计算给定范围内的矩阵中的值数

问题描述

我有矩阵

A=[2 3 4 5 6 7;
   7 6 5 4 3 2]

我想计算多少个元素的值大于3且小于6.

I want to count how many number of elements have a value greater than 3 and less than 6.

推荐答案

我可以想到几种方法:

count = numel(A( A(:)>3 & A(:)<6 ))      %# (1)

count = length(A( A(:)>3 & A(:)<6 ))     %# (2)

count = nnz( A(:)>3 & A(:)<6 )           %# (3)

count = sum( A(:)>3 & A(:)<6 )           %# (4)

Ac = A(:);
count = numel(A( Ac>3 & Ac<6 ))          %# (5,6,7,8)
%# prevents double expansion
%# similar for length(), nnz(), sum(),
%# in the same order as (1)-(4)

count = numel(A( abs(A-(6+3)/2)<3/2 ))   %# (9,10,11,12)
%# prevents double comparison and &
%# similar for length(), nnz(), sum()
%# in the same order as (1)-(4)

所以，让我们测试一下最快的方法是什么.测试代码:

So, let's test what the fastest way is.Test code:

A = randi(10, 50);
tic
for ii = 1:1e5

    %# method is inserted here

end
toc

结果(最好的5次运行，都在几秒钟之内):

results (best of 5 runs, all in seconds):

%# ( 1): 2.981446
%# ( 2): 3.006602
%# ( 3): 3.077083
%# ( 4): 2.619057
%# ( 5): 3.011029
%# ( 6): 2.868021
%# ( 7): 3.149641
%# ( 8): 2.457988
%# ( 9): 1.675575
%# (10): 1.675384
%# (11): 2.442607
%# (12): 1.222510

因此，看来count = sum(( abs(A(:)-(6+3)/2)<3/2 ));是前往此处的最佳方法.

So it seems that count = sum(( abs(A(:)-(6+3)/2)<3/2 )); is the best way to go here.

就个人而言:我没想到比较会比Matlab中的算术慢-有人知道对此的解释吗?

On a personal note: I did not expect comparison to be slower than arithmetic in Matlab -- does anyone know an explanation for this?

Plus:为什么nnz与sum相比这么慢?我想现在我知道比较比算术要慢，所以这很有道理.

Plus: why is nnz so slow compared to sum? I guess that makes sense now that I know comparison is slower than arithmetic...

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