来自Dive Into Python:

>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> li.index("example")
5

如果你只是想知道一个元素是否包含在列表中:

>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> 'example' in li
True
>>> 'damn' in li
False

来自Dive Into Python:

>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> li.index("example")
5

假设你想在numpy数组中找到一个值， 我想这样做可能有用:

Numpy.where(arr=="value")[0]

列表的索引方法将为您完成此工作。如果想要保证顺序，请先使用sorted()对列表进行排序。Sorted接受cmp或key参数来指示如何进行排序:

a = [5, 4, 3]
print sorted(a).index(5)

Or:

a = ['one', 'aardvark', 'a']
print sorted(a, key=len).index('a')

我通过改编一些教程发现了这一点。感谢谷歌，也感谢你们所有人;)

def findall(L, test):
    i=0
    indices = []
    while(True):
        try:
            # next value in list passing the test
            nextvalue = filter(test, L[i:])[0]

            # add index of this value in the index list,
            # by searching the value in L[i:] 
            indices.append(L.index(nextvalue, i))

            # iterate i, that is the next index from where to search
            i=indices[-1]+1
        #when there is no further "good value", filter returns [],
        # hence there is an out of range exeption
        except IndexError:
            return indices

一个非常简单的用法:

a = [0,0,2,1]
ind = findall(a, lambda x:x>0))

[2, 3]

附言:抱歉我的英语不好