如何从列表中删除具有三个重复字母的单词？

Question

So I have two lists:

list1 = ["sch", "ggg", "bbb", "hello", "bye"]
list2 = ["s","c","h","b","g"]

and the output should look like this:

["hello","bye"]

Answer 1

Assuming you mean consecutive repeating characters - using a regular expression:

import re

words = ["sch", "ggg", "bbb", "hello", "bye"]
chars = ["s","c","h","b","g"]

pattern = "[{}]{{3,}}".format("".join(chars))

filtered = [word for word in words if not re.search(pattern, word)]

print(filtered)

Output:

['hello', 'bye']
>>> 

Answer 2

list1 = ["sch", "ggg", "bbb", "hello", "bye"]
list2 = ["s","c","h","b","g"]

list3=list1[3:]
print(list3)

Output: ['hello', 'bye']

Answer 3

Depending on whether you want to exclude words that have 3 consecutive instances of the letters or simply 3 instances anywhere in the word you can use one of these list comprehensions to filter list1:

list1 = ["sch", "ggg", "bbb", "hello", "bye","bags"]
list2 = ["s","c","h","b","g"]

# exclude 3 non-consecutive
list3 = [ w for w in list1 if sum(c in list2 for c in w)<3 ]
print(list3) # ['hello', 'bye']

# exclude 3 consecutive
list3 = [ w for w in list1 if "111" not in "".join("01"[c in list2] for c in w) ]
print(list3) # ['hello', 'bye', 'bags']

note: I added "bags" to your example data to illustrate the distinction