您必须设置一个条件，以检查正在搜索的元素是否在列表中

if 'your_element' in mylist:
    print mylist.index('your_element')
else:
    print None

我发现这两种解决方案更好，我自己尝试过

>>> expences = [2200, 2350, 2600, 2130, 2190]
>>> 2000 in expences
False
>>> expences.index(2200)
0
>>> expences.index(2350)
1
>>> index = expences.index(2350)
>>> expences[index]
2350

>>> try:
...     print(expences.index(2100))
... except ValueError as e:
...     print(e)
... 
2100 is not in list
>>> 

使用字典，首先处理列表，然后将索引添加到其中

from collections import defaultdict

index_dict = defaultdict(list)    
word_list =  ['foo','bar','baz','bar','any', 'foo', 'much']

for word_index in range(len(word_list)) :
    index_dict[word_list[word_index]].append(word_index)

word_index_to_find = 'foo'       
print(index_dict[word_index_to_find])

# output :  [0, 5]

a = ["foo","bar","baz",'bar','any','much']

indexes = [index for index in range(len(a)) if a[index] == 'bar']

me = ["foo", "bar", "baz"]
me.index("bar") 

您可以将此应用于列表中的任何成员以获取其索引

对于一个可比的

# Throws ValueError if nothing is found
some_list = ['foo', 'bar', 'baz'].index('baz')
# some_list == 2

自定义谓词

some_list = [item1, item2, item3]

# Throws StopIteration if nothing is found
# *unless* you provide a second parameter to `next`
index_of_value_you_like = next(
    i for i, item in enumerate(some_list)
    if item.matches_your_criteria())

按谓词查找所有项的索引

index_of_staff_members = [
    i for i, user in enumerate(users)
    if user.is_staff()]