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

I know about islower and isupper, but can you check whether or not that character is a letter?
For Example:

>>> s = 'abcdefg'
>>> s2 = '123abcd'
>>> s3 = 'abcDEFG'
>>> s[0].islower()
True

>>> s2[0].islower()
False

>>> s3[0].islower()
True

Is there any way to just ask if it is a character besides doing .islower() or .isupper()?

Answers:

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Method 1

You can use str.isalpha().

For example:

s = 'a123b'

for char in s:
    print(char, char.isalpha())

Output:

a True
1 False
2 False
3 False
b True

Method 2

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)
>>>

Method 3

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.")

Method 4

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.

Method 5

This works:

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

Method 6

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.")

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0