Borrar filtros
Borrar filtros

Double conversion to use prctile function

1 visualización (últimos 30 días)
Alexander Guillen
Alexander Guillen el 15 de Sept. de 2022
Comentada: Star Strider el 16 de Sept. de 2022
Instead of evaluating column by column . I include a for loop so I can use the cell2mat function to convert from double to whatever data type the prctile function uses. However, after the for loop, I get an error message:
"Error using double
Conversion to double from cell is not possible.
Error in prctile (line 55)
x = double(x);
Error in E99 (line 45)
P = prctile(Model_mat(1,:),(99))"
But if I individually convert the column with the cell2mat function- the percentile function works fine.
% This is column by column
Model_mat1 = cell2mat(Model(1,:));
Model_mat2 = cell2mat(Model(2,:));
Model_mat3 = cell2mat(Model(3,:));
Model_mat4 = cell2mat(Model(4,:));
Model_mat5 = cell2mat(Model(5,:));
Model_mat6 = cell2mat(Model(6,:));
Model_mat7 = cell2mat(Model(7,:));
Model_mat8 = cell2mat(Model(8,:));
Model_mat9 = cell2mat(Model(9,:));
Model_mat10 = cell2mat(Model(10,:));
% This is the for loop
for i = 1:length(Model)
%Model_mat{i,:} = cell2mat(Model(i,:));
%Model_mat{i,:} = int64(Model(i,:));
Model_mat{i} = cell2mat(Model(i,:));
end
% This iis the syntax for the percentile
P = prctile(Model_mat(1,:),(99))
Any help would be appreciated.
  1 comentario
Rik
Rik el 15 de Sept. de 2022
What exactly is your question? You seem to have a cell array of some sort, but you want the percentile function to do that conversion for you?
It would all be easier to understand if you attach your data to the question.

Iniciar sesión para comentar.

Respuesta aceptada

Star Strider
Star Strider el 15 de Sept. de 2022
I am not certain what ‘Model’ is, however if it resembles this array, indexing into it as a cell array is straightforward —
Model = mat2cell(randn(10), ones(1,10),10)
Model = 10×1 cell array
{[ 0.3559 0.8681 0.5171 -1.9330 1.6688 -1.1222 -0.5230 0.4412 -2.6904 -0.3102]} {[ -0.4756 -0.0911 -0.4101 -1.2918 -0.8179 -0.6191 -0.2680 1.0212 0.5873 -2.4094]} {[ 1.9852 0.2609 0.5998 -1.2972 -0.9622 1.9963 -0.4795 -0.4450 1.4728 -1.6797]} {[ 1.7691 -0.3796 0.8273 0.1806 -0.7725 0.8785 0.3614 0.4027 -1.0187 0.6261]} {[ 0.0359 1.2318 0.4570 1.4907 1.2762 0.3159 0.4088 -0.2682 0.9570 -0.1869]} {[-1.3947 -0.2354 2.6897 -0.0101 -1.2078 -0.1153 -1.3919 -1.0017 -1.7950 -2.0804]} {[ 1.1646 -0.1035 -0.6771 0.6717 -0.3764 -0.5092 -1.9939 -0.9529 -0.5511 0.5268]} {[ -0.1635 1.9250 0.8539 0.1680 0.1291 0.5319 0.2604 -1.3130 -0.1796 1.1728]} {[ 0.0794 0.6675 0.2235 1.7531 -0.9750 -0.3654 -0.1841 -0.5366 0.1033 0.6050]} {[ 0.4417 -0.3038 0.6669 2.1762 -0.8061 -0.5986 2.7666 0.6785 -0.6522 -1.3446]}
for k = 1:size(Model,1)
P(k) = prctile(Model{k,:},99); % Loop Approach
end
P
P = 1×10
1.6688 1.0212 1.9963 1.7691 1.4907 2.6897 1.1646 1.9250 1.7531 2.7666
P = prctile(cell2mat(Model),99,2) % Matrix Approach
P = 10×1
1.6688 1.0212 1.9963 1.7691 1.4907 2.6897 1.1646 1.9250 1.7531 2.7666
The loop approach would likely be appropriate if all the rows are not the same lengths. Otherwise, the matrix approach is likely more efficient.
.
  2 comentarios
Alexander Guillen
Alexander Guillen el 16 de Sept. de 2022
Thank you so much. I frequently look for MATLAB Answers for various applications, and often see you helping users. I want to say that your help is very valuable to the community.
Star Strider
Star Strider el 16 de Sept. de 2022
As always, my pleasure!
I very much appreciate your compliment!

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Matrices and Arrays en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by