Bob is a lazy man. 

He needs you to create a method that can determine how many ```letters``` and ```digits``` are in a given string.



Example:

"hel2!lo" --> 6

"wicked .. !" --> 6

"!?..A" --> 1

Source: codewars | Difficulty: EASY
Tags: String algorithms

Fundamentals

Starter code: def count_letters_and_digits(s):
	

Solutions:
def count_letters_and_digits(s):
	return isinstance(s, str) and sum(map(str.isalnum, s))


def count_letters_and_digits(s):
	return sum(map(str.isalnum, s))


import re

def count_letters_and_digits(s):
	if s == None or s == []:
		return 0
	return len(re.findall('[A-Za-z0-9]', s))


def count_letters_and_digits(s):
	count = 0
	for x in s:
		if x.isalpha() or x.isnumeric():
			count += 1
	return count


import re

def count_letters_and_digits(s):
	try:
		list = re.findall('[A-Za-z0-9]', s)
		return len(list)
	except:
		return 0


def count_letters_and_digits(s):
	numerics = 0
	letters = 0
	other = 0
	for i in s:
		if i.isnumeric():
			numerics += 1
		elif i.isalpha():
			letters += 1
		else:
			other += 1
	return numerics + letters


def count_letters_and_digits(s):
	return len([item for item in s if item.isalpha() or item.isdigit()])


def count_letters_and_digits(s):
	return bool(s) and sum(map(str.isalnum, s))


import re

def count_letters_and_digits(s):
	return len(re.sub('[\\W_]', '', s))


import re

def count_letters_and_digits(s):
	s = re.sub('[^\\w]', '', s)
	s = s.replace('_', '')
	return len(s)