Azzera filtri
Azzera filtri

rounding down using floor

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marcus peixoto
marcus peixoto il 1 Set 2020
Risposto: Dana il 1 Set 2020
so im using:
floor(10^decimal*(total/AVERAGE_COUNT))/10^decimal;
where:
decimal = 4, AVERAGE_COUNT = 9, total = 8.9289
this entry gives me 0.9920, however the answer should be 0.9921
furthermore, when I just do
total/AVERAGE_COUNT;
it gives me 0.9921 which is correct but i need to use the floor function for the rest of my data.
how do i get around this
  3 Commenti
marcus peixoto
marcus peixoto il 1 Set 2020
I'm on R2020a on Windows
per isakson
per isakson il 1 Set 2020
On my R2018b, Win10 this script
%%
format long
decimal = 4; AVERAGE_COUNT = 9; total = 8.9289;
x = floor(10^decimal*(total/AVERAGE_COUNT))/10^decimal
outputs
x =
0.992100000000000
>>

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Risposte (1)

Dana
Dana il 1 Set 2020
I'm on R2020a on Windows 10 and I also get 0.9921.
Is it possible that total does not equal exactly 8.9289? For example, if total actually equals 8.92889 (which would usually display as 8.9289 in the command window), then you'll end up with 0.9920.

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