Summary
When using function argument validation that depends on multiple
arguments in a (Repeating) arguments block, the current argument is passed
to the validation function normally while other arguments are passed as
partially-populated
cell arrays. This contasts with how things work in non-(Repeating) arguments blocks. Is this the expected behavior or a bug?
Scenario
Consider the function dummy1 below, which uses the custom argument validation
function
mustBeEqualSize to ensure that the arguments x and y have the same size.
Since the validation of y depends on the value of x, I'm calling this
"cross-argument" validation*.
*If there's a better term for this,
please comment or edit.
function dummy1(x, y)
arguments
x (1,:)
y (1,:) {mustBeEqualSize(x,y)}
end
% Do something with x and y.
end
function mustBeEqualSize(a, b)
% Validates that function arguments have the same size.
if ~isequal(size(a), size(b))
eid = 'Size:notEqual';
msg = "Arguments must have the same size.";
throwAsCaller(MException(eid, msg))
end
end
As written, this form of argument validation works as expected:
dummy1(1:3, 4:6) % arguments have same size; validation passes (okay)
dummy1(1:3, 4:7) % arguments have different size; validation fails (okay)
In both cases, when mustBeEqualSize is called during argument
validation for y, both a and b are received as 1xN double arrays,
which matches my expectations:
% Inside call to mustBeEqualSize(x,y), when x=1:3, y=4:6 in dummy1
a =
1 2 3
b =
4 5 6
The problem arises when dummy1 is modified to accept repeating
arguments by adding (Repeating) to the arguments block:
function dummy2(x, y)
arguments (Repeating)
x (1,:)
y (1,:) {mustBeEqualSize(x,y)}
end
% Do something with each pair of x-y arguments.
% In this body, both x and y will be 1xN cell arrays, where N is the
% number of argument groups passed.
end
Now when we call dummy2(1:3, 4:6), argument validation fails. Using the
debugger, I found that when mustBeEqualSize is called during the
validation of y, a is received as a 1x1 cell array while b remains a 1x3 double array:
% Inside call to mustBeEqualSize(x,y), when x=1:3, y=4:6 in dummy2
a =
1×1 cell array
{[1 2 3]}
b =
4 5 6
The issue is even more obvious when more repeating argument are used:
dummy2(1:3, 4:6, 1:3, 4:6, 1:3, 4:6) % 3 argument groups
results in
a =
1×3 cell array
{[1 2 3]} {0×0 double} {0×0 double}
b =
4 5 6
It seems that during the validation of y in dummy2, y takes on the value of the
current argument being validated while x is a (parially populated)
cell array buffer that MATLAB allocated to hold all of the x-arguments
that were passed.
This of course breaks the cross-argument validation, since only the
argument currently being validated actually presents itself as a single
argument, while other arguments present themsevles as cell array
buffers.
Question
Is the mismatch between how cross-argument validation works for (Repeating) versus non-(Repeating) arguments a bug, or is cross-argument validation with (Repeating) arguments not supported in MATLAB? If this difference in behavior is expected, is there any way to make cross-argument validation work with (Repeating) arguments?
The MATLAB docs say the following about argument validation with
Repeating arguments
In the function, each repeating argument becomes a cell array with the
number of elements equal to the number of repeats passed in the function
call. The validation is applied to each element of the cell array.
which doesn't seem to shed any light on how cross-argument validation
should work with repeating arguments.
Tested using MATLAB R2021a (9.10.0.1602886).
Note: This was an edge case of the design we didn't consider. I've made the relevant devs teams aware and they'll consider fixing it in a future release.
In the short term, you could try grabbing the last non 0x0 double element of the cell array:
function mustBeEqualSize(x,y)
if iscell(x)
% Get last non-empty element of `x`
notEmptyDouble = #(e)isa(e,'double') && ~isequal(size(e), [0, 0]);
idx = find(cellfun(notEmptyDouble, x), 1, 'last');
x = x{idx};
end
% check equality
if ~isequal(size(x), size(y))
error("size mismatch")
end
end
This will work on all non-empty arrays, but unfortunately given the empty type used, will not work for dummy2([],[]).
Also, I want to note, that whatever validator ends up working for this use case, consider if it will break the function if/when this issue is fixed. I.e. if this function needs to handle both double arrays and cell arrays, could be problematic in the future. However if this function only needs double arrays, it can be constrained using
arguments(Repeating)
a (1,:) double
b (1,:) double {mustBeEqualSize(x,y)}
end
This combined with the iscell check in the validator, should future-proof this function.
Related
How can I access both arguments of ismember when it is used inside splitapply?
slitapply only returns scalar values for each group, so in order to compute nonscalar values for each group (as returned by the first argument of ismemebr), one has to enclose the anonymous function (in this case ismember) inside curly brackets {} to return a cell array.
But now, when I provide two output arguments to splitapply, I get an error:
Output argument "varargout{2}" (and maybe others) not assigned during call to
"#(x,y) {ismember(x,y)}"
ADD 1
I can create another function, say, ismember2cell which would apply ismember and turn outputs into cell arrays:
function [a, b] = ismember2cell(x,y)
[a,b] = ismember(x,y);
a = {a};
b = {b};
end
but maybe there is a solution which doesn't require this workaround.
