How to round the decimals?

193 vues (au cours des 30 derniers jours)
Isti
Isti le 3 Mai 2012
Commenté : Walter Roberson le 18 Mai 2021
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 :)
  10 commentaires
Mahaveer Singh
Mahaveer Singh le 18 Mai 2021
ans=round(X,2)
Walter Roberson
Walter Roberson le 18 Mai 2021
Right, these days round() in MATLAB supports passing in the number of decimal digits. When the question was originally asked, that option was not available.
Also, some of the users were needing to work in Simulink, but the round block https://www.mathworks.com/help/simulink/slref/roundingfunction.html does not support giving a number of decimal digits.

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Réponse acceptée

Jos (10584)
Jos (10584) le 11 Fév 2014
Modifié(e) : Stephen23 le 11 Nov 2015
A = [pi exp(1) 1/7]
Ndecimals = 2
f = 10.^Ndecimals
A = round(f*A)/f
  1 commentaire
Marc Lalancette
Marc Lalancette le 13 Oct 2015
Divide by f, not A.

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Plus de réponses (7)

Walter Roberson
Walter Roberson le 3 Mai 2012
Computationally it cannot be done: binary floating point arithmetic is not able to exactly represent most multiples of 0.01.

Steven Lord
Steven Lord le 7 Nov 2016
As of release R2014b you can use the round function in MATLAB to round to a specific number of decimal places.

Vladimir Melnikov
Vladimir Melnikov le 29 Avr 2020
Modifié(e) : Vladimir Melnikov le 29 Avr 2020
the easiest way:
round (X,N)
e.g:
>> round(0.12345,1)
ans = 0.100000000000000
>> round(0.12345,2)
ans = 0.120000000000000
>> round(0.12345,3)
ans = 0.123000000000000
also read
>> doc round

Andrei Bobrov
Andrei Bobrov le 3 Mai 2012
use roundn from Mapping Toolbox
roundn(X,-2)
  1 commentaire
Vladimir Melnikov
Vladimir Melnikov le 29 Avr 2020
roundn(1.12345,-1)
ans = 1.100000000000000
>> roundn(1.12345,-2)
ans = 1.120000000000000
>> roundn(1.12345,-3)
ans = 1.123000000000000

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Wayne King
Wayne King le 3 Mai 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 commentaires
Isti
Isti le 3 Mai 2012
thanks :)
Jos (10584)
Jos (10584) le 11 Fév 2014
Use round instead of ceil!

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Prateek Sahay
Prateek Sahay le 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 commentaire
Walter Roberson
Walter Roberson le 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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Jason Garcia
Jason Garcia le 7 Fév 2019
Modifié(e) : Jason Garcia le 7 Fév 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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