Question

我有一个字符串“Hello I am going to I with hello am”。我想找出一个单词在字符串中出现的次数。示例hello发生2次。我尝试过只打印字符的方法 -

def countWord(input_string):
    d = {}
    for word in input_string:
        try:
            d[word] += 1
        except:
            d[word] = 1

    for k in d.keys():
        print "%s: %d" % (k, d[k])
print countWord("Hello I am going to I with Hello am")

我想学习如何找到字数。

Answer 1

如果您想查找单个单词的计数，请使用count：

input_string.count("Hello")

使用collections.Counter和split()来计算所有字词：

from collections import Counter

words = input_string.split()
wordCount = Counter(words)

Answer 2

Counter from collections是你的朋友：

>>> from collections import Counter
>>> counts = Counter(sentence.lower().split())

Answer 3

from collections import *
import re

Counter(re.findall(r"[\w']+", text.lower()))

使用re.findall比split更通用，因为否则您无法考虑收缩，例如“不要”和“我会”等等。

演示（使用您的示例）：

>>> countWords("Hello I am going to I with hello am")
Counter({'i': 2, 'am': 2, 'hello': 2, 'to': 1, 'going': 1, 'with': 1})

如果您希望进行许多这样的查询，这只会执行一次O（N）工作，而不是O（N *＃查询）工作。

Answer 4

单词出现次数的向量称为bag-of-words。

Scikit-learn为计算它提供了一个很好的模块，sklearn.feature_extraction.text.CountVectorizer。例如：

import numpy as np
from sklearn.feature_extraction.text import CountVectorizer

vectorizer = CountVectorizer(analyzer = "word",   \
                             tokenizer = None,    \
                             preprocessor = None, \
                             stop_words = None,   \
                             min_df = 0,          \
                             max_features = 50) 

text = ["Hello I am going to I with hello am"]

# Count
train_data_features = vectorizer.fit_transform(text)
vocab = vectorizer.get_feature_names()

# Sum up the counts of each vocabulary word
dist = np.sum(train_data_features.toarray(), axis=0)

# For each, print the vocabulary word and the number of times it 
# appears in the training set
for tag, count in zip(vocab, dist):
    print count, tag

输出：

2 am
1 going
2 hello
1 to
1 with

部分代码来自此Kaggle tutorial on bag-of-words。

仅供参考：How to use sklearn's CountVectorizerand() to get ngrams that include any punctuation as separate tokens?

Answer 5

将Hello和hello视为相同的字词，不论其情况如何：

>>> from collections import Counter
>>> strs="Hello I am going to I with hello am"
>>> Counter(map(str.lower,strs.split()))
Counter({'i': 2, 'am': 2, 'hello': 2, 'to': 1, 'going': 1, 'with': 1})

Answer 6

这是一种替代的，不区分大小写的方法

sum(1 for w in s.lower().split() if w == 'Hello'.lower())
2

通过将字符串和目标转换为小写来匹配。

ps：关注下面@DSM指出的"am ham".count("am") == 2 str.count()问题：）

Answer 7

您可以将字符串划分为元素并计算其数字

count = len(my_string.split())

Answer 8

您可以使用Python正则表达式库re查找子字符串中的所有匹配项并返回该数组。

import re

input_string = "Hello I am going to I with Hello am"

print(len(re.findall('hello', input_string.lower())))

<强>打印

Answer 9

def countSub(pat,string):
    result = 0
    for i in range(len(string)-len(pat)+1):
          for j in range(len(pat)):
              if string[i+j] != pat[j]:
                 break
          else:   
                 result+=1
    return result

如何查找字符串中单词的计数？

9 个答案: