How to count the number of words in a sentence, ignoring numbers, punctuation and whitespace?
Question:
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.
Answers:
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)
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!
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")
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
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.
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)
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)
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)
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!
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")
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
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.
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)
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)