计算Python中多维数组中数组的出现 [英] count occurrences of arrays in multidimensional arrays in python

问题描述

我有以下类型的数组:

a = array([[1,1,1],
           [1,1,1],
           [1,1,1],
           [2,2,2],
           [2,2,2],
           [2,2,2],
           [3,3,0],
           [3,3,0],
           [3,3,0]])

我想计算每种类型的数组(例如

I would like to count the number of occurrences of each type of array such as

[1,1,1]:3, [2,2,2]:3, and [3,3,0]: 3 

如何在python中实现呢?是否可以不使用for循环并计入字典?它必须快，并且应少于0.1秒左右.我查看了Counter，numpy bincount等.但是，这些是针对单个元素而不是数组.

How could I achieve this in python? Is it possible without using a for loop and counting into a dictionary? It has to be fast and should take less than 0.1 seconds or so. I looked into Counter, numpy bincount, etc. But, those are for individual element not for an array.

谢谢.

推荐答案

您可以使用 np.unique 给我们提供职位每个唯一行的开头，也有一个可选参数return_counts来给我们计数.因此，实现看起来像这样-

You could convert those rows to a 1D array using the elements as two-dimensional indices with np.ravel_multi_index. Then, use np.unique to give us the positions of the start of each unique row and also has an optional argument return_counts to give us the counts. Thus, the implementation would look something like this -

def unique_rows_counts(a):

    # Calculate linear indices using rows from a
    lidx = np.ravel_multi_index(a.T,a.max(0)+1 )

    # Get the unique indices and their counts
    _, unq_idx, counts = np.unique(lidx, return_index = True, return_counts=True)

    # return the unique groups from a and their respective counts
    return a[unq_idx], counts

样品运行-

In [64]: a
Out[64]: 
array([[1, 1, 1],
       [1, 1, 1],
       [1, 1, 1],
       [2, 2, 2],
       [2, 2, 2],
       [2, 2, 2],
       [3, 3, 0],
       [3, 3, 0],
       [3, 3, 0]])

In [65]: unqrows, counts = unique_rows_counts(a)

In [66]: unqrows
Out[66]: 
array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 0]])
In [67]: counts
Out[67]: array([3, 3, 3])

基准化

假设您可以使用numpy数组或集合作为输出，那么可以对到目前为止提供的解决方案进行基准测试，就像这样-

Benchmarking

Assuming you are okay with either numpy arrays or collections as outputs, one can benchmark the solutions provided thus far, like so -

函数定义:

import numpy as np
from collections import Counter

def unique_rows_counts(a):
    lidx = np.ravel_multi_index(a.T,a.max(0)+1 )
    _, unq_idx, counts = np.unique(lidx, return_index = True, return_counts=True)
    return a[unq_idx], counts

def map_Counter(a):
    return Counter(map(tuple, a))    

def forloop_Counter(a):      
    c = Counter()
    for x in a:
        c[tuple(x)] += 1
    return c

时间:

In [53]: a = np.random.randint(0,4,(10000,5))

In [54]: %timeit map_Counter(a)
10 loops, best of 3: 31.7 ms per loop

In [55]: %timeit forloop_Counter(a)
10 loops, best of 3: 45.4 ms per loop

In [56]: %timeit unique_rows_counts(a)
1000 loops, best of 3: 1.72 ms per loop

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