Numpy提供了numpy.searchsorted函数：http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.searchsorted.html
示例：

>>> import numpy as np
>>> sorted = np.argsort(big_list)
>>> r = np.searchsorted(big_list, small_list, side='right',sorter=sorted)
>>> l  = np.searchsorted(big_list, small_list, side='left',sorter=sorted)
>>> for b, e in zip(l, r):
...     inds = sorted[b:e]

在您的情况下，您可能会受益于预排序您的大数组。下面的例子演示了如何将时间从~ 45秒减少到2秒（在我的笔记本电脑上）（对于一组特定长度的数组5e 6和1 e3）。显然，如果数组大小相差很大，那么这个解决方案就不是最优的。例如，默认解决方案的复杂度为O（bigN*smallN），但对于我建议的解决方案，复杂度为O（（bigN+smallN）*log（bigN））

import numpy as np, numpy.random as nprand, time, bisect

bigN = 5e6
smallN = 1000
maxn = 1e7
nprand.seed(1)  
bigArr = nprand.randint(0, maxn, size=bigN)
smallArr = nprand.randint(0, maxn, size=smallN)

# brute force 
t1 = time.time()
for i in range(len(smallArr)):
    inds = np.where(bigArr == smallArr[i])[0]
t2 = time.time()
print "Brute", t2-t1

# not brute force (like nested loop with index scan)
t1 = time.time()
sortedind = np.argsort(bigArr)
sortedbigArr = bigArr[sortedind]
for i in range(len(smallArr)):
    i1 = bisect.bisect_left(sortedbigArr, smallArr[i])
    i2 = bisect.bisect_right(sortedbigArr, smallArr[i])
    inds = sortedind[i1:i2]
t2=time.time()
print "Non-brute", t2-t1

输出：
电话：+86-527 - 8530121
非暴力1.57193303108

到目前为止，我看不出有任何需要numpy;你可以使用defaultdict，只要你的内存足够，如果观察的数量不是太多，应该是数百万。

big_list = [0,1,0,2,3,2,5,6,7,5,6,4,5,3,4,3,5,6,5]
small_list = [0,1,2,3,4]

from collections import defaultdict

dicto = defaultdict(list) #dictionary stores all the relevant coordinates
                          #so you don't have to search for them later

for ind, ele in enumerate(big_list):
    dicto[ele].append(ind)

结果：

>>> for ele in small_list:
...     print dicto[ele]
... 
[0, 2]
[1]
[3, 5]
[4, 13, 15]
[11, 14]

这应该能给予你一些速度。