Rescale an axis in other steps without creating vector

2013 12 3
4 个回答
更新时间：2013 12 3
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Hello,
I have the following problem: I have for example 1000 measurement points but in 0.02 second steps. In the plot I want to have not a scale from 0 to 1000 but from 0 to 20 (=1000*0.02). As I don't know the length of my data points, I cannot create a vector in 0.02 steps for the x-axis because then Matlab returns that the vectors must be the same lengths.
Is there a possibility, just to divide the x-axis by 50?
Thanks, Leo

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回答（4 个）

sixwwwwww
sixwwwwww 2013-12-3
编辑：sixwwwwww 2013-12-3

0 个投票

You can control your x-axis by plotting your data against your time as follows:
y = rand(1, 1000); % your data
x = linspace(0, 0.02, numel(y)); % your time series
plot(x, y) % your plot of data against time series
or you can use 'xlim' for this purpose. see
doc xlim
for more information
Leonard
Leonard 2013-12-3
编辑：Leonard 2013-12-3

0 个投票

That's exactly my point: With this I can't do it.
I don't want the first 20 measurement points. I want to have all 1000 measurement points, but the x-axis shall contain the numbers 0 to 20 (instead of 0 to 1000).
I hope, you now get, what I mean. Thx
sixwwwwww
sixwwwwww 2013-12-3
编辑：sixwwwwww 2013-12-3

0 个投票

Try this:
y = rand(1, 1000); % your data
x = linspace(0, 20, numel(y)); % your time series
plot(x, y) % your plot of data against time series
It will plot 1000 points within range of 0 to 20
Leonard
Leonard 2013-12-3

0 个投票

OK, this wasn't the solution either but I found out a way for myself now. Looks like this (for your interest or anybody else who needs it):
plot(0.02:0.02:numel(y)/50,y)
y is the function.

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sixwwwwww
sixwwwwww 2013-12-3
whats the difference in your code and solution I provided? I am also generating the same values of x = 0.02 0.04 ...
you can run my code and then you can see workspace varaiable x for confirmation. Also I appreciate you in a way that you found it yourself as well and it is a better way of learning. Good luck!
Leonard
Leonard 2013-12-3
In your case you have determined that the length is 20 seconds. But in my case this value is variable. So sometimes it's 20 seconds, sometimes 50 seconds or whatever other value.
But I just figured out, that your way could be also possible if I just choose the end point as a variable as well like this:
x = linspace(0, numel(y)/50, numel(y));
Thanks for your help again!
sixwwwwww
sixwwwwww 2013-12-3
In my case 20 is not fix. in the following line of code you can just replace 20 by any value then it will also work
x = linspace(0, 20, numel(y));
Good luck!

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Leonard
2013-12-3

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sixwwwwww
sixwwwwww
2013-12-3

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