Filter löschen
Filter löschen

Sorting a cell array

1 Ansicht (letzte 30 Tage)
Yongmin
Yongmin am 17 Jun. 2015
Kommentiert: Yongmin am 17 Jun. 2015
Hello, I have the following cell(data, 1x3):
data = [data{1} data{2} data{3}]
data{1} data{2} data{3}
A 15 1400
A 3 250
B 10 1100
C 13 1350
B 7 600
Now I would like to sort the data and make a new cell as follows:
newcell = [newcell{1} newcell{2} newcell{3}]
newcell{1} newcell{2} newcell{3}
A [15; 3] [1400; 250]
B [10; 7] [1100; 600]
C [13] [1350]
In other words, I want to sort and save the data of a cell with respect to A, B and C. Would you please show me how to do that with or without using for loop etc?
Thanks in advance.
  2 Kommentare
Geoff Hayes
Geoff Hayes am 17 Jun. 2015
Yongmin - why don't you want to use a for loop?
Yongmin
Yongmin am 17 Jun. 2015
Bearbeitet: Yongmin am 17 Jun. 2015
I guess the problem can be solved without the loop. If we find idx matrix describing the elements for A, B, and C, can we use the idx matrix for sorting data{2} and data{3}? But I don't know how to do it exactly.
If we use a for loop, would you please explain how to do it?
Thanks for your comment.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Guillaume
Guillaume am 17 Jun. 2015
data = {'A' 15 1400; 'A' 3 250; 'B' 10 1100; 'C' 13 1350; 'B' 7 600}
[header, ~, row] = unique(data(:, 1));
catcol = @(col) arrayfun(@(r) [data{r == row, col}], 1:numel(header), 'UniformOutput', false)';
newcell = [header, catcol(2), catcol(3)]
would be one way to do it without a loop. It could be made even more generic with an additional arrayfun looping over the columns, but at this point, you'd be better off using a for loop to make it clearer.
  3 Kommentare
Guillaume
Guillaume am 17 Jun. 2015
[header, ~, row] = unique(data(:, 1));
newcell = [header, cell(numel(header), size(data, 2)-1)];
for r = 1:numel(header)
for c = 2:size(data, 2)
newcell{r, c} = [data{r == row, c}];
end
end
Yongmin
Yongmin am 17 Jun. 2015
Thanks a lot for your kind answer. It will help to develop codes for my problem.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Tags

Produkte

Community Treasure Hunt

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

Start Hunting!

Translated by