计算与元组中的模式匹配的元素 [英] Counting elements matching a pattern in a tuple of tuples

问题描述

我有一个矩阵m，我要在其中计算零的数量.

I have a matrix m where I would like to calculate the number of zeros.

m=((2,0,2,2),(4,4,5,4),(0,9,4,8),(2,2,0,0))

我当前的代码如下:

def zeroCount(M):
    return [item for row in M for item in row].count(0)
    # list of lists is flattened to form single list, and number of 0 are counted

有什么办法可以更快地做到这一点?目前，我花了0.4秒的时间在4 x 4矩阵上执行20,000次功能，其中矩阵很可能包含零，而不是不包含零.

Is there any way to do this quicker? Currently, I'm taking 0.4s to execute the function 20,000 times on 4 by 4 matrices, where the matrices are equally likely to contain zeros as they are not.

其他一些问题可能是一些开始的地方(但是我无法使其工作比我的代码快):对可迭代的非零元素进行计数.

Some possible places to start (but which I could not make to work faster than my code) are these other questions: counting non-zero elements in numpy array, finding the indices of non-zero elements, and counting non-zero elements in iterable.

推荐答案

到目前为止最快的方法:

The fastest so far:

def count_zeros(matrix):
    total = 0
    for row in matrix:
        total += row.count(0)
    return total

---

对于2D元组，您可以使用生成器表达式:

def count_zeros_gen(matrix):
    return sum(row.count(0) for row in matrix)

时间比较:

%timeit [item for row in m for item in row].count(0) # OP
1000000 loops, best of 3: 1.15 µs per loop

%timeit len([item for row in m for item in row if item == 0]) # @thefourtheye
1000000 loops, best of 3: 913 ns per loop

%timeit sum(row.count(0) for row in m) 
1000000 loops, best of 3: 1 µs per loop

%timeit count_zeros(m)
1000000 loops, best of 3: 775 ns per loop

对于基线:

def f(m): pass
%timeit f(m)
10000000 loops, best of 3: 110 ns per loop

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