python – 计算与元组元组中的模式匹配的元素

发布时间：2020-12-20 12:24:27 所属栏目：Python 来源：网络整理

导读：我有一个矩阵m,我想计算零的数量. m=((2,2,2),(4,4,5,4),(0,9,8),(2,0)) 我目前的代码如下： 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 有没

我有一个矩阵m,我想计算零的数量.

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

我目前的代码如下：

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

有没有办法更快地做到这一点？目前,我花了0.4秒在4乘4矩阵上执行20,000次函数,其中矩阵同样可能包含零,因为它们不是.

一些可能的起点(但我无法比我的代码更快地工作)是其他问题：counting non-zero elements in numpy array,finding the indices of non-zero elements和counting non-zero elements in iterable.

解决方法

到目前为止最快的：

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

对于2D元组,你可以use a generator expression：

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

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!