how to use function "find" over matrices

5 vues (au cours des 30 derniers jours)
parham kianian
parham kianian le 9 Avr 2020
Commenté : parham kianian le 10 Avr 2020
Suppose x = rand(1e5,1e5);
I want to find the lowest number in each column of x without using for loop.
Is it possible?

Réponse acceptée

Stephen23
Stephen23 le 10 Avr 2020
"find the first value in each column which lis ess than 0.25 but greater than 0.2."
>> x = rand(13,7)
x =
0.833676 0.654529 0.869031 0.922756 0.586565 0.278136 0.595271
0.335144 0.936490 0.552751 0.676447 0.924786 0.813253 0.388776
0.171351 0.992074 0.426227 0.814150 0.599205 0.378885 0.470132
0.368957 0.162066 0.044178 0.911514 0.431260 0.111011 0.348894
0.787866 0.150796 0.783209 0.406310 0.116503 0.232302 0.350849
0.676606 0.782741 0.251472 0.223849 0.872576 0.665249 0.287961
0.176415 0.750830 0.958001 0.274026 0.107420 0.716966 0.612980
0.030644 0.103396 0.297286 0.256401 0.902245 0.486087 0.812681
0.563573 0.414845 0.615615 0.335131 0.589437 0.396942 0.780523
0.994748 0.314337 0.721215 0.946815 0.446822 0.252527 0.593235
0.438298 0.516228 0.978322 0.183097 0.011558 0.731435 0.948024
0.496606 0.172242 0.224708 0.339960 0.425773 0.730056 0.809002
0.234744 0.195880 0.086287 0.702632 0.708232 0.489843 0.558111
>> y = x>0.2 & x<0.25;
>> [row,col] = find(y & cumsum(y,1)==1)
row =
13
12
6
5
col =
1
3
4
6

Plus de réponses (1)

David Hill
David Hill le 9 Avr 2020
  3 commentaires
David Hill
David Hill le 10 Avr 2020
x = rand(1e4);
a = x.*(x>.2&x<.25);
b=arrayfun(@(y)a(find(a(:,y),1),y),1:size(a,2));
parham kianian
parham kianian le 10 Avr 2020
Thank you David. It works well. But the method suggested by Stephen is a little easier.

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