How to scatter/plot a vector vs a cell array ?

Hello,
I have a vector w such that
w=0:0.1:5
I use this vector to get values saved in a cell array. such that for each element of w, I get 1,2, or 3 values stored in one cell as a vector.
as an example:
w=[0, 0.1, 0.2, 0.3, ....]
cellArray= {[3, -4, 5], [1], [-2], [3,3], ...}
I want to plot this such that w is the x axis and the corresponding values on the y axis?

4 comentarios

Ahmed Anas
Ahmed Anas el 12 de Mzo. de 2020
What you want is at x=0 , y varies at 3, -4 , 5 , at x= 0.1 y =1, at x=0.2, y=-2, and so on
and you want to plot all these on the same figure?
Nora Khaled
Nora Khaled el 12 de Mzo. de 2020
Yes, this how I want to do it.
x=0, y=3, -4 , 5 it should be 3 points on the same graph.
x=0.1, y=1
..
Ahmed Anas
Ahmed Anas el 12 de Mzo. de 2020
Editada: Ahmed Anas el 12 de Mzo. de 2020
clc
clear all
w=[0, 0.1, 0.2, 0.3]
cellArray= {[3, -4, 5], [1], [-2], [3,3]}
X=0;
t(1)=1;
for i=2:size(cellArray,2)
t(i)=t(i-1)+size(cell2mat(cellArray(i-1)),2);
end
for i=1:size(cellArray,2)
y=X;
X=size(cell2mat(cellArray(i)),2);
X=y+X;
G(i)=X;
end
for i=1:size(cellArray,2)
New_X_Matrix(t(i):G(i))=w(i);
end
New_X_Matrix=New_X_Matrix
New_Y_Matrix=cell2mat(cellArray)
plot(New_X_Matrix,New_Y_Matrix,'ro-','linewidth',2)
Check this code, it will work accordingly and hope you will understand this, it is somehow generalized code.
New_X_Matrix is your desired x matrix
New_Y_Matrix is the y coordinates matrix
Nora Khaled
Nora Khaled el 15 de Mzo. de 2020
Thank you so much!
The code works perfectly.

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 Respuesta aceptada

Adam Danz
Adam Danz el 12 de Mzo. de 2020
w=[0, 0.1, 0.2, 0.3];
cellArray= {[3, -4, 5], [1], [-2], [3,3]};
figure()
hold on % important
arrayfun(@(i)plot(w(i),cellArray{i}, '-o'), 1:numel(w))

3 comentarios

Fego Etese
Fego Etese el 13 de Mzo. de 2020
Hello Adam Danz, sorry to call your attention like this but I'm in dire need of help at the moment. Please help me with this cross correlation question.
Thank you
Nora Khaled
Nora Khaled el 15 de Mzo. de 2020
Thank you so much!
Adam Danz
Adam Danz el 15 de Mzo. de 2020
Glad I could help!

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el 12 de Mzo. de 2020

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el 15 de Mzo. de 2020

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