How can I check if character in a string is a letter? (Python)

str.isalpha() 

Return true if all characters in the string are alphabetic and there is at least one character, false otherwise. Alphabetic characters are those characters defined in the Unicode character database as “Letter”, i.e., those with general category property being one of “Lm”, “Lt”, “Lu”, “Ll”, or “Lo”. Note that this is different from the “Alphabetic” property defined in the Unicode Standard.

In python2.x:

>>> s = u'a1中文' >>> for char in s: print char, char.isalpha() ... a True 1 False 中 True 文 True >>> s = 'a1中文' >>> for char in s: print char, char.isalpha() ... a True 1 False � False � False � False � False � False � False >>> 

In python3.x:

>>> s = 'a1中文' >>> for char in s: print(char, char.isalpha()) ... a True 1 False 中 True 文 True >>> 

This code work:

>>> def is_alpha(word): ... try: ... return word.encode('ascii').isalpha() ... except: ... return False ... >>> is_alpha('中国') False >>> is_alpha(u'中国') False >>> >>> a = 'ａ' >>> b = 'a' >>> ord(a), ord(b) (65345, 97) >>> a.isalpha(), b.isalpha() (True, True) >>> is_alpha(a), is_alpha(b) (False, True) >>> 

I found a good way to do this with using a function and basic code. This is a code that accepts a string and counts the number of capital letters, lowercase letters and also 'other'. Other is classed as a space, punctuation mark or even Japanese and Chinese characters.

def check(count): lowercase = 0 uppercase = 0 other = 0 low = 'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z' upper = 'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z' for n in count: if n in low: lowercase += 1 elif n in upper: uppercase += 1 else: other += 1 print("There are " + str(lowercase) + " lowercase letters.") print("There are " + str(uppercase) + " uppercase letters.") print("There are " + str(other) + " other elements to this sentence.") 

This works:

any(c.isalpha() for c in 'string') 

data = "abcdefg hi j 12345"

digits_count = 0 letters_count = 0 others_count = 0 for i in userinput: if i.isdigit(): digits_count += 1 elif i.isalpha(): letters_count += 1 else: others_count += 1 print("Result:") print("Letters=", letters_count) print("Digits=", digits_count) 

Output:

Please Enter Letters with Numbers: abcdefg hi j 12345 Result: Letters = 10 Digits = 5 

By using str.isalpha() you can check if it is a letter.

Easier way that considers only letters a-z and A-Z:

myString = "Hi!" for char in myString: if "a" <= char <= "z" or "A" <= char <= "Z": print(f"{char}: letter") else: print(f"{char}: not letter") 

Which gives output:

H: letter i: letter !: not letter 

Note that if "char" has multiple characters, this will only consider the first character.

If you want to test only a-z and A-Z (i.e. exclude 中 as valid), you can do:

def validate_word(word): """ Only allow a-z and A-Z only """ valid_chars = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'} if set(word).difference(valid_chars): print('there\'s at least 1 char that\'s not allowed') 

Haven't timed it but pretty using a set would be faster than iterating chars

This works:

word = str(input("Enter string:")) notChar = 0 isChar = 0 for char in word: if not char.isalpha(): notChar += 1 else: isChar += 1 print(isChar, " were letters; ", notChar, " were not letters.")