计算元组中与模式匹配的元素

我有一个矩阵m我想计算零的数量。

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吗？目前，我花费 0.4 秒在 4 x 4 矩阵上执行该函数 20,000 次，其中矩阵包含零的可能性与不包含零的可能性相同。

一些可能的起点（但我无法使其比我的代码更快地工作）是以下其他问题：计算 numpy 数组中的非零元素 https://stackoverflow.com/questions/3797158/counting-non-zero-elements-within-each-row-and-within-each-column-of-a-2d-numpy, 查找非零元素的索引 https://stackoverflow.com/questions/23159791/find-the-indices-of-non-zero-elements-and-group-by-values, and 计算可迭代中的非零元素 https://stackoverflow.com/questions/3393431/how-to-counting-not-0-elements-in-an-iterable.

迄今为止最快：

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

对于 2D 元组你可以使用生成器表达式 https://stackoverflow.com/a/23299944/4279:

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

时间对比：

%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