numpy:累积多重计数 [英] numpy: cumulative multiplicity count

问题描述

我有一个可能有重复的整数排序数组.我想计算连续的相等值，当一个值与前一个值不同时从零开始.这是用一个简单的python循环实现的预期结果:

I have a sorted array of ints which might have repetitions. I would like to count consecutive equal values, restarting from zero when a value is different from the previous one. This is the expected result implemented with a simple python loop:

import numpy as np

def count_multiplicities(a):
    r = np.zeros(a.shape, dtype=a.dtype)
    for i in range(1, len(a)):
        if a[i] == a[i-1]:
            r[i] = r[i-1]+1
        else:
            r[i] = 0
    return r

a = (np.random.rand(20)*5).astype(dtype=int)
a.sort()

print "given sorted array: ", a
print "multiplicity count: ", count_multiplicities(a)

输出:

given sorted array:  [0 0 0 0 0 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4]
multiplicity count:  [0 1 2 3 4 0 1 2 0 1 2 3 0 1 2 3 0 1 2 3]

如何使用 numpy 以有效的方式获得相同的结果?数组很长，但重复次数很少(比如不超过十次).

How can I get the same result in an efficient way using numpy? The array is very long, but the repetitions are just a few (say no more than ten).

在我的特殊情况下，我也知道值从零开始，并且连续值之间的差异是 0 或 1(值之间没有间隙).

In my special case I also know that values start from zero and that the difference between consecutive values is either 0 or 1 (no gaps in values).

推荐答案

这里有一个 基于cumsum的矢量化方法-

Here's one cumsum based vectorized approach -

def count_multiplicities_cumsum_vectorized(a):      
    out = np.ones(a.size,dtype=int)
    idx = np.flatnonzero(a[1:] != a[:-1])+1
    out[idx[0]] = -idx[0] + 1
    out[0] = 0
    out[idx[1:]] = idx[:-1] - idx[1:] + 1
    np.cumsum(out, out=out)
    return out

样品运行 -

In [58]: a
Out[58]: array([0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4])

In [59]: count_multiplicities(a) # Original approach
Out[59]: array([0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 0, 1, 2])

In [60]: count_multiplicities_cumsum_vectorized(a)
Out[60]: array([0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 0, 1, 2])

运行时测试 -

In [66]: a = (np.random.rand(200000)*1000).astype(dtype=int)
    ...: a.sort()
    ...: 

In [67]: a
Out[67]: array([  0,   0,   0, ..., 999, 999, 999])

In [68]: %timeit count_multiplicities(a)
10 loops, best of 3: 87.2 ms per loop

In [69]: %timeit count_multiplicities_cumsum_vectorized(a)
1000 loops, best of 3: 739 µs per loop

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