MATLAB Answers

Isti
1

How to round the decimals?

Asked by Isti
on 3 May 2012
Latest activity Edited by Jason Garcia on 7 Feb 2019
I have a number X = 0.135678
Then i just want to round it become 0.14. What to do?
Use round(X) will only give "0".
Thanks before :)

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Vida
on 11 Feb 2014
For many reasons, for one, you have many decimal points and you export to excel the data for distribution and others want to manipulate data further. You want your data to be represented 2 decimal points because it makes sense to look at. And also you want it to be precise so any further manipulation of data would not be compromised.
Kamania Ray on 6 Nov 2015
And because a lot of us are taking CS 1371 at Tech and out TAs like us to do this.
It probably helps to understand the visualization of data as being quite distinct from its storage in memory.
Certainly it is nice to see values with two decimal digits in Excel, but this is related to displaying of data (i.e. formatting in Excel), and not to how the data is stored in memory. Changing the formatting of a cell in Excel does not change its stored value, even if it displays different numbers of decimal digits. Similarly in MATLAB it makes more sense to keep all the numeric precision possible while calculating and storing a value, and considering displaying of that value as a separate task.
Unless there is a good reasons to change the stored data value then both MATLAB and Excel can handle those extra digits without breaking sweat: rounding does not save memory or make computations faster.

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8 Answers

Answer by Jos (10584)
on 11 Feb 2014
Edited by Stephen Cobeldick on 11 Nov 2015
 Accepted Answer

A = [pi exp(1) 1/7]
Ndecimals = 2
f = 10.^Ndecimals
A = round(f*A)/f

  1 Comment

Divide by f, not A.

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Junaid
Answer by Junaid
on 3 May 2012

This also works.
X = 0.135678;
Y = sprintf('%.2f',X)
You can set as many decimals as you want.

  6 Comments

Daniel
on 4 Mar 2015
Thanks!
Joep
on 11 Nov 2015
Very unconventional way to use. Converting from str2num and otherwise are "slow" process. It works but it is a real meshing solution. I prefer
round(X, n)
X your number witch you want to round, n the decicmals (- behind the comma/dot)
Stevilinsko on 21 Nov 2016
Joeps way only works with Matlab Versions starting at R2014b :/

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Answer by Walter Roberson
on 3 May 2012

Computationally it cannot be done: binary floating point arithmetic is not able to exactly represent most multiples of 0.01.

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Steven Lord
Answer by Steven Lord
on 7 Nov 2016

As of release R2014b you can use the round function in MATLAB to round to a specific number of decimal places.

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Wayne King
Answer by Wayne King
on 3 May 2012

One way here is:
X = 0.135678;
format bank;
X
Another way is:
format; %just returning the formatting
X = ceil(X*100)/100;
Probably the last way is the best because you don't have to mess with the formatting.

  2 Comments

Isti
on 3 May 2012
thanks :)
Jos (10584)
on 11 Feb 2014
Use round instead of ceil!

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Andrei Bobrov
Answer by Andrei Bobrov
on 3 May 2012

use roundn from Mapping Toolbox
roundn(X,-2)

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Answer by Prateek Sahay on 7 Nov 2016

If you want to round 1.556876 to three decimal places then multiply it with 1000 and the use round command and then again divide it by 1000. X=1.556876 X=X*1000 Means now X=1556.876 round(x) Means now X=1556.9 X=X/1000 Means now X=1.5569

  1 Comment

Walter Roberson
on 7 Nov 2016
Note that the result of the round() would be 1557 not 1556.9
Note that the result will not be exact. There is no way to represent exactly 1.557 in binary floating point. The closest it gets is 1.556999999999999939603867460391484200954437255859375
This will display as 1.557 in most output modes, but it will not be exactly that value.

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Answer by Jason Garcia on 7 Feb 2019
Edited by Jason Garcia on 7 Feb 2019

Maybe not exactly what you're looking for, but if you are looking for ceiling or floor measurements the below is a fun way to specifiy directly how you want to bin the array/value.
X = rand(100,1); %Rand 100 elmnt vector w/ range 0-1.
n = 100; %Use 100 for the nearest tenth.
cX = discretize(X,[0:1/n:1],[0+1/n:1/n:1]); %Rounds X UP to nearest 1/N.
%OR
fX = discretize(X,[0:1/n:1],[0:1/n:1-1/n]); %Rounds X DOWN to nearest 1/N.

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