Matlab矩阵中每行的最后一个正元素

Question

I would like to find the last positive element per row in a matrix X of dimension wxy in Matlab.

rng default;  

%if w=1, then the code below works
A=sort(randn(1,20), 'descend');
idx=find(A>=0, 1, 'last');

%if w>1, how can I proceed?
A=sort(randn(8000,20),2, 'descend');
%idx=?
%I am expecting idx=[12;5;8;...]

Could you help me with a very efficient code?

Answer 1

The general answer to your title is harder than the answer to your specific case. In your example you seem to be asking for the "last positive element per row in a Matlab matrix in which each row is sorted in descending order ". This is equivalent to asking for "the smallest positive value in each row" which can be accomplished without sorting:

function [val, ind] = smallest_positive(A, dim)

if nargin < 2, A = A(:); dim = 1; end
A(A <= 0) = inf;
[val, ind] = min(A, [], dim);

Example of use:

>> A = randn(3, 8)

A =

    0.7990    0.2120   -0.7420    0.3899   -0.5596    0.7812   -0.2656    0.9863
    0.9409    0.2379    1.0823    0.0880    0.4437    0.5690   -1.1878   -0.5186
   -0.9921   -1.0078   -0.1315   -0.6355   -0.9499   -0.8217   -2.2023    0.3274

>> [val, ind] = smallest_positive(A, 2)

val =

    0.2120
    0.0880
    0.3274


ind =

     2
     4
     8

Note that this returns the "last" positive value on each row only in the sense of "the value that would be last if you were to do the sort ". If you literally want the last positive value on each row, whether sorted or not, then Divakar's answer is the way to go.

Answer 2

rng default;  

A=sort(randn(8000,20),2, 'descend');
idx = sum(A>=0, 2);

Actually you don't need to sort.

A = randn(8000,20);
idx = sum(A>=0, 2);

Answer 3

The most simple solution I can think of (given A is sorted):

[~,idx] = min(A>=0,[],2);   % Returns the first indices of the zero entries
idx = idx-1;               % Get the indices before the first zero entries

Answer 4

Based on the question title I am solving for a generic case and making no assumption about the inputs being sorted or otherwise.

The idea here is to flip the rows and compare against zero and then get argmax along each row and then compensate for the flipping by subtracting from row length -

[~,idx] = max(a(:,end:-1:1)>=0,[],2);
out = size(a,2) - idx + 1

To get the corresponding elements, simply get the linear indices and index -

a_last = a((out-1)*size(a,1) + [1:size(a,1)]')

Sample run -

>> a
a =
    1.6110    0.0854   -0.8777    0.6078    0.0544   -0.4089    0.0675
    0.7708    1.6510    0.1572   -0.7475    0.0218   -0.8292    1.0934
   -0.4704    1.2351    1.2660    2.2117   -0.3616   -0.9500   -0.7682
    0.8539   -0.5427   -1.0213    0.2489   -1.6312    0.0723    0.1284
    1.5050    1.4430    1.1947    0.2846   -1.2621    0.5518    1.4290
    0.1785    1.1087   -0.0225    1.1447    0.2316   -0.2431   -1.2750
    0.3089    1.5716   -1.9958    0.0015    1.5448   -0.0750    0.4965
    0.3593    0.8143    0.4389   -0.2541    0.1558   -0.2965    0.7111
>> [~,idx] = max(a(:,end:-1:1)>=0,[],2);
>> out = size(a,2) - idx + 1
out =
     7
     7
     4
     7
     7
     5
     7
     7
>> a((out-1)*size(a,1) + [1:size(a,1)]')
ans =
    0.0675
    1.0934
    2.2117
    0.1284
    1.4290
    0.2316
    0.4965
    0.7111