How to flip an ordered X-axis that is half half separated for data?

Hi, I have a plot that I want its X-axis to be reversed. My data on X-axis is showing as

[1 2 3 4 5 6 7 8 9 10 11 12] I want to reverse it in a way that it represents [6 5 4 3 2 1 12 11 10 9 8 7] (reversing it from the middle for two seperate parts). How is it possible? I tried set(gca, 'XDir','reverse') but it reverses it to [12 11 10 9 8 7 6 5 4 3 2 1]. I cannot share the code or data due to confidentiality. Thanks in advance.

Answer 1

[Dorna]

Ok, so what I needed to do was

A = Data((1:20),(1:5))

B = flip(A,1)

C = Data((21:40),(1:5))

D = flip(C,1)

And then

New = cat(1,B,D)

[Star Strider]

O.K.

That seems to be essentially what I wrote, although in one column not five. You can use my approach to plot it.

Answer 2

[VBBV]

Use xticklabels function to modify the order of axis labels

x1 = {'6','5','4','3','2','1'};
x2 ={'12','11','10','9','8','7'};
plot(rand(1,12))
xticks(1:12)
xticklabels([x1,x2])

[Dorna]

@VBBV will also flip the data? I think it will only flip the axis label. Plus i prefer something that does not require me to actually work with the plots as they are very complex. Something like set gca

[VBBV]

can you explain why you DONT want to actually work with plot data but still want to flip ordered x-axis values ?

v = [16 5 9 4 2 11 7 14];
v2 = v([5:8 1:4])     % Extract and swap the halves of v

v2 = 1×8

2 11 7 14 16 5 9 4

% flip the values for vector v
v2 = flip(v)

v2 = 1×8

14 7 11 2 4 9 5 16

% another way 
v2 = v([8:-1:5 , 4:-1:1])

v2 = 1×8

14 7 11 2 4 9 5 16

How can I make the V2 to be [14 7 11 2 4 9 5 16], can I use prime on 5:8'?

You can use the flip function to get the desired result as shown above

Answer 3

[Star Strider]

One approach that changes the x-axis and the data —

xv = [1 2 3 4 5 6 7 8 9 10 11 12];                      % Independent Variable
yv = (xv/max(xv)).^2;                                   % Dependent Variable

xvi = [6 5 4 3 2 1 12 11 10 9 8 7];                     % Indexing Vector

figure
plot(xv, yv)
grid
xticks(xv)
title('Original')

Vectors = [xv(xvi); yv(xvi)]

Vectors = 2×12

6.0000 5.0000 4.0000 3.0000 2.0000 1.0000 12.0000 11.0000 10.0000 9.0000 8.0000 7.0000 0.2500 0.1736 0.1111 0.0625 0.0278 0.0069 1.0000 0.8403 0.6944 0.5625 0.4444 0.3403

figure
plot(xv, yv(xvi))
grid
xticks(xv)
xticklabels(xv(xvi))
title('Reversed & Concatenated')

It may be challenging to elliminate the connecting line between 1 and 12 here. If necessary, that would require putting a NaN value at the beginning or end of the ‘xv’ and ‘yv’ vectors, and adjusting ‘xvi’ accordingly.

.

[Dorna]

@Star Strider

Imagine we have this

v = [16 5 9 4 2 11 7 14];

v2 = v([5:8 1:4]) % Extract and swap the halves of v

v2 =

2 11 7 14 16 5 9 4

How can I make the V2 to be [14 7 11 2 4 9 5 16], can I use prime on 5:8'?

[Star Strider]

I do not understand what ‘using prime’ means.

My code is based on the reversed vector being able to be used as an index of some sort into the original vectors.

v = [16 5 9 4 2 11 7 14];
v2 = v([5:8 1:4])     % Extract and swap the halves of v

v2 = 1×8

2 11 7 14 16 5 9 4

If individual values of the dependent variable were mapped to individual values of ‘v’ my code would work for ‘v2’.

.

[Dorna]

@Star Strider thank you so much. I think what I needed was what I posted as answer but I do really appreciate your input.

[Dorna]

I wish there was an easier alternative to the solution that I posted.