How to count the number of words in a sentence, ignoring numbers, punctuation and whitespace?

How would I go about counting the words in a sentence? I'm using Python.

For example, I might have the string:

string = "I     am having  a   very  nice  23!@$      day. "

That would be 7 words. I'm having trouble with the random amount of spaces after/before each word as well as when numbers or symbols are involved.

str.split() without any arguments splits on runs of whitespace characters:

>>> s = 'I am having a very nice day.'
>>> 
>>> len(s.split())
7

From the linked documentation:

If sep is not specified or is None, a different splitting algorithm is applied: runs of consecutive whitespace are regarded as a single separator, and the result will contain no empty strings at the start or end if the string has leading or trailing whitespace.

You can use regex.findall():

import re
line = " I am having a very nice day."
count = len(re.findall(r'\w+', line))
print (count)

This is a simple word counter using regex. The script includes a loop which you can terminate it when you're done.

#word counter using regex
import re
while True:
    string =raw_input("Enter the string: ")
    count = len(re.findall("[a-zA-Z_]+", string))
    if line == "Done": #command to terminate the loop
        break
    print (count)
print ("Terminated")

boon kwee

s = "I     am having  a   very  nice  23!@$      day. "
sum([i.strip(string.punctuation).isalpha() for i in s.split()])

The statement above will go through each chunk of text and remove punctuations before verifying if the chunk is really string of alphabets.

Ok here is my version of doing this. I noticed that you want your output to be 7, which means you dont want to count special characters and numbers. So here is regex pattern:

re.findall("[a-zA-Z_]+", string)

Where [a-zA-Z_] means it will match any character beetwen a-z (lowercase) and A-Z (upper case).

About spaces. If you want to remove all extra spaces, just do:

string = string.rstrip().lstrip() # Remove all extra spaces at the start and at the end of the string
while "  " in string: # While  there are 2 spaces beetwen words in our string...
    string = string.replace("  ", " ") # ... replace them by one space!

Anto

How about using a simple loop to count the occurrences of number of spaces!?

txt = "Just an example here move along" 
count = 1
for i in txt:
if i == " ":
   count += 1
print(count)

Darrell White

    def wordCount(mystring):  
        tempcount = 0  
        count = 1  

        try:  
            for character in mystring:  
                if character == " ":  
                    tempcount +=1  
                    if tempcount ==1:  
                        count +=1  

                    else:  
                        tempcount +=1
                 else:
                     tempcount=0

             return count  

         except Exception:  
             error = "Not a string"  
             return error  

    mystring = "I   am having   a    very nice 23!@$      day."           

    print(wordCount(mystring))

output is 8

import string 

sentence = "I     am having  a   very  nice  23!@$      day. "
# Remove all punctuations
sentence = sentence.translate(str.maketrans('', '', string.punctuation))
# Remove all numbers"
sentence = ''.join([word for word in sentence if not word.isdigit()])
count = 0;
for index in range(len(sentence)-1) :
    if sentence[index+1].isspace() and not sentence[index].isspace():
        count += 1 
print(count)

来源：https://stackoverflow.com/questions/19410018/how-to-count-the-number-of-words-in-a-sentence-ignoring-numbers-punctuation-an

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