Question

vowel = ['a','e','i','o','u']
my_list = ['hbyl','hawk','ibzj','hcxk','gawk']

my_list中的单词。如果单词没有一个元音。从列表中删除单词

for word in my_list:
    if word does not include at least one vowel:
        my_list.remove(word)

print (my_list)
['hawk','ibzj','gawk']

Answer 1

这应该有效！：）

vowel = ['a','e','i','o','u'] 
my_list = ['hbyl','hawk','ibzj','hcxk','gawk']
vowset = set(vowel)
[word for word in my_list if vowset.intersection(word)]

Answer 2

您不应在迭代时删除元素。相反，构造一个新列表，最好使用列表理解：

vowel = ['a','e','i','o','u']
my_list = ['hbyl','hawk','ibzj','hcxk','gawk']

new_my_list = [word for word in my_list if any(v in word for v in vowel)]

本质上，列表理解所说的是“仅当该单词中有任何元音时，才包含任何给定的单词。”

Answer 3

您可以将函数filterfalse()与设置的方法isdisjoint()一起使用：

from itertools import filterfalse

vowel = ['a','e','i','o','u']
my_list = ['hbyl','hawk','ibzj','hcxk','gawk']

vowel= set(vowel)
list(filterfalse(lambda x: vowel.isdisjoint(x), my_list))
# ['hawk', 'ibzj', 'gawk']

如何从python列表中删除没有元音的单词

3 个答案: