how to use function "find" over matrices

Suppose x = rand(1e5,1e5);
I want to find the lowest number in each column of x without using for loop.
Is it possible?

 採用された回答

Stephen23
Stephen23 2020 年 4 月 10 日

1 投票

"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

その他の回答 (1 件)

David Hill
David Hill 2020 年 4 月 9 日

0 投票

min(x);

3 件のコメント

parham kianian
parham kianian 2020 年 4 月 9 日
編集済み: parham kianian 2020 年 4 月 9 日
Let me to clarify the question.
Suppose it is aimed to find the first value in each column which lis ess than 0.25 but greater than 0.2.
How can I do that?
David Hill
David Hill 2020 年 4 月 10 日
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 2020 年 4 月 10 日
Thank you David. It works well. But the method suggested by Stephen is a little easier.

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