK可以让你进行各种有趣的文本分析，如果需要的话。以下是一个快速示例，演示如何找出在文本中出现超过3次的前10个二元组：
bigram_measures = nltk.collocations.BigramAssocMeasures()
bigram_finder   = nltk.collocations.BigramCollocationFinder.from_words(all_words)
bigram_finder.apply_freq_filter(3)
bigram_finder.nbest(bigram_measures.pmi, 10)

Out[28]: 
[('my', 'fellow'),
 ('both', 'sides'),
 ('can', 'do'),
 ('dare', 'not'),
 ('let', 'us'),
 ('we', 'dare'),
 ('do', 'for'),
 ('let', 'both'),
 ('we', 'shall'),
 ('a', 'call')]

请参考NLTK文档了解更多信息和如何快速创建文本中最频繁词汇图表的示例。

2
之前不知道有一个Counter对象可以完成这样的任务。以下是我当时的做法，与你的方法类似。您可以对同一字典的表示进行排序。
#Takes a list and returns a descending sorted dict of words and their counts
def countWords(a_list):
    words = {}
    for i in range(len(a_list)):
        item = a_list[i]
        count = a_list.count(item)
        words[item] = count
    return sorted(words.items(), key = lambda item: item[1], reverse=True)

一个例子：
>>>countWords("the quick red fox jumped over the lazy brown dog".split())
[('the', 2), ('brown', 1), ('lazy', 1), ('jumped', 1), ('over', 1), ('fox', 1), ('dog', 1), ('quick', 1), ('red', 1)]
2

我几天前写了一个类似的程序。程序使用两个参数：文件名（必需）和 N（可选）。

from collections import Counter
import re
import sys

if sys.version_info <(2,7):
    Sys.exit("Must use Python 2.7 or greater")

if len(sys.argv)<2:
    sys.exit('Usage: python %s filename N'%sys.argv[0])

n=0
if len(sys.argv)>2:
    try:
        n=int(sys.argv[2])
        if n<=0:
            raise ValueError
    except ValueError:
        sys.exit("Invalid value for N: %s.\nN must be an integer greater than 0"%sys.argv[2])

filename=sys.argv[1]
try:
        with open(filename,"r") as input_text:
            wordcounter=Counter()
            for line in input_text:
                 wordcounter.update(re.findall("\w+",line.lower()))
        if n==0:
            n=len(wordcounter)

        for word, frequency in wordcounter.most_common(n):
            print("%s %d" % (word, frequency))

except IOError:
        sys.exit("Cannot open file: %s"% filename)
0
这个问题涉及到几个步骤：
  • 清除标点符号。

  • 根据频率对数组进行排序。

    def wordCount(self,nums):
  nums = "Hello, number of transaction which happened, for,"
  nums=nums.lower().translate(None,string.punctuation).split()
  d = {}
  for i in nums:
    if i not in d:
    d[i] = 1
  else:
    d[i] = d[i]+1
 sorted_d = (sorted(d.items(), key = operator.itemgetter(1), reverse = True)

for key,val in sorted_d:
 print key,val
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