Return all values of a struct parameter within a cell array.

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I have a cell array (21x1) with multiple embedded structs such that the information I want is found at myCell{1,1}.Pose.Position.X.(value_here). Each cell array has identical structure. Is there a way to return the all X values in this 21x1 cell array without having to use a for loop? I want to use the ':' operator somewhere but can't figure it out (I'm no matlab expert either, so could be something obvious I'm missing or maybe it's just not possible).
  3 Commenti
Gregory Beale
Gregory Beale il 4 Feb 2021
Modificato: Gregory Beale il 4 Feb 2021
I’m reading in ROS messages from .bag files using the ROS toolbox. I use the ‘readmessages’ function and they are automatically put into cell arrays.
Matt J
Matt J il 4 Feb 2021
Is there a way to return the all X values in this 21x1 cell array without having to use a for loop?
Is there a reason you think you need to avoid for-loops when your array is so small? I doubt you will notice differences in speed between a for-loop and any of the Answers below.

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Risposte (2)

Matt J
Matt J il 4 Feb 2021
tmp=[myCell{:}];
tmp=[tmp.Pose];
tmp=[tmp.Position];
X=[tmp.X];

Cam Salzberger
Cam Salzberger il 4 Feb 2021
Hello Gregory,
Matt J's way works, by creating multiple struct arrays, working your way down the nested fields. It's functional, but not very fast, since you are creating new arrays each time, unless you are going to be accessing all of the data at that level.
On the other hand, this is kind perfect for cellfun:
x = cellfun(@(c) c.Pose.Position.X, myCell);
This works great when the field you are accessing always contains scalar data. If it contains vector data, or text, you'd have to use 'UniformOutput',false, which would output another cell array.
If you are concerned about performance, it may be faster to create a local function to handle the access of all data within a deeply-nested message at once. Something like:
[x, y, z] = cellfun(@extractPose, myCell);
function [x, y, z] = extractPose(msg)
% Extract pose coordinates from geometry_msgs/PoseStamped message
pos = msg.Pose.Position;
x = pos.X;
y = pos.Y;
z = pos.Z;
end
This is the case for deeply-nested message objects. I'm not as certain if it holds true for message structs.
-Cam

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