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
有没有办法更快地完成这项工作？目前，我在4乘4矩阵上执行函数需要0.4s来执行函数20,000次，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。
答案 0 :(得分：4)
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
答案 1 :(得分：3)
这是我的答案。
reduce(lambda a, b: a + b, m).count(0)
时间：
%timeit count_zeros(m) #@J.F. Sebastian
1000000 loops, best of 3: 813 ns per loop
%timeit len([item for row in m for item in row if item == 0]) #@thefourtheye
1000000 loops, best of 3: 974 ns per loop
%timeit reduce(lambda a, b: a + b, m).count(0) #Mine
1000000 loops, best of 3: 1.02 us per loop
%timeit countzeros(m) #@frostnational
1000000 loops, best of 3: 1.07 us per loop
%timeit sum(row.count(0) for row in m) #@J.F. Sebastian
1000000 loops, best of 3: 1.28 us per loop
%timeit [item for row in m for item in row].count(0) #OP
1000000 loops, best of 3: 1.53 us per loop
@ thefourtheye是最快的。这是因为函数调用很少。
@ J.F。塞巴斯蒂安是我环境中最快的。我不知道为什么......
答案 2 :(得分：2)
您的解决方案的问题在于，您必须再次迭代列表以获得计数O（N）。但是len函数可以在O（1）中得到计数。
使用此
可以更快地完成此操作def zeroCount(M):
return len([item for row in M for item in row if item == 0])
答案 3 :(得分：2)
检查一下：
from itertools import chain, filterfalse # ifilterfalse for Python 2
def zeroCount(m):
total = 0
for x in filterfalse(bool, chain(*m)):
total += 1
return total
Python 3.3.3上的性能测试：
from timeit import timeit
from itertools import chain, filterfalse
import functools
m = ((2,0,2,2),(4,4,5,4),(0,9,4,8),(2,2,0,0))
def zeroCountOP():
return [item for row in m for item in row].count(0)
def zeroCountTFE():
return len([item for row in m for item in row if item == 0])
def zeroCountJFS():
return sum(row.count(0) for row in m)
def zeroCountuser2931409():
# `reduce` is in `functools` in Py3k
return functools.reduce(lambda a, b: a + b, m).count(0)
def zeroCount():
total = 0
for x in filterfalse(bool, chain(*m)):
total += 1
return total
print('Original code ', timeit(zeroCountOP, number=100000))
print('@J.F.Sebastian ', timeit(zeroCountJFS, number=100000))
print('@thefourtheye ', timeit(zeroCountTFE, number=100000))
print('@user2931409 ', timeit(zeroCountuser2931409, number=100000))
print('@frostnational ', timeit(zeroCount, number=100000))
以上给出了这些结果：
Original code 0.244224319984056
@thefourtheye 0.22169152169497108
@user2931409 0.19247795242092186
@frostnational 0.18846473728790825
@J.F.Sebastian 0.1439318853410907
@ J.F.Sebastian的解决方案是胜利者，我的是一名亚军（大约慢了20％）。
Python 2和Python 3的综合解决方案：
import sys
import itertools
if sys.version_info < (3, 0, 0):
filterfalse = getattr(itertools, 'ifilterfalse')
else:
filterfalse = getattr(itertools, 'filterfalse')
def countzeros(matrix):
''' Make a good use of `itertools.filterfalse`
(`itertools.ifilterfalse` in case of Python 2) to count
all 0s in `matrix`. '''
counter = 0
for _ in filterfalse(bool, itertools.chain(*matrix)):
counter += 1
return counter
if __name__ == '__main__':
# Benchmark
from timeit import repeat
print(repeat('countzeros(((2,0,2,2),(4,4,5,4),(0,9,4,8),(2,2,0,0)))',
'from __main__ import countzeros',
repeat=10,
number=100000))
答案 4 :(得分：1)
使用numpy：
import numpy
m=((2,0,2,2),(4,4,5,4),(0,9,4,8),(2,2,0,0))
numpy_m = numpy.array(m)
print numpy.sum(numpy_m == 0)
以上如何运作？首先，您的“矩阵”将转换为numpy数组（numpy.array(m)）。然后，检查每个条目是否为零（numpy_m == 0）。这产生二进制数组。对此二进制数组求和可得出原始数组中的零元素数。
请注意，对于较大的矩阵，numpy显然是有效的。与普通的python代码相比，4x4可能太小而无法看到大的性能差异，尤其是如果你正在初始化如上所述的python“矩阵”。
答案 5 :(得分：0)
一个笨拙的解决方案是：
import numpy as np
m = ((2,0,2,2),(4,4,5,4),(0,9,4,8),(2,2,0,0))
mm = np.array(m)
def zeroCountSmci():
return (mm==0).sum() # sums across all axes, by default