how to consider only the integer part discarding the exponent part

2 vues (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
.

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