How to count digits, letters, spaces for a string in Python?

Question

I am trying to make a function to detect how many digits, letter, spaces, and others for a string.

Here's what I have so far:

def count(x):
    length = len(x)
    digit = 0
    letters = 0
    space = 0
    other = 0
    for i in x:
        if x[i].isalpha():
            letters += 1
        elif x[i].isnumeric():
            digit += 1
        elif x[i].isspace():
            space += 1
        else:
            other += 1
    return number,word,space,other

But it's not working:

>>> count(asdfkasdflasdfl222)
Traceback (most recent call last):
  File "<pyshell#4>", line 1, in <module>
    count(asdfkasdflasdfl222)
NameError: name 'asdfkasdflasdfl222' is not defined

What's wrong with my code and how can I improve it to a more simple and precise solution?

Answer 1

Here's another option:

s = 'some string'

numbers = sum(c.isdigit() for c in s)
letters = sum(c.isalpha() for c in s)
spaces  = sum(c.isspace() for c in s)
others  = len(s) - numbers - letters - spaces

Answer 2

Following code replaces any nun-numeric character with '', allowing you to count number of such characters with function len.

import re
len(re.sub("[^0-9]", "", my_string))

Alphabetical:

import re
len(re.sub("[^a-zA-Z]", "", my_string))

More info - https://docs.python.org/3/library/re.html

Answer 3

You shouldn't be setting x = []. That is setting an empty list to your inputted parameter. Furthermore, use python's for i in x syntax as follows:

for i in x:
    if i.isalpha():
        letters+=1
    elif i.isnumeric():
        digit+=1
    elif i.isspace():
        space+=1
    else:
        other+=1

Answer 4

To count the number of letters in a string:

def iinn():        # block for numeric strings
      
   b=(input("Enter numeric letter string "))

   if b.isdigit():
       print (f"you entered {b} numeric string" + "\n")
            
   else:
    
       letters = sum(c.isalpha() for c in b)
       print (f"in the string {b} are {letters} alphabetic letters")
       print("Try again...","\n")        
    
   while True:
       iinn()

A block of numeric strings will be in inverse order:

numbers = sum(c.isdigit() for c in b)

Answer 5

# Write a Python program that accepts a string and calculate the number of digits 
# andletters.

    stre =input("enter the string-->")
    countl = 0
    countn = 0
    counto = 0
    for i in stre:
        if i.isalpha():
            countl += 1
        elif i.isdigit():
            countn += 1
        else:
            counto += 1
    print("The number of letters are --", countl)
    print("The number of numbers are --", countn)
    print("The number of characters are --", counto)

Answer 6

If you want a simple solution, use list comprehension, then get len of that list:

len([ch for ch in text if ch.isdigit()])

This can be applied to isalpha() in a similar fashion

Answer 7

There are 2 errors is this code:

1) You should remove this line, as it will reqrite x to an empty list:

x = []

2) In the first "if" statement, you should indent the "letter += 1" statement, like:

if x[i].isalpha():
    letters += 1

Answer 8

Ignoring anything else that may or may not be correct with your "revised code", the issue causing the error currently quoted in your question is caused by calling the "count" function with an undefined variable because your didn't quote the string.

• count(thisisastring222) looks for a variable called thisisastring222 to pass to the function called count. For this to work you would have to have defined the variable earlier (e.g. with thisisastring222 = "AStringWith1NumberInIt.") then your function will do what you want with the contents of the value stored in the variable, not the name of the variable.
• count("thisisastring222") hardcodes the string "thisisastring222" into the call, meaning that the count function will work with the exact string passed to it.

To fix your call to your function, just add quotes around asdfkasdflasdfl222 changing count(asdfkasdflasdfl222) to count("asdfkasdflasdfl222").

As far as the actual question "How to count digits, letters, spaces for a string in Python", at a glance the rest of the "revised code" looks OK except that the return line is not returning the same variables you've used in the rest of the code. To fix it without changing anything else in the code, change number and word to digit and letters, making return number,word,space,other into return digit,letters,space,other, or better yet return (digit, letters, space, other) to match current behavior while also using better coding style and being explicit as to what type of value is returned (in this case, a tuple).

Answer 9

sample = ("Python 3.2 is very easy") #sample string  
letters = 0  # initiating the count of letters to 0
numeric = 0  # initiating the count of numbers to 0

        for i in sample:  
            if i.isdigit():      
                numeric +=1      
            elif i.isalpha():    
                letters +=1    
            else:    
               pass  
letters  
numeric  

Answer 10

def match_string(words):
    nums = 0
    letter = 0
    other = 0
    for i in words :
        if i.isalpha():
            letter+=1
        elif i.isdigit():
            nums+=1
        else:
            other+=1
    return nums,letter,other

x = match_string("Hello World")
print(x)
>>>
(0, 10, 2)
>>>

Answer 11

This is a mistake in your calling. You are calling the code with the argument (asdfkasdflasdfl222) which is interpreted as a variable. However, you should call it with the string "asdfkasdflasdfl222".

Answer 12

INPUT :

1

26

sadw96aeafae4awdw2 wd100awd

import re

a=int(input())
for i in range(a):
    b=int(input())
    c=input()

    w=re.findall(r'\d',c)
    x=re.findall(r'\d+',c)
    y=re.findall(r'\s+',c)
    z=re.findall(r'.',c)
    print(len(x))
    print(len(y))
    print(len(z)-len(y)-len(w))

OUTPUT :

4

1

19

The four digits are 96, 4, 2, 100 The number of spaces = 1 number of letters = 19