Question

我有两个列表，其中一个是大量的（数百万个元素），另外几个。我想做以下

bigArray=[0,1,0,2,3,2,,.....]

smallArray=[0,1,2,3,4]

for i in len(smallArray):
  pts=np.where(bigArray==smallArray[i])
  #Do stuff with pts...

上述工作，但很慢。有没有办法在不诉诸C语言的情况下更有效地做到这一点？

Answer 1

在您的情况下，您可以从预先分配您的大阵列中受益。下面的示例演示如何将时间从约45秒减少到2秒（在我的笔记本电脑上）（对于阵列5e6与1e3的一组特定长度）。显然，如果阵列大小差别很大，那么解决方案将不是最佳选择。例如。使用默认解决方案，复杂度为O（bigN * smallN），但对于我建议的解决方案，它是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：

Brute 42.5278530121

非暴力1.57193303108

Answer 2

Numpy提供函数numpy.searchsorted：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]

Answer 3

到目前为止，我认为没有必要为numpy;你可以使用defaultdict，只要记忆力充足，如果观察次数不是太多，就应该这样。

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]

这应该会给你一些速度。

Numpy数组：有效地找到匹配的索引

3 个答案: