# Help with if statement in a calculation loop

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Ivan Mich 2021 年 4 月 12 日
コメント済み: Rik 2021 年 4 月 12 日
Hello,
I have a problem with a code. I want to calculate some numbers but I want to "limit" my results via a loop.
I want values <1 to be =1, values with >10 to be 10 and the others to be calculated from this equation:
x= x*1.3+0.5. I wrote this code but it is not use
x= 5 + randn * 0.5
if x < 1
x == 1
elseif x > 10
x == 10
else x > 3
x == 1.3* x -0.75
end
Where is the problem?
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Ivan Mich 2021 年 4 月 12 日
I am sorry but running this code stil values up to 10 exist.
It is no use

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### 採用された回答

Stephen 2021 年 4 月 12 日
The simplest and most efficient solution is to use MIN and MAX:
x = 5 + randn(5,7)*1.5;
x = x*1.3 + 0.5
x = 5×7
7.6491 8.4519 8.8820 8.7385 9.8065 8.6945 8.5660 5.7131 13.1416 6.3650 9.1859 6.9860 9.4199 5.4750 8.8947 7.3018 6.4618 7.2714 9.2859 8.7553 5.7112 8.6327 5.6802 3.2153 9.5393 7.2815 7.1600 11.3087 6.1828 6.8579 9.3617 8.8780 4.6976 3.9478 7.3776
x = min(max(x,1),10) % this is all you need.
x = 5×7
7.6491 8.4519 8.8820 8.7385 9.8065 8.6945 8.5660 5.7131 10.0000 6.3650 9.1859 6.9860 9.4199 5.4750 8.8947 7.3018 6.4618 7.2714 9.2859 8.7553 5.7112 8.6327 5.6802 3.2153 9.5393 7.2815 7.1600 10.0000 6.1828 6.8579 9.3617 8.8780 4.6976 3.9478 7.3776

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### その他の回答 (1 件)

Rik 2021 年 4 月 12 日

The problem is that you assume Matlab will process each element of x separately. Matlab will only do that if you use a loop.
An alternative is to use logical indexing to process x as an array.
L=x<1;
x(L)=1;
L=x>10;
x(L)=10;
L=x>3;
x(L)=1.3* x(L) -0.75;
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Rik 2021 年 4 月 12 日
In that case you should use the code Stephen suggested.
If you have a piecewise function you can use the code I suggested.

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