# Moving graph along x-axis

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Sokratis Panagiotidis le 17 Avr 2022
Hi,
I am trying to plot 3 different vectors that showcase the same function but with different starting points from the y-axis.
Now what I wanna do is move the yellow and the red graph so they overlap with the blue one, to see wether my determined data actually is correct. I have another picture here to further show what exactly I mean:
Now I have tried to do something similar but with the y-axis but unfortunately I couldn't make it work.
What would be a clever way to move the graphs without having to change the values of the vectors?
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Torsten le 17 Avr 2022
Now what I wanna do is move the yellow and the red graph so they overlap with the blue one, to see wether my determined data actually is correct.
Better work with the data, not with the graphics.
Sokratis Panagiotidis le 19 Avr 2022
@Torsten I mean yeah that's exactly what I am trying to do. I only used the graphics to better show what I was talking about.

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### Réponse acceptée

Tala le 17 Avr 2022
Modifié(e) : Tala le 18 Avr 2022
I would subtitute the other 2 array in the main one and plot that. something lik ethis:
id1 = find(blue==yellow(1)); % the index on the blue curve that matches the first value of yellow
id2 = find(blue==red(1)); % the index on the blue curve that matches the first value of red
now use the id values to create a new array. newblue contain the values from blue, yellow and red curves, and you can just plot newblue.
newblue=blue;
newblue(id1:end)=yellow;
newblue(id2:end)=red;
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Tala le 19 Avr 2022
Modifié(e) : Tala le 19 Avr 2022
You basically need to move your curves along th X axis; in other words you need to plot(x+dx,y). To define dx you need to clean up your data. in the code below, I removed all nans and chose the last value (because there is no "jump") as the ref point and moved all the curves. As you see you have a better match towards the tail of the curves because the ref point is the last point.
You could clean up your data and get rid of the jump in the start of blue curve and use the first element as your ref point, then you get a better match towards the beginning. In that case just change "end" to "1"
you could take the avarage of first and last, or even interpolate over a unique x aeeay, and move each point individually using this method
X1=data.x1;Y1=data.y1;X2=data.x2;Y2=data.y2;X3=data.x3;Y3=data.y3;
% removing nans
TF = find(~isnan(Y1));
Y1N=Y1(TF);
X1N=X1(TF);
TF = find(~isnan(Y2));
Y2N=Y2(TF);
X2N=X2(TF);
TF = find(~isnan(Y3));
Y3N=Y3(TF);
X3N=X3(TF);
% now plot arrays with no nans
plot(X1N,Y1N)
hold on
plot(X2N,Y2N)
plot(X3N,Y3N)
% defining dx and moving your plots
d13=(X1N(end)-X3N(end));
plot(X3N+d13,Y3N);
d12=(X1N(end)-X2N(end));
plot(X2N+d12,Y2N);
Sokratis Panagiotidis le 23 Avr 2022
Hello,
You method worked thanks a lot!
But here's one thing I'd like to say:
Your method removes all NaN's, but let's say I have a vector
[0 0 NaN 0 0 NaN NaN 0 NaN 0 0 NaN 1 0 0 0 0 0]
it would give me
[0 0 0 0 0 0 0 1 0 0 0 0 0] (I put in a 1 to make it more clear).
Now that itself is already great but I believe the values before the last NaN are useless, as I cannot work with them if I remove something after it, so Imma go and look into a method to first remove everything up to the last NaN, so that I get
[1 0 0 0 0 0] and have therefore a much shorter vector but at least I won't have gaps in between two values, but I'll also try to find a way to adapt it into a automated loop, so that it works for a N-amount of vectors and it would additionally find the longer vector (here blue) and valculate dx for the rest rather than having to do it for every sinle one manually (as I will sometimes be working with up to 20 vectors or even more).

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### Plus de réponses (1)

Matt J le 17 Avr 2022
Modifié(e) : Matt J le 17 Avr 2022
Let's call your 3 data sets (x0,y0), (x1,y1), (x2,y2) where (x0,y0) are your target data points. Then I might try something like this:
x1_shifted=x1-mean(x1)+mean(interp1(y0,x0,y1));
x2_shifted=x2-mean(x2)+mean(interp1(y0,x0,y2));
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