Mixing
Inverse of 1D vector Matlab [closed]
I have 1D vector. For example: y=[0.2 0.9 1.0 1.0]. I can plot it with plot(y) to get a graph of y(x) where x values are just indices [1, 2, 3, 4].
Now, instead of x values being just indices, I want to map them to [0,1] range: x = linspace(0,1,length(y)). I get: x=[0 0.3333 0.6667 1.000].
I can now make a graph with plot(x,y):
Now, however, I want an inverse graph, so I make a plot with plot(y,x):
I want to be able to now use plot(x) to get the same shape as above. However, if I use plot(x), as expected, I just get a straight line.
How to transform x in such a way that plot(x) will give the same shape as plot(y,x)?
Upd.: If I try just 1./x:
matlab plot
asked Nov 19 '18 at 12:59
Valeria
10410
closed as unclear what you're asking by Wolfie, Dev-iL, M-M, Makyen, Paul Roub Nov 20 '18 at 14:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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I have 1D vector. For example: y=[0.2 0.9 1.0 1.0]. I can plot it with plot(y) to get a graph of y(x) where x values are just indices [1, 2, 3, 4].
Now, instead of x values being just indices, I want to map them to [0,1] range: x = linspace(0,1,length(y)). I get: x=[0 0.3333 0.6667 1.000].
I can now make a graph with plot(x,y):
Now, however, I want an inverse graph, so I make a plot with plot(y,x):
I want to be able to now use plot(x) to get the same shape as above. However, if I use plot(x), as expected, I just get a straight line.
How to transform x in such a way that plot(x) will give the same shape as plot(y,x)?
Upd.: If I try just 1./x:
matlab plot
asked Nov 19 '18 at 12:59
Valeria
10410
closed as unclear what you're asking by Wolfie, Dev-iL, M-M, Makyen, Paul Roub Nov 20 '18 at 14:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Why don't you want to useplot(y,x)? I almost never useplot(x), even when I want integer indices (I'd useplot(1:numel(x),x)), because of controlabillity of the function.
– Adriaan
Nov 19 '18 at 13:07
@Adriaan Because I am not going to plot it. The plot was used only as an example, because I don't know how to better explain what I mean by "inverse of 1D vector". In general, I need to use is as a transformation vector and multiply it element-wise with another 1D vector.
– Valeria
Nov 19 '18 at 13:09
1
Then1./xwould suffice. The plot just creates confusion, since thexlocations of correspondingyvalues also get changed (e.g. the 2nd value of the first plot is atx==1/3, whereas in the second plot it's atx==0.9), which means you need 2 numbers per point, as opposed to the single one you're apparently after, judging your comment.
– Adriaan
Nov 19 '18 at 13:17
@Adriaan if I just get1./x, this gives me completely different y-axis values from the ones I am looking for (added to the original question).
– Valeria
Nov 19 '18 at 13:25
fliplrfor row-vectors then? That's not the inverse, but rather 'read in opposite direction'. The images don't help a thing; all they do is confuse me. Can you just show in numbers what you want; i.e. first yourx,ypair as you have it now, then what you get using your method, and finally what you want to obtain? Just in numbers?
– Adriaan
Nov 19 '18 at 13:29
|
show 1 more comment
I have 1D vector. For example: y=[0.2 0.9 1.0 1.0]. I can plot it with plot(y) to get a graph of y(x) where x values are just indices [1, 2, 3, 4].
Now, instead of x values being just indices, I want to map them to [0,1] range: x = linspace(0,1,length(y)). I get: x=[0 0.3333 0.6667 1.000].
I can now make a graph with plot(x,y):
Now, however, I want an inverse graph, so I make a plot with plot(y,x):
I want to be able to now use plot(x) to get the same shape as above. However, if I use plot(x), as expected, I just get a straight line.
How to transform x in such a way that plot(x) will give the same shape as plot(y,x)?
Upd.: If I try just 1./x:
matlab plot
asked Nov 19 '18 at 12:59
Valeria
10410
I have 1D vector. For example: y=[0.2 0.9 1.0 1.0]. I can plot it with plot(y) to get a graph of y(x) where x values are just indices [1, 2, 3, 4].
Now, instead of x values being just indices, I want to map them to [0,1] range: x = linspace(0,1,length(y)). I get: x=[0 0.3333 0.6667 1.000].
I can now make a graph with plot(x,y):
Now, however, I want an inverse graph, so I make a plot with plot(y,x):
I want to be able to now use plot(x) to get the same shape as above. However, if I use plot(x), as expected, I just get a straight line.
How to transform x in such a way that plot(x) will give the same shape as plot(y,x)?
