Using index data to fill a matrix

14 vues (au cours des 30 derniers jours)
DrEamonn
DrEamonn le 15 Fév 2019
Commenté : Jan le 15 Fév 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.

Réponse acceptée

Jan
Jan le 15 Fév 2019
Modifié(e) : Jan le 15 Fév 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 commentaires
Jan
Jan le 15 Fév 2019
Modifié(e) : Jan le 15 Fév 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 le 15 Fév 2019
You are welcome. I've fixed the typo now.
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