如何让python程序检查字符串中是否有字母

Question

I am trying to make the python program check if at least one letter is in a string?

import string

s = ('hello')

if string.ascii_lowercase in s:
    print('yes')
else:
    print('no')

It always just prints no

Answer 1

Well, string.ascii_lowercase is equal to 'abcdefghijklmnopqrstuvwxyz' . That doesn't look like it's contained in hello , right?

What you should do instead is to go over the letters in ascii_lowercase and check if any of them are in your string s.

import string

s = ('hello')

if any([letter in s for letter in string.ascii_lowercase]):
    print('yes')
else:
    print('no')

Wonderfully smart people in the comments have pointed out that you can drop the [ ] brackets that would usually create a list, turning our list comprehension into something called a generator . This would prevent the need to check every single letter in ascii_lowercase and make our code a little bit faster - as it stands, the whole list is generated and then checked. With the generator, the letters are checked only up to e , as that's in 'hello' .

I was able to shave off a whole nanosecond this way! Still, straight up going through the whole list should be fine as well for most cases and is certainly simpler.

Answer 2

An efficient way to check if some string s contains any character from some alphabet:

alphabet = frozenset(string.ascii_lowercase)
any(letter in alphabet for letter in s)

Key points:

• Avoid linear search by storing the alphabet in a set instead of a more general iterable that doesn't allow fast (O(1)) check of elements
• Loop over the input, not the target alphabet, because the alphabet is probably a finite set of constant size, and allow even very large inputs efficiently, without linear searching and excessive memory use (putting input in a set instead of the alphabet)
• Avoid unnecessary list creation (and wasted memory) by using a generator expression

Here are some inferior alternatives.

Linear search over string.ascii_lowercase :

any(letter in string.ascii_lowercase for letter in s)

Linear search over string.ascii_lowercase , and a useless list creation:

any([letter in string.ascii_lowercase for letter in s])

Linear search over the input, very poor performance in the worst case when the input is very long and does not contain any character from the alphabet:

any(letter in s for letter in string.ascii_lowercase)

Answer 3

Currently you are checking whether the whole string string.ascii_lowercase is in s .

You have to check every single character of string.ascii_lowercase instead.

The naive solution would look like this:

>>> s = 'hello'
>>> for letter in string.ascii_lowercase:
...     if letter in s:
...         print('yes')
...         break
... else:
...     print('no')
... 
yes

Here, the else block will only execute if the loop was not broken by the break statement.

A shorthand for the for loop would be to use the any builtin paired with a generator-expression:

>>> contained = any(letter in s for letter in string.ascii_lowercase)
>>> print('yes' if contained else 'no')
yes

Finally, you can improve the runtime of both implementations by using the set of characters from s , ie s = set(s) . This will ensure that every in check is performed in constant time rather than iterating over s for every letter that is searched.

edit: Here's another short one:

>>> if set(s).intersection(string.ascii_lowercase):
...     print('yes')
... else:
...     print('no')
... 
yes

This uses the fact that an empty set (the possible result of the intersection) will be treated as False in the if check.

(It has the slight drawback that the computation of the intersection does not stop once a single shared letter letter is found.)

Answer 4

Make a set of each string and check the size of their intersection

def share_letter(s1, s2):
    return bool(set(s1).intersection(s2))

Answer 5

This is another option:

import string

s = ('hello')
alpha = string.ascii_lowercase

if any(i in alpha for i in s):
    print('yes')
else:
    print('no')

Or maybe quicker:

import string

s = ('hello')
alpha = string.ascii_lowercase

for l in s:
    if l in alpha:
        print("yes")
        break
    print("no")

Answer 6

string.ascii_lowercase is a string that contains all the lower case alphabets, ie abcdefghijklmnopqrstuvwxyz .

So, in the if condition, if string.ascii_lowercase in s you are checking if the string contains a substring abcdefghijklmnopqrstuvwxyz .

You can try this,

if any(e in string.ascii_lowercase for e in s):
    ...

The expression inside any is a generator, thus it stop checking at the first match.

Another way to do this is,

if any(e.islower() for e in s):
    ...