Sometimes when I use the interp1 function in MATLAB, it throws an error saying my vectors need to be monotonically increasing, and other times it doesn't.
For example, let's say I have 3 vectors.
A = [286, 295, 298, 301, 304, 308, 310, 324, 330, 335];
B = [31000, 30950, 30875, 30775, 30650, 30500, 30425, 29900, 29675, 29450];
C = [290, 291, 292, 293, 294, 295, 296, 297, 298, 299];
And I want to run
D = interp1(A,B,C);
This function will return successfully even though B is not monotonically increasing. Does the monotonicity only apply to the first and third vectors passing into the equation?
What the error message actually means
The error is actually a little mis-leading in this case and is caused by all values in A not being unique (not strictly monotonic). The error (which is less useful) actually propagates up from griddedInterpolant which is used by many interpolation functions and therefore has a generic error message.
Why it only applies to some inputs
With interp1 you are essentially attempting to construct an estimate of a function f(x) using x locations provided by the user as well as their corresponding values (f(x)). In your example, A contains the location of each data point (x) and B contains the values of your function at each of those points (f(x)). It is only necessary that the locations (values in A) are unique so that you don't have multiple values in B for the same value of A. If you did, interp1 doesn't know how to cope with that.
The ordering of A (the monotonically increasing part of the error) doesn't matter because interp1 will automatically sort A to be increasing (it also re-arranges B so that the values still correspond to A).*
C is simply the locations at which you want to sample the interpolant. You can request the value of the function at the same point a million times with no issue. interp1 will simply return the corresponding value for each location in C so there are no constraints on the values or ordering of C.
A = [1 3 2]; % Not monotically increasing but DOES need to be unique values
B = [1 2 1]; % Can have any value and can repeat values but each
% value corresponds with each element in A
C = [3 3 1 1 2 2]; % Can be any order and can repeat values
% ERROR FREE!
interp1(A, B, C)
% 2 2 1 1 1 1
*If you do want the ordering of your A and B points to be respected, then you'll want to parameterize your input data in a different way as suggested in this answer
Related
Here is an example of convolution given:
I have two questions here:
Why is the vector 𝑥 padded with two 0s on each side? As, the length of kernel ℎ is 3. If 𝑥 is padded with one 0 on each side, the middle element of convolution output would be within the range of the length of 𝑥, why not one 0 on each side?
Explain the following output to me:
>> x = [1, 2, 1, 3];
>> h = [2, 0, 1];
>> y = conv(x, h, 'valid')
y =
3 8
>>
What is valid doing here in the context of the previously shown mathematics on vectors 𝑥 and ℎ?
I can't speak as to the amount of zero padding that is proper .... That being said, any zero padding is making up data that is not there. This isn't necessarily wrong, but you should be aware that the values computing this information may be biased. Sometimes you care about this, sometimes you don't. Introducing 1 zero (in this case) would leave the middle kernel value always in the data, but why should that be a stopping criteria? Importantly, adding on 2 zeros still leaves one multiplication of values that are actually present in the data and the kernel (the x[0]*h[0] and x[3]*h[2] - using 0 based indexing). Adding on a 3rd zero (or more) would just yield zeros in the output since 3 is the length of the kernel. In other words zero padding will always yield an output that is partially based on the actual data (but not completely) for any zero padding from n=1 to n = length(h)-1 (in this case either 1 or 2).
Even though zero padding with length 2 or 1 still has multiplications based on real data, some values are summed over "fake" data (those multiplied with a padded zero). In this case Matlab gives you 3 options for how you want the data returned. First, you can get the full convolution, which includes values that are biased because they include adding in 0 values that aren't really in the data. Alternatively you can get same, which means the length of the output is the length of the data y = [4 3 8 1]. This corresponds to 1 zero but note that for longer kernels you could technically get other lengths between full and same, Matlab just doesn't return those for you.
Finally, and probably most important to understand out of all this, you have the valid option. In your example only 2 samples of the output are computed from summations that occur only from multiplications over real data (i.e. from multiplying samples of the kernel with samples from x and not from zeros). More specifically:
y[2] = h[2]*x[0] + h[1]*x[1] + h[2]*x[2] = 3 //0 based indexing like example
y[3] = h[2]*x[1] + h[1]*x[2] + h[2]*x[3] = 8
Note none of the other y values are computed with only h and x, they all involve a padded zero which is not necessarily indicative of the real data. For example:
y[4] = h[2]*x[2] + h[1]*x[3] + h[2]*0 <= padded zero
I am a complete newbie to MATLAB and the first task I have is to create a volcano plot. I have been using the documentation to understand about it and get started.
