Round the elements column of the matrix to the same digit than it is in the other column

I round elements one of column of matrix to 2 significant digits:

Result(:,5) = round(Result(:,5),2,'significant');

I wold like to round the elements another column of the same matrix to exactly same digits than it is in 'significant' digit column.

Is there any way to read back the number of digits from the 'significant' digit column to use it in other column?

Answer 1

[Matt J]

One way,

Result(:,5) = round(Result(:,5),2,'significant');
numDigits=strlength(extractAfter(string(Result(:,5)),'.'));
numDigits(isnan(numDigits))=0

[Laszlo Sleisz]

If the rounded number is integer, I get NaN value.

[Matt J]

You could post-correct with,

numDigits(isnan(numDigits))=0

[Matt J]

Laszlo Sleisz's comment moved here:

I got this error message:

Result = round(Result,numDigits);

Error using round

The second input must be a real integer scalar.

Error in kiertekeles (line 32)

Result = round(Result,numDigits);

[Matt J]

Yes, unfortunately, while Result may be a vector, round() will not take a vector input for numDigits. You will have to loop along the column and round one entry at a time, or abbreviate that to,

Result = arrayfun(@round, Result, numDigits)

[Matt J]

Laszlo Sleisz's comment moved here:

Thank you, now it works.

Is it possible to format the number to N digits to the right of the decimal point, delete the unnecessary zeros.

For example:

In first cell I have 20.600 and I would like to see 20.6.

[Matt J]

Thank you, now it works.

You are welcome, but please Accept-click the answer to indicate so.

Is it possible to format the number to N digits to the right of the decimal point, delete the unnecessary zeros.

Only if the array is converted to a string. If you plan to work with Result as numeric data, the only control you have over how it is displayed is through the format() command.

[Laszlo Sleisz]

I think in this case I should use Cell array, because as I know not possible to create one matrix with different data type.

Answer 2

[Stephen23]

S = 2;
A = 32*rand(5,7)-13

A = 5×7

-8.9001 -12.7270 9.5712 14.7717 6.2782 3.5775 10.7267 16.4892 2.1012 12.5711 0.2720 8.0913 -11.5669 -9.3372 -5.9647 -12.7967 3.1765 5.9842 12.7080 15.4362 -4.9144 -8.0932 -12.6045 6.2981 14.2722 -9.1178 8.2538 -3.3988 15.7385 0.4380 2.9713 -6.0512 2.5839 -9.1609 -4.6357

round(A(:,5),S,'significant') % for comparison

ans = 5×1

6.3000 8.1000 13.0000 -9.1000 2.6000

P = S-1-floor(log10(abs(A(:,5))));
B = round(A.*10.^P)./(10.^P)

B = 5×7

-8.9000 -12.7000 9.6000 14.8000 6.3000 3.6000 10.7000 16.5000 2.1000 12.6000 0.3000 8.1000 -11.6000 -9.3000 -6.0000 -13.0000 3.0000 6.0000 13.0000 15.0000 -5.0000 -8.1000 -12.6000 6.3000 14.3000 -9.1000 8.3000 -3.4000 15.7000 0.4000 3.0000 -6.1000 2.6000 -9.2000 -4.6000