Using index data to fill a matrix

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DrEamonn
DrEamonn el 15 de Feb. de 2019
Comentada: Jan el 15 de Feb. de 2019
I have a matrix which I have extracted from an Excel file that contains [row_location, column_location, data]. The row and column are non conituous, so the matrix looks like this:
ExtractedData =
[8] [93] [0.1981]
[8] [94] [0.2209]
[8] [95] [0.2276]
[8] [96] [0.2276]
[9] [49] [0.2773]
[9] [50] [0.3941]
[9] [51] [0.3256]
[9] [52] [0.3135]
[9] [53] [0.2343]
[9] [54] [0.3337]
I know how big the full matrix would be from
MaxRow = max(ExtractedData(:, 1))
MaxCol = max(ExtractedData(:, 2))
So I create a NaN matrix of the size
DataToPlot = NaN(MaxRow MaxCol)
& then I would like to populate the correct entry in the DataToPlot matrix with the data from the spreadsheet i.e.
DataToPlot(8, 93) = 0.1981
DataToPlot(8, 94) = 0.2209
DataToPlot(8, 95) = 0.2276
etc
I can see that I want to use a loop to step through the ExtractedData one row at a time, but then I'm unsure of how to extract & use the index data to populate the correct location with the data.
The end goal is to generate a surface plot of the data so if there are any shortcuts or 'best practice' ways of going from extacted data to surface plot I'd appreciate a steer in the right direction.

Respuesta aceptada

Jan
Jan el 15 de Feb. de 2019
Editada: Jan el 15 de Feb. de 2019
This works with a loop easily:
D = ExtractedData; % Easier to read...
MaxRow = max(D(:, 1));
MaxCol = max(D(:, 2));
ToPlot = NaN(MaxRow, MaxCol);
for k = 1:size(D, 1)
ToPlot(D(k, 1), D(k, 2)) = D(k, 3);
end
But it is smarter to do this by sub2ind (link):
MaxSize = max(D(:, 1:2), [], 1); % Get size as vector
ToPlot = NaN(MaxSize);
index = sub2ind(MaxSize, D(:, 1), D(:, 2)); % [EDITED, Typo: D(:,1) -> D(:,2)]
ToPlot(index) = D(:, 3);
  3 comentarios
Jan
Jan el 15 de Feb. de 2019
Editada: Jan el 15 de Feb. de 2019
[MOVED from section for answers] DrEamonn wrote:
I think I fixed the problem - it was a resize function elsewhere in my code to remove a text header
Thanks for you help Jan
Eamonn
Jan
Jan el 15 de Feb. de 2019
You are welcome. I've fixed the typo now.
Please use the section from comments to post a comment.

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