How can I check if character in a string is a letter? (Python)
PythonStringFindPython Problem Overview
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()
?
Python Solutions
Solution 1 - Python
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
Solution 2 - 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)
>>>
Solution 3 - Python
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.")
Solution 4 - Python
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.
Solution 5 - Python
This works:
any(c.isalpha() for c in 'string')
Solution 6 - Python
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.")