How to reduce precision of data greator than zero only? (Matlab R2019a)

Question

taimour sadiq 2021-3-16

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https://ww2.mathworks.cn/matlabcentral/answers/774172-how-to-reduce-precision-of-data-greator-than-zero-only-matlab-r2019a

评论： taimour sadiq 2021-3-16

Hello... Lets suppose i have a data stored in a variable

A = [1.23323      45.24343       0.1       0.001        455.34543         0.0001 ]

i want to shrink pricision of data only greator than zero... i tryed

A(A>0) - str2double(compose('%.2f',A(A>0)))

so the result should be

A = [1.23     45.24      0.1       0.001        455.34          0.0001]       %% <<<< Desired Results

but its not woking, it applying %.2f on whole data....and giving >>>

A = [1.23 45.24 0.10 0.00 455.34 0.00]

Kindly Guide. Thanks

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Walter Roberson 2021-3-16

so the result should be

No, 0.001 is greater than 0, so it should have its precision reduced.

taimour sadiq 2021-3-16

ooops....Yes you r right... my basic mistake ...

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Answer 1

Walter Roberson 2021-3-16

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/774172-how-to-reduce-precision-of-data-greator-than-zero-only-matlab-r2019a#answer_648972

在 MATLAB Online 中打开

format long g
A = [1.23323      -45.24343       0.1       0.001        455.34543         -0.0001 ]
A = 1×6
                   1.23323                 -45.24343                       0.1                     0.001                 455.34543                   -0.0001
mask = A>0;
A(mask) = round(A(mask),2);
A
A = 1×6
                      1.23                 -45.24343                       0.1                         0                    455.35                   -0.0001

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Walter Roberson 2021-3-16

在 MATLAB Online 中打开

It is possible that what you really want is

format long g
A = [1.23323      -45.24343       0.1       0.001        455.34543         -0.0001 ]
A = 1×6
                   1.23323                 -45.24343                       0.1                     0.001                 455.34543                   -0.0001
mask = abs(A)>0.01;
A(mask) = round(A(mask),2);
A
A = 1×6
                      1.23                    -45.24                       0.1                     0.001                    455.35                   -0.0001