how to consider only the integer part discarding the exponent part

1 vue (au cours des 30 derniers jours)
ANUSAYA SWAIN
ANUSAYA SWAIN le 2 Juin 2023
Modifié(e) : Star Strider le 2 Juin 2023
Suppose i have a= [1 23 56]*10^(-9);
i want to acccess only the integer part that is 1 23 and 56.
i want to find the max(a). That is the output should be 56.

Réponses (2)

Pramil
Pramil le 2 Juin 2023
Modifié(e) : Pramil le 2 Juin 2023
What you can do is multiply "a" with 10^9 first and then proceed to find max(a). Like this :
a = [1 23 56]*10^(-9); % given array
a_int = int64(a*10^9); % convert to integer array
max_a = max(a_int); % find maximum value
disp(max_a);
56

Star Strider
Star Strider le 2 Juin 2023
Modifié(e) : Star Strider le 2 Juin 2023
This appears to be a reasonably robust approach —
a = [1 23 56]*10^(-9);
b = a.*10.^ceil(-log10(abs(a))+1)
b = 1×3
10.0000 23.0000 56.0000
max_a = max(b)
max_a = 56.0000
a = [1 23 56]*10^(-6);
b = a.*10.^ceil(-log10(abs(a))+1)
b = 1×3
10 23 56
a = [1 23 56]*10^(10);
b = a.*10.^ceil(-log10(abs(a))+1)
b = 1×3
10 23 56
a = [1 23 56]*10^(-1);
b = a.*10.^ceil(-log10(abs(a))+1)
b = 1×3
10.0000 23.0000 56.0000
EDIT — Corrected typographical error
.

Catégories

En savoir plus sur Entering Commands dans Help Center et File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by