Upd.: If I try just 1./x:
matlab plot
matlab plot
asked Nov 19 '18 at 12:59
Valeria
10410
asked Nov 19 '18 at 12:59
Valeria
10410
edited Nov 19 '18 at 13:24
asked Nov 19 '18 at 12:59
Valeria
10410
asked Nov 19 '18 at 12:59
Valeria
10410
asked Nov 19 '18 at 12:59
Valeria
10410
10410
closed as unclear what you're asking by Wolfie, Dev-iL, M-M, Makyen, Paul Roub Nov 20 '18 at 14:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Wolfie, Dev-iL, M-M, Makyen, Paul Roub Nov 20 '18 at 14:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Why don't you want to useplot(y,x)? I almost never useplot(x), even when I want integer indices (I'd useplot(1:numel(x),x)), because of controlabillity of the function.
– Adriaan
Nov 19 '18 at 13:07
@Adriaan Because I am not going to plot it. The plot was used only as an example, because I don't know how to better explain what I mean by "inverse of 1D vector". In general, I need to use is as a transformation vector and multiply it element-wise with another 1D vector.
– Valeria
Nov 19 '18 at 13:09
1
Then1./xwould suffice. The plot just creates confusion, since thexlocations of correspondingyvalues also get changed (e.g. the 2nd value of the first plot is atx==1/3, whereas in the second plot it's atx==0.9), which means you need 2 numbers per point, as opposed to the single one you're apparently after, judging your comment.
– Adriaan
Nov 19 '18 at 13:17
@Adriaan if I just get1./x, this gives me completely different y-axis values from the ones I am looking for (added to the original question).
– Valeria
Nov 19 '18 at 13:25
fliplrfor row-vectors then? That's not the inverse, but rather 'read in opposite direction'. The images don't help a thing; all they do is confuse me. Can you just show in numbers what you want; i.e. first yourx,ypair as you have it now, then what you get using your method, and finally what you want to obtain? Just in numbers?
– Adriaan
Nov 19 '18 at 13:29
|
show 1 more comment
Why don't you want to useplot(y,x)? I almost never useplot(x), even when I want integer indices (I'd useplot(1:numel(x),x)), because of controlabillity of the function.
– Adriaan
Nov 19 '18 at 13:07
@Adriaan Because I am not going to plot it. The plot was used only as an example, because I don't know how to better explain what I mean by "inverse of 1D vector". In general, I need to use is as a transformation vector and multiply it element-wise with another 1D vector.
– Valeria
Nov 19 '18 at 13:09
1
Then1./xwould suffice. The plot just creates confusion, since thexlocations of correspondingyvalues also get changed (e.g. the 2nd value of the first plot is atx==1/3, whereas in the second plot it's atx==0.9), which means you need 2 numbers per point, as opposed to the single one you're apparently after, judging your comment.
– Adriaan
Nov 19 '18 at 13:17
@Adriaan if I just get1./x, this gives me completely different y-axis values from the ones I am looking for (added to the original question).
– Valeria
Nov 19 '18 at 13:25
fliplrfor row-vectors then? That's not the inverse, but rather 'read in opposite direction'. The images don't help a thing; all they do is confuse me. Can you just show in numbers what you want; i.e. first yourx,ypair as you have it now, then what you get using your method, and finally what you want to obtain? Just in numbers?
– Adriaan
Nov 19 '18 at 13:29
Why don't you want to use
plot(y,x)? I almost never use plot(x), even when I want integer indices (I'd use plot(1:numel(x),x)), because of controlabillity of the function.– Adriaan
Nov 19 '18 at 13:07
Why don't you want to use
plot(y,x)? I almost never use plot(x), even when I want integer indices (I'd use plot(1:numel(x),x)), because of controlabillity of the function.– Adriaan
Nov 19 '18 at 13:07
@Adriaan Because I am not going to plot it. The plot was used only as an example, because I don't know how to better explain what I mean by "inverse of 1D vector". In general, I need to use is as a transformation vector and multiply it element-wise with another 1D vector.
– Valeria
Nov 19 '18 at 13:09
@Adriaan Because I am not going to plot it. The plot was used only as an example, because I don't know how to better explain what I mean by "inverse of 1D vector". In general, I need to use is as a transformation vector and multiply it element-wise with another 1D vector.
– Valeria
Nov 19 '18 at 13:09
1
1
Then
1./x would suffice. The plot just creates confusion, since the x locations of corresponding y values also get changed (e.g. the 2nd value of the first plot is at x==1/3, whereas in the second plot it's at x==0.9), which means you need 2 numbers per point, as opposed to the single one you're apparently after, judging your comment.– Adriaan
Nov 19 '18 at 13:17
Then
1./x would suffice. The plot just creates confusion, since the x locations of corresponding y values also get changed (e.g. the 2nd value of the first plot is at x==1/3, whereas in the second plot it's at x==0.9), which means you need 2 numbers per point, as opposed to the single one you're apparently after, judging your comment.– Adriaan
Nov 19 '18 at 13:17
@Adriaan if I just get
1./x, this gives me completely different y-axis values from the ones I am looking for (added to the original question).– Valeria
Nov 19 '18 at 13:25
@Adriaan if I just get
1./x, this gives me completely different y-axis values from the ones I am looking for (added to the original question).– Valeria
Nov 19 '18 at 13:25
fliplr for row-vectors then? That's not the inverse, but rather 'read in opposite direction'. The images don't help a thing; all they do is confuse me. Can you just show in numbers what you want; i.e. first your x,y pair as you have it now, then what you get using your method, and finally what you want to obtain? Just in numbers?– Adriaan
Nov 19 '18 at 13:29
fliplr for row-vectors then? That's not the inverse, but rather 'read in opposite direction'. The images don't help a thing; all they do is confuse me. Can you just show in numbers what you want; i.e. first your x,y pair as you have it now, then what you get using your method, and finally what you want to obtain? Just in numbers?– Adriaan
Nov 19 '18 at 13:29
|
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1 Answer
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I have managed to find a solution, so for anybody who also need its:
x = linspace(0,1,length(y));
% not needed in this toy example, but can be required for a bigger vector:
[y_unique, idx] = unique(y);
inv_y = interp1(y_unique,x(idx),x);
answered Nov 19 '18 at 13:45
Valeria
10410
1
Note that this is only possible ifyis strictly monotoníc. Theuniqueremoves duplicate points, making it applicable to any monotonic function.
