检查NumPy数组中是否存在值的最有效方法是什么？

Question

我有一个非常大的NumPy数组

1 40 3
4 50 4
5 60 7
5 49 6
6 70 8
8 80 9
8 72 1
9 90 7
.... 

我想检查数组的第一列中是否存在值.我有一堆本土方式(例如遍历每一行并检查),但考虑到数组的大小,我想找到最有效的方法.

谢谢!

Answer 1

怎么样

if value in my_array[:, col_num]:
    do_whatever

编辑:我认为__contains__这是以与@ detly的版本相同的方式实现的

Answer 2

对我来说最明显的是:

np.any(my_array[:, 0] == value)

Answer 3

要检查多个值,可以使用numpy.in1d(),这是python关键字的元素功能版本.如果您的数据已排序,则可以使用numpy.searchsorted():

import numpy as np
data = np.array([1,4,5,5,6,8,8,9])
values = [2,3,4,6,7]
print np.in1d(values, data)

index = np.searchsorted(data, values)
print data[index] == values

Answer 4

迷人.我需要提高一系列循环的速度,这些循环必须以同样的方式执行匹配的索引确定.所以我决定在这里解决所有的解决方案,以及一些riff.

以下是我对Python 2.7.10的速度测试:

import timeit
timeit.timeit('N.any(N.in1d(sids, val))', setup = 'import numpy as N; val = 20010401020091; sids = N.array([20010401010101+x for x in range(1000)])')

18.86137104034424

timeit.timeit('val in sids', setup = 'import numpy as N; val = 20010401020091; sids = [20010401010101+x for x in range(1000)]')

15.061666011810303

timeit.timeit('N.in1d(sids, val)', setup = 'import numpy as N; val = 20010401020091; sids = N.array([20010401010101+x for x in range(1000)])')

11.613027095794678

timeit.timeit('N.any(val == sids)', setup = 'import numpy as N; val = 20010401020091; sids = N.array([20010401010101+x for x in range(1000)])')

7.670552015304565

timeit.timeit('val in sids', setup = 'import numpy as N; val = 20010401020091; sids = N.array([20010401010101+x for x in range(1000)])')

5.610057830810547

timeit.timeit('val == sids', setup = 'import numpy as N; val = 20010401020091; sids = N.array([20010401010101+x for x in range(1000)])')

1.6632978916168213

timeit.timeit('val in sids', setup = 'import numpy as N; val = 20010401020091; sids = set([20010401010101+x for x in range(1000)])')

0.0548710823059082

timeit.timeit('val in sids', setup = 'import numpy as N; val = 20010401020091; sids = dict(zip([20010401010101+x for x in range(1000)],[True,]*1000))')

0.054754018783569336

非常令人惊讶!数量级差异的订单!

总而言之,如果您只是想知道某个列表中是否存在某些内容:

• 19s N.any(N.in1d(numpy array))
• 15s x in(list)
• 8s N.any(x == numpy数组)
• 6s x in(numpy数组)
• .1s x in(set或a dictionary)

如果你想知道列表中的某些内容(顺序很重要):

• 12s N.in1d(x,numpy数组)
• 2s x ==(numpy数组)