我該如何平均分塊列？

Question

說，例如，我有以下矩陣：

[1 2 3 4 1 2 3 4 

1 2 3 4 1 2 3 4 

1 2 3 4 1 2 3 4] 

我怎麼能編程平均起來，也就是說，每四列，這樣我就結束了 有：

[2.5 2.5 

2.5 2.5 

2.5 2.5] 

？

感謝，

湯姆

Answer 1

可以沒有循環使用reshape完成：

A = [1 2 3 4 1 2 3 4 
    1 2 3 4 1 2 3 4 
    1 2 3 4 1 2 3 4]; %// data matrix 
N = 4; %// number of columns to average 
result = squeeze(mean(reshape(A,size(A,1), N, []), 2)); 

---

如果您有圖像處理工具箱，你也可以使用blockproc：基於sum

result = blockproc(A, [size(A,1) N], @(x) mean(x.data, 2)) 

Answer 2

一個簡單的方法是濫用filter：

A = [1 2 3 4 1 2 3 4 ; 
    1 2 3 4 1 2 3 4 ; 
    1 2 3 4 1 2 3 4 ] 

N = 4; 

B = filter(ones(1,N)/N,1,A,[],2) 
C = B(:,N:N:end) 

給出：

C = 

    2.5000 2.5000 
    2.5000 2.5000 
    2.5000 2.5000 

這似乎不是所有的整形頗有幾分更快擠壓行動;）雖然沒有證據。

Answer 3

一種方法和reshape -

reshape(sum(reshape(A,size(A,1),N,[]),2)/N,size(A,1),[]) 

---

這裏給出了迄今發佈的解決方案的一些基準。

基準測試代碼 -

m = 1000; %// No. of rows in input 
n = 4000; %// No. of cols in input 
A = rand(m,n); %// Input with random data 
N = 4; %// number of columns to average; 
num_runs = 200; %// No. of iterations 

disp('----------------- With sum+reshape'); 
tic 
for k1 = 1:num_runs 
    out1 = reshape(sum(reshape(A,size(A,1),N,[]),2)/N,size(A,1),[]); 
end 
toc,clear out1 

disp('----------------- With mean+reshape'); 
tic 
for k1 = 1:num_runs 
    out2 = squeeze(mean(reshape(A,size(A,1), N, []), 2)); 
end 
toc,clear out2 

disp('----------------- With blockproc'); 
tic 
for k1 = 1:num_runs 
    out3 = blockproc(A, [size(A,1) N], @(x) mean(x.data, 2)); 
end 
toc,clear out3 

disp('----------------- With filter'); 
tic 
for k1 = 1:num_runs 
    B = filter(ones(1,N)/N,1,A,[],2); 
    out4 = B(:,N:N:end); 
end 
toc,clear out4 B 

結果 -

----------------- With sum+reshape 
Elapsed time is 4.011844 seconds. 
----------------- With mean+reshape 
Elapsed time is 3.945733 seconds. 
----------------- With blockproc 
Elapsed time is 59.018457 seconds. 
----------------- With filter 
Elapsed time is 31.220875 seconds.