how to consider only the integer part discarding the exponent part

Question

ANUSAYA SWAIN 2023-6-2

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https://ww2.mathworks.cn/matlabcentral/answers/1977254-how-to-consider-only-the-integer-part-discarding-the-exponent-part

编辑： Star Strider 2023-6-2

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.

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Answer 1

Pramil 2023-6-2

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https://ww2.mathworks.cn/matlabcentral/answers/1977254-how-to-consider-only-the-integer-part-discarding-the-exponent-part#answer_1249034

编辑：Pramil 2023-6-2

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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

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Answer 2

Star Strider 2023-6-2

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https://ww2.mathworks.cn/matlabcentral/answers/1977254-how-to-consider-only-the-integer-part-discarding-the-exponent-part#answer_1249039

编辑：Star Strider 2023-6-2

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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