I tried to run it on dummy values -
a=[1 2 3]
b=[4.6 2.7 4.5]
c=[0.05 0.33 0.45]
And then I ran -
SigStructure = mavolcanoplot(a, b, c)
My understanding is that a represents the gene expression values for condition 1, b for condition 2, and c is the list of p-values for the 3 data points in a and b.
However running this code gives me the error -
Index exceeds matrix dimensions.
Error in mavolcanoplot (line 127)
appdata.effect = X(paramStruct.goodVals) - Y(paramStruct.goodVals);
Error in volc (line 4)
SigStructure = mavolcanoplot(a, b, c)
Can anyone explain where I am going wrong?
You're encountering an issue because you are using row vectors.
Inside the mavolcanoplot function (you can see the file by typing edit mavolcanoplot in the command window) there is a local function for checking the inputs, called check_inputdata.
Your data passes all of the validation checks, and then encounters this section:
% Here, 'X' and 'Y' are the local names for your inputs 'a' and 'b'
% Below code is directly from mavolcanoplot.m:
% Handle the matrix input. Use its mean values per row
if size(X, 2) > 1
X = mean(X,2);
end
if size(Y, 2) > 1
Y = mean(Y,2);
end
This reduces your inputs to their mean. Later in the main function (line 127) you encounter the error as described, where paramStruct.goodVals is a 3 element array, which is now trying to index a 1 element array and fails!
This is basically a lesson in debugging and reading the documentation, which states
DataX, DataY: If a [...] matrix, each row is a gene, each column is a sample and an average expression value is calculated for each gene.
You should use one of these equivalent methods to create column vector inputs
a=[1 2 3].'; % Using transpose (.') to create a column vector from a row vector
b=[4.6; 2.7; 4.5]; % Creating a column vector using the semi-colon operator to end each row
c=[0.05
0.33
0.45]; % Using actual code layout to create a column vector
In Matlab R2016a, I have a large set of small X-vectors and Y-vectors which are paired (e.g. 10,000 1x3 X-vectors paired with 10,000 1x3 Y vectors). For each {X,Y} pair, I want to calculate a 2-scalar-argument function for every pairwise combination of the elements in X and Y, (so in my example I would get 10,000 3x3 matrices).
I thought I could use bsxfun to perform these calculations, but it doesn't work when I try to do some simple tests. bsxfun(#(x,y) x*y,[1 2],[1 2]') returns:
ans =
1 2
2 4
Which is what I would expect. However, bsxfun(#(x,y) 1,[1 2],[1 2]') returns:
Error using bsxfun
Specified function handle produces invalid output dimensions. The function handle
must be a binary elementwise function.
Which makes no sense. The function handle is a binary elementwise function that always returns the scalar 1, so bsxfun should give the same result as ones(2,2), unless I'm not understanding how bsxfun works.
The inputs to the function handle that are passed to bsxfun are not scalars. In versions prior to R2016b, the inputs are either scalar or they are the same size.
FUNC can also be a handle to any binary element-wise function not listed
above. A binary element-wise function in the form of C = FUNC(A,B)
accepts arrays A and B of arbitrary but equal size and returns output
of the same size. Each element in the output array C is the result
of an operation on the corresponding elements of A and B only. FUNC must
also support scalar expansion, such that if A or B is a scalar, C is the
result of applying the scalar to every element in the other input array.
In releases since R2016b, they do not have to be equal sizes, but should be compatible sizes
In the example you have shown, the first input to the function handle is a scalar and the second is a vector (y) and the function is evaluated for every element of x and the output is expected to be the size of y
In the case you've posted, the call to bsxfun is essentially the equivalent of:
x = [1 2];
y = [1 2].';
yourfunc = #(x,y)x * y;
for k = 1:numel(x)
output(:,k) = yourfunc(x(k), y)
end
If you want to return a 1 for every entry, you need to replace your function with something that yields the appropriately sized output.
bsxfun(#(x,y)ones(max(size(x), size(y))), [1 2], [1 2]')
How you formulate the function handle really depends upon your specific problem
I have 2 m x n matrices and would like to calculate canonical correlations for segments within a given window length. For example, if my window length is 100, I'd like to have the correlation coefficients for
canoncorr(X(1:100,:),Y(1:100,:))
canoncorr(X(101:200,:),Y(101:200,:))
canoncorr(X(201:300,:),Y(201:300,:))
...
and so on all accumulated into one matrix. I am only interested in the correlation coefficient r.
I am trying the following:
win=100;
r=[];
for i=1:win:size(X,1)-win-2
[A,B,r(i,:)] = canoncorr(X(i:i+win,:),Y(i:i+win,:));
end
However, my resulting matrix does not only save the values from row 1, 101, 201 etc. but also fills the rows between 1 and 101 and so on with zeroes.
If I try
[A,B,r(i:i+win,:)] = canoncorr(X(i:i+win,:),Y(i:i+win,:));
then subscripted assignment dimensions mismatch.
