how to use function "find" over matrices

1 Ansicht (letzte 30 Tage)
parham kianian
parham kianian am 9 Apr. 2020
Kommentiert: parham kianian am 10 Apr. 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?

Akzeptierte Antwort

Stephen23
Stephen23 am 10 Apr. 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

Weitere Antworten (1)

David Hill
David Hill am 9 Apr. 2020
min(x);
  3 Kommentare
David Hill
David Hill am 10 Apr. 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 am 10 Apr. 2020
Thank you David. It works well. But the method suggested by Stephen is a little easier.

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