– Cris Luengo
Nov 19 '18 at 14:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I have managed to find a solution, so for anybody who also need its:
x = linspace(0,1,length(y));
% not needed in this toy example, but can be required for a bigger vector:
[y_unique, idx] = unique(y);
inv_y = interp1(y_unique,x(idx),x);
answered Nov 19 '18 at 13:45
Valeria
10410
1
Note that this is only possible ifyis strictly monotoníc. Theuniqueremoves duplicate points, making it applicable to any monotonic function.
– Cris Luengo
Nov 19 '18 at 14:03
add a comment |
I have managed to find a solution, so for anybody who also need its:
x = linspace(0,1,length(y));
% not needed in this toy example, but can be required for a bigger vector:
[y_unique, idx] = unique(y);
inv_y = interp1(y_unique,x(idx),x);
answered Nov 19 '18 at 13:45
Valeria
10410
1
Note that this is only possible ifyis strictly monotoníc. Theuniqueremoves duplicate points, making it applicable to any monotonic function.
– Cris Luengo
Nov 19 '18 at 14:03
add a comment |
I have managed to find a solution, so for anybody who also need its:
x = linspace(0,1,length(y));
% not needed in this toy example, but can be required for a bigger vector:
[y_unique, idx] = unique(y);
inv_y = interp1(y_unique,x(idx),x);
answered Nov 19 '18 at 13:45
Valeria
10410
I have managed to find a solution, so for anybody who also need its:
x = linspace(0,1,length(y));
% not needed in this toy example, but can be required for a bigger vector:
[y_unique, idx] = unique(y);
inv_y = interp1(y_unique,x(idx),x);
answered Nov 19 '18 at 13:45
Valeria
10410
answered Nov 19 '18 at 13:45
Valeria
10410
answered Nov 19 '18 at 13:45
Valeria
10410
answered Nov 19 '18 at 13:45
Valeria
10410
10410
1
Note that this is only possible ifyis strictly monotoníc. Theuniqueremoves duplicate points, making it applicable to any monotonic function.
– Cris Luengo
Nov 19 '18 at 14:03
add a comment |
1
Note that this is only possible ifyis strictly monotoníc. Theuniqueremoves duplicate points, making it applicable to any monotonic function.
– Cris Luengo
Nov 19 '18 at 14:03
1
1
Note that this is only possible if
y is strictly monotoníc. The unique removes duplicate points, making it applicable to any monotonic function.– Cris Luengo
Nov 19 '18 at 14:03
Note that this is only possible if
y is strictly monotoníc. The unique removes duplicate points, making it applicable to any monotonic function.– Cris Luengo
Nov 19 '18 at 14:03
add a comment |
Why don't you want to use
plot(y,x)? I almost never useplot(x), even when I want integer indices (I'd useplot(1:numel(x),x)), because of controlabillity of the function.– Adriaan
Nov 19 '18 at 13:07
@Adriaan Because I am not going to plot it. The plot was used only as an example, because I don't know how to better explain what I mean by "inverse of 1D vector". In general, I need to use is as a transformation vector and multiply it element-wise with another 1D vector.
– Valeria
Nov 19 '18 at 13:09
1
Then
1./xwould suffice. The plot just creates confusion, since thexlocations of correspondingyvalues also get changed (e.g. the 2nd value of the first plot is atx==1/3, whereas in the second plot it's atx==0.9), which means you need 2 numbers per point, as opposed to the single one you're apparently after, judging your comment.– Adriaan
Nov 19 '18 at 13:17
@Adriaan if I just get
1./x, this gives me completely different y-axis values from the ones I am looking for (added to the original question).– Valeria
Nov 19 '18 at 13:25
fliplrfor row-vectors then? That's not the inverse, but rather 'read in opposite direction'. The images don't help a thing; all they do is confuse me. Can you just show in numbers what you want; i.e. first yourx,ypair as you have it now, then what you get using your method, and finally what you want to obtain? Just in numbers?– Adriaan
Nov 19 '18 at 13:29