Double conversion to use prctile function

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Alexander Guillen
Alexander Guillen 2022 年 9 月 15 日
コメント済み: Star Strider 2022 年 9 月 16 日
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 件のコメント
Rik
Rik 2022 年 9 月 15 日
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.

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Star Strider
Star Strider 2022 年 9 月 15 日
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 件のコメント
Star Strider
Star Strider 2022 年 9 月 16 日
As always, my pleasure!
I very much appreciate your compliment!

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