How to find the count of a word in a string

Question

I have a string "Hello I am going to I with hello am". I want to find how many times a word occur in the string. Example hello occurs 2 time. I tried this approach that only prints characters -

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")

I want to learn how to find the word count.

Answer 1

If you want to find the count of an individual word, just use count:

input_string.count("Hello")

Use collections.Counter and split() to tally up all the words:

from collections import Counter

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

Answer 2

from collections import *
import re

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

Using re.findall is more versatile than split, because otherwise you cannot take into account contractions such as "don't" and "I'll", etc.

Demo (using your example):

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

If you expect to be making many of these queries, this will only do O(N) work once, rather than O(N*#queries) work.

Answer 3

Counter from collections is your friend:

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

Answer 4

The vector of occurrence counts of words is called bag-of-words.

Scikit-learn provides a nice module to compute it, sklearn.feature_extraction.text.CountVectorizer. Example:

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

Output:

2 am
1 going
2 hello
1 to
1 with

Part of the code was taken from this Kaggle tutorial on bag-of-words.

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

Answer 5

Here is an alternative, case-insensitive, approach

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

It matches by converting the string and target into lower-case.

ps: Takes care of the "am ham".count("am") == 2 problem with str.count() pointed out by @DSM below too :)

Answer 6

Considering Hello and hello as same words, irrespective of their cases:

>>> 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 7

You can divide the string into elements and calculate their number

count = len(my_string.split())

Answer 8

You can use the Python regex library re to find all matches in the substring and return the array.

import re

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

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

Prints:

2

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