Python：如果字符串 1 包含在字符串 2 中，如何检查两个具有重复字母的字符串？

Question

def can_spell_with(target_word, letter_word):
    valid = True
    target_word1 = [x.lower() for x in target_word]
    letter_word1 = [x.lower() for x in letter_word]
    for c in target_word1:
        if c not in letter_word1:
            valid = False
        else:
            valid = True
    return valid
print(can_spell_with('elL','HEllo'))
# True
print(can_spell_with('ell','helo'))
# False

In the code above: I am trying to figure out how to return True if the letter_word contains the target_word.

So 'ell' in 'helo' would return False But 'ell' in 'hello' would return True

Answer 1

Try to use in :

def can_spell_with(a, b):
     return (a.upper() in b.upper())
print(can_spell_with('Ell','HEllo'))
>>>True
print(can_spell_with('ell','helo'))
>>>False

Answer 2

You can use in but you first have to make the cases the same:

def can_spell_with(target_word, letter_word):
    return target_word.lower() in letter_word.lower()

Answer 3

You're doing a case-insensitive search. No need to do a list comprehension. Just use lower() or upper() to convert all characters to lower or uppercase and use in :

def can_spell_with(target_word, letter_word):
    return target_word.lower() in letter_word.lower()

Alternatively you could do a case insensitive regex search:

import re

def can_spell_with(target_word, letter_word):
    return re.search(target_word, letter_word, re.IGNORECASE) is not None

Answer 4

target_word = target_word.lower() 
letter_word = letter_word.lower()

lenn = len(target_word)     
valid = False               

for i in range(len(letter_word)-lenn):   
    if letter_word[i:i+lenn] == target_word:   
        valid = True
return valid

Answer 5

this will check the whether word1 in word2 irrespective of lowercase or uppercase

def func(word1, word2):
    first_char = word1[0]
    for index, char in enumerate(word2):
        if char==first_char:
            if word2[index:len(word1)+1]==word1:
                return True
    return False