How to create a For loop for a linspace?

I have this code:
for j = 1:20
c(j) = linspace(1,10,length(j));
end
But, the result is a vector "c" with all the values "10":
c= 10 10 10 10 10 10 10....
How can I do to solve the problem? Thnaks

 採用された回答

Star Strider
Star Strider 2017 年 6 月 28 日

1 投票

Since in every iteration, ‘j’ is a scalar, the length of ‘j’ will always be 1.
I would just use:
c = linspace(1, 10, 20);
or if you actually want varying-length vectors, save ‘c’ as a cell array:
for j = 1:20
c{j} = linspace(1,10,j);
end

5 件のコメント

Michela Longhi
Michela Longhi 2017 年 6 月 28 日
The problem is that I have to use the linespace in a "scatter" function:
for i = 1:length(temp_discesa_i)
pause(0.1)
c = linspace(1, 10, length(temp_discesa_i));
scatter(temp_discesa_i(i), spazio_discesa_i(i),30,c,'filled')
pause(0.01);
end
and Matlab give me an error: "Error using scatter (line 93) CData must be an RGB triplet, an M-by-1 vector of the same length as X, or an M-by-3 matrix."
Steven Lord
Steven Lord 2017 年 6 月 28 日
The loop variable i is a scalar. So you're creating a scatter plot of one point with x=temp_discesa_i(i) and y=spazio_discesa_i(i) each time. In that case, the color input must be an RGB triplet or a scalar that is mapped into the colormap. In order to achieve what I think you want to do with this code, create the linspace vector prior to the for loop and index into c the same way you index into temp_discesa_i and spazio_discesa_i.
But creating a scatter plot for one point at a time seems like a bit of overkill. Are you sure you don't want to simply create a line once and update its XData, YData, MarkerFaceColor, and MarkerEdgeColor properties in the loop? Or would scatter plotting all the data at once be acceptable?
Star Strider
Star Strider 2017 年 6 月 28 日
@Steven — Thank you.
I am not certain what the objective is. My suggestion is to define the colormap before the loop, as:
c = colormap(jet(length(temp_discesa_i)));
then do the plotting as:
scatter(temp_discesa_i(i), spazio_discesa_i(i),30,c(i,:),'filled')
awaiting further details.
Michela Longhi
Michela Longhi 2017 年 6 月 28 日
Yes!!!! It's now working. Thank you to all.
Star Strider
Star Strider 2017 年 6 月 28 日
Our pleasure!

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