检查字符串是否包含 set 中的任何字符

Question

Actually, i'm working on a task from SPOJ. How to check if a string contains any char from set, but first char from string where a char from set occur can not be deleted.

Fe Have a string

word = "anAconda_elEphant"

and a set of vowels:

vowels = set('aeiouyAEIOUY')

I want in result a string

word = "ancnd_lphnt"

This should return True when occurence of any char in set is equal 1. I know that argument for a method .count() must be str, not set.

if word.count(vowels) == 1:
   for char in word[char_pos:]:
        if char in vowels:
            char.replace('')

Answer 1

just use a regular expression

import re
word = "anAconda_elEphant"
# use a  "lookbehind" to make sure there is at least one character in front of this character...
print(re.sub("(?<=.)[aeiouyAEIOUY]",'',word))
# 'ancnd_lphnt'

as mentioned if you expect it to skip the first match of the set as opposed to just the first letter you will need a different solution

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',"bace"))
# 'bc' # a is not the FIRST letter so it is replaced

the easiest is to split it into two steps first split the string on the first match

word = "bace"
splitted_string = re.split("(.*?[aeiouyAEIOUY])",word,1)
# you will notice we have an extra empty string at the beginning of our matches ... so we can skip that
lhs,rhs = splitted_string[1:]
# now just run a simple re.sub on our rhs and rejoin the halves
print(lhs + re.sub("[aeiouyAEIOUY]",'',rhs))
# results in "bac"

Answer 2

You can use a for loop as below. The idea is to build a list, and use a flag to mark when you meet a character from vowels .

word = "anAconda_elEphant"
vowels = set('aeiouyAEIOUY')

flag = False

L = []
for ch in word:
    if (ch not in vowels) or (not flag):
        L.append(ch)
    if ch in vowels:
        flag = True

word = ''.join(L)

print(word)

ancnd_lphnt

Answer 3

print(vowels.intersection(word))

https://docs.python.org/3/library/stdtypes.html#frozenset.intersection