rounding elements in matrix if > or < to 0.75

example:
3.32==> .032 is < 0.75 ===> 3
0.78===>0.78 is >0.75==> 1

1 commentaire

Dyuman Joshi
Dyuman Joshi le 18 Juil 2023
What should be the result when the decimal part is equal to 0.75?

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Réponses (2)

Cris LaPierre
Cris LaPierre le 17 Juil 2023
Modifié(e) : Cris LaPierre le 17 Juil 2023
Currently, round uses 0.5 as the threshold. The quickest solution would be to modify your data by subtracting 0.25 to artificially move the threshold. Now, 0.75 become 0.5 and is still rounded up. 1.74 becomes 1.49, and is rounded down. In both cases, you end up with the value you want.
x = [3.32;0.78];
round(x-0.25)
ans = 2×1
3 1

5 commentaires

aldo
aldo le 17 Juil 2023
" is still rounded up. 1.74 becomes 1.54"
do you mean "is still rounded up. 1.74 becomes (1.74-.25) 1,49"
thanks
aldo
aldo le 17 Juil 2023
Déplacé(e) : Cris LaPierre le 17 Juil 2023
fixed the code you wrote But I wrote wrong I want to put a decimal input and then the code runs: Ex: .75.. .82 ... .25 and if <x I round down otherwise up
aldo
aldo le 17 Juil 2023
Déplacé(e) : Cris LaPierre le 17 Juil 2023
do you think this code is correct?
function ris=Test_round(k)
x = [3.32;0.78;1.25;0.20];
ris=round(x-(k-0.5));
end
matrix K
input:dec
x=0.75
x=.25
x=.91
Cris LaPierre
Cris LaPierre le 17 Juil 2023
You can determine if it works just by testing it. Do you get the expected results?
aldo
aldo le 17 Juil 2023
ah ok...I'll have to do more appropriate numerical tests I'm bad at math :(

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Hi @aldo,
you could use this as another way to round your numbers according to what you desire by doing the following
x=[0.33 0.45 1.56 1.77];
x=floor(x+0.25)
x = 1×4
0 0 1 2
Hope this helps!

Question posée :

le 17 Juil 2023

Commenté :

le 18 Juil 2023

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