What am I doing wrong?
i goes from 1, 101, 201, ...
so, please check for 101th row and see if they are zero.
you can also spy(r) to see matrix elements.
I need help figuring out how to code the following problem. Any help would be greatly appreciated!
Create a function that will take a vector/array input for x (1 by n) and a scalar input for a, and produce the output defined by the following equation:
y(x,a)=((xsin(ax-2))/(sqrt(1+(ax)^2)
-π ≤ x ≤ π
a={.5 1 1.5 2}
The equation must be vectorized in terms of x and the output from the function is the array y which has the same dimension as the array x.
Write a script that calls this function to compute y(x,a) for the range of x defined above and each value of the parameter a. Results should be stored in a solution matrix using a different row of the solution matrix for each value of a.
So far for my function I have:
function [y] = part1(a,x)
y=((x*sin(a*x-2))/(sqrt(1+(a*x).^2)));
end
I'm not sure how to output this into the solution matrix
For my script I have:
%%
clear,clc
a={0.5 1 1.5 2};
x=-pi:0.1:pi;
for
part1(x,a)
end
I'm getting the following errors when I run this now:
Undefined function 'mtimes' for input arguments of type 'cell'.
Error in part1 (line 4)
y=((x*sin(a*x-2))/(sqrt(1+(a*x).^2)));
Error in labtest2 (line 8)
y(i,:)=part1(x,a(i));
EDIT
I've made some changes and am still getting some errors that I cannot resolve.
Here is my full code for function followed by full code for script:
Function
function [y] = part1(x,a)
nx=numel(x);
na=numel(a);
y=((x.*sin(a.*x-2))./(sqrt(1+(a.*x).^2)));
size(y)=[nx na]
end
Script
%%
clear,clc
a={0.5 1 1.5 2};
x=-pi:0.1:pi;
for i = 1:length(a)
y(i,:)=part1(x,a(i));
end
Errors
Undefined function 'times' for input arguments of type 'cell'.
Error in part1 (line 6)
y=((x.*sin(a.*x-2))./(sqrt(1+(a.*x).^2)));
Error in labtest2 (line 8)
y(i,:)=part1(x,a(i));
The reason you're getting Undefined function 'times' for input arguments of type 'cell' is because your variable a is a cell array. You need to change your assignment of a from
a={0.5 1 1.5 2};
to
a=[0.5 1 1.5 2];
which will make it just a normal array. Alternatively, you need to reference it with cell array notation: a{i} instead of a(i).
You're almost there. Note that you've written
function [y] = part1(a,x)
but you call it in your script as
part1(x,a)
so you should probably correct that.
A few things jump out at me:
You never assign the output of part1(x,a) to anything. You're told that
Results should be stored in a solution matrix using a different row of the solution matrix for each value of a.
What I take this to mean is that the 1st row corresponds to part1() evaluated for the 1st element of a. Since we're operating on x which is a vector, that row will have multiple columns. Your output is indeed a matrix. In your case, length(-pi:0.1:pi) == 63, therefore size(y) == [4 63], where y is your output matrix.
Your for loop is backwards. You're told to accept scalar a and vector x. Therefore, your script should be something like:
a = 0.5:0.5:2;
x = -pi:0.1:pi;
for i = 1:length(a)
y(i,:) = part1(x, a(i));
end
Note the use of length and the : operator. I want to iterate between 1 to length(a) (in this case, length(a) == 4) so that I can use the current a(i) value as an index into my output matrix, y. The : operator in y(i,:) signifies "The ith row and all columns of y will take the value output by part1(x,a(i))."
Your function needs to be changed up for element-by-element operations. Notice that, for instance, x*sin(a*x-2) works for scalar x but not vectors. This is because x is a vector and sin(a*x-2) is also a vector (since the sin call will operate element-by-element and a is a scalar). Trying to multiply two vectors together will result in errors since MATLAB will try to perform a matrix multiplication. Resolve this by replacing * with .*. This way it is unambiguous that you are going to multiply these two vectors element-by-element. You'll also need to change / to ./.
On another note, thank you for attempting to do your homework before asking SO for help. We've been getting a huge influx of questions from students that have made no attempt to do their own work before dumping it on us, so it's refreshing that we regulars of the MATLAB tag get to actual help out instead of telling people to do their own work.
Finally got the whole thing worked out.
Function
function [y] = part1(x,a)
y=((x.*sin(a.*x - 2))./(sqrt(1 + (a.*x).^2)));
end
Script
%%
clear all;
clc;
close all;
x=[-pi:.1:pi];
a=[.5:.5:2];
for i=1:length(a)
y(i,:)=part1(x,a(i));
plot(x,y)
end
Sol=[y]