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Uneven data multiplication in Matlab

2 visualizaciones (últimos 30 días)
Neesha
Neesha el 17 de Jun. de 2014
Respondida: Andrei Bobrov el 18 de Jun. de 2014
I have following
Data.numberOfDishes
Data.numOfLocations
As follows
numberOfDishes numberOfLocations
3 [1, 2, 4]
4 [9, 8, 5]
6 [11, 1, 9]
7 [2, 3, 4]
9 [7, 7, 7]
I want to have resulting column 'total' by doing numberOfDishes/each element of numberOfLocations
For example, so that 'total' for first row would be [3/1, 3/2, 3/4]
i know '.' would not work. mrdivide does not seem an option. Can i do this without having to create numberOfDishes of the same size as numOfLocations

Respuesta aceptada

Star Strider
Star Strider el 17 de Jun. de 2014
This works:
NumberOfDishes = [3 4 6 7 9];
NumberOfLocations = [1, 2, 4; 9, 8, 5; 11, 1, 9; 2, 3, 4; 7, 7, 7];
for k1 = 1:length(NumberOfDishes)
Total(k1,:) = NumberOfDishes(k1)./NumberOfLocations(k1,:);
end
% To express them as fractions:
Total = rats(Total)
  4 comentarios
Image Analyst
Image Analyst el 18 de Jun. de 2014
Let's compare the "vectorized" approach with the brute force "for loop" approach:
NumberOfDishes = [3; 4; 6; 7; 9]
NumberOfLocations = [1, 2, 4; 9, 8, 5; 11, 1, 9; 2, 3, 4; 7, 7, 7]
tic
for k1 = 1:length(NumberOfDishes)
Total(k1,:) = NumberOfDishes(k1)./NumberOfLocations(k1,:);
end
toc
% To express them as fractions:
Total
Total = rats(Total)
% Use a "vectorized" approach.
tic
t = repmat(NumberOfDishes(:), [1, size(NumberOfLocations, 2)]) ./ NumberOfLocations
toc
Results in command window:
Elapsed time is 0.000032 seconds.
Elapsed time is 0.000402 seconds.
Hmmmm.... Looks like the for loop is 12 times faster.
Star Strider
Star Strider el 18 de Jun. de 2014
Interesting!
I think the delay with the vectorised approach is the overhead in repmat.

Iniciar sesión para comentar.

Más respuestas (1)

Andrei Bobrov
Andrei Bobrov el 18 de Jun. de 2014
out = bsxfun(@rdivide,NumberOfDishes,NumberOfLocations);

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