One potentially faster option is to just do what splitapply is already doing under the hood by splitting your data into cell arrays (using functions like mat2cell or accumarray) and then using cellfun to apply your function across them. Using cellfun will allow you to easily capture multiple outputs (such as from ismember). For example:
% Sample data:
A = [1 2 3 4 5];
B = [1 2 1 5 5];
G = [1 1 1 2 2]; % Group index
% Group data into cell arrays:
cellA = accumarray(G(:), A(:), [], #(x) {x(:).'}); % See note below about (:).' syntax
cellB = accumarray(G(:), B(:), [], #(x) {x(:).'});
% Apply function:
[Lia, Locb] = cellfun(#ismember, cellA, cellB, 'UniformOutput', false);
NOTE: My sample data are row vectors, but I had to use the colon operator to reshape them into column vectors when passing them to accumarray (it wants columns). Once distributed into a cell array, each piece of the vector would still be a column vector, and I simply wanted to keep them as row vectors to match the original sample data. The syntax (:).' is a colon reshaping followed by a nonconjugate transpose, ensuring a row vector as a result no matter the shape of x. In this case I probably could have just used .', but I've gotten into the habit of never assuming what the shape of a variable is.
I cannot find a global solution, but the accepted answer of this post helps me to define a helper function for your problem:
function varargout = out2cell(varargin)
[x{1:nargout}]=feval(varargin{:});
varargout = num2cell(x);
I think that you may succeed in calling
splitapply(#(x,y) out2cell(#ismember, x, y), A, B);
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 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]
A = {'A'; 'E'; 'A'; 'F'};
B = {'A';'B';'C';'D';'E'; 'F'};
I am trying to get for each string in cell array A, the index that matches that string in cell array B. A will have repeated values, B will not.
find(ismember(B, A) == 1)
outputs
1
5
6
but I want to get
1
5
1
6
preferably in a one liner. I can't use strcmp instead of ismember either as the vectors are different sizes.
The vectors will actually contain date strings, and I need the index not a logical index matrix, I'm interested in the number not to use it for indexing.
How do I do it?
You flip the arguments to ismember, and you use the second output argument:
[~,loc]=ismember(A,B)
loc =
1
5
1
6
The second output tells you where the elements of A are in B.
If you are working with very strict limits to how many lines you can have in your code, and are in no position to fire the manager who imposed such limitations, you may want to access the second output of ismember directly. In order to do this, you can create the following helper function that allows to directly access the i-th output of a function
function out = accessIthOutput(fun,ii)
%ACCESSITHOUTPUT returns the i-th output variable of the function call fun
%
% define fun as anonymous function with no input, e.g.
% #()ismember(A,B)
% where A and B are defined in your workspace
%
% Using the above example, you'd access the second output argument
% of ismember by calling
% loc = accessIthOutput(#()ismember(A,B),2)
%# get the output
[output{1:ii}] = fun();
%# return the i-th element
out = output{ii};
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.
Closed 10 years ago.
Over the years, reading others code, I encountered and collected some examples of MATLAB syntax which can be at first unusual and counterintuitive. Please, feel free to comment or complement this list. I verified it with r2006a.
MATLAB always returns first output argument of a function (if it has at least one) into its caller workspace, also unexpectedly if function is being called without returning arguments like myFunc1(); myFunc2(); the caller workspace still would contain first output of myFunc2(); as "invisible" ans variable. It could play an important role if ans is a reference object - it would remain alive.
set([], 'Background:Color','red')
MATLAB is very forgiving sometimes. In this case, setting properties to an array of objects works also with nonsense properties, at least when the array is empty. Such arrays usually come from harray = findobj(0,'Tag','NotExistingTag')
myArray([1,round(end/2)])
This use of end keyword may seem unclean but is sometimes very handy instead of using length(myArray).
any([]) ~= all([])
Surprisigly any([]) returns false and all([]) returns true. And I always thought that all is stronger then any.
EDIT:
with not empty argument all() returns true for a subset of values for which any() returns true (e.g. truth table). This means that any() false implies all() false. This simple rule is being violated by MATLAB with [] as argument.
Loren also blogged about it.
Select(Range(ExcelComObj))
Procedural style COM object method dispatch. Do not wonder that exist('Select') returns zero!
[myString, myCell]
MATLAB makes in this case an implicit cast of string variable myString to cell type {myString}. It works, also if I would not expect it to do so.
[double(1.8), uint8(123)] => 2 123
Another cast example. Everybody would probably expect uint8 value being cast to double but Mathworks have another opinion. Without a warning this behavior is very dangerous.
a = 5;
b = a();
It looks silly but you can call a variable with round brackets. Actually it makes sense because this way you can execute a function given its handle.
Syntax Foo(:) works not only on data but also with functions if called as Bar.Foo(:), in this scenario the function input argument is passed as char colon ':'.
For example let Bar.Foo = #(x) disp(x)
Now calling Bar.Foo(:) prints char ':' in the MATLAB Command Window.
This strange feature works with all MATLAB 7 versions without warnings.
a = {'aa', 'bb'
'cc', 'dd'};
Surprsisingly this code neither returns a vector nor rises an error but defins matrix, using just code layout. It is probably a relict from ancient times.
EDIT: very handy feature, see the comment by gnovice.
set(hobj, {'BackgroundColor','ForegroundColor'},{'red','blue'})
This code does what you probably expect it to do. That function set accepts a struct as its second argument is a known fact and makes sense, and this sintax is just a cell2struct away.
Equvalence rules are sometimes unexpected at first. For example 'A'==65 returns true (although for C-experts it is self-evident). Similarly isequal([],{}) retuns, as expected, false and isequal([],'') returns true.
The string-numeric equivalence means that all string functions can be used also for numeric arrays, for example to find indices of a sub-array in a large array:
ind = strfind( [1 2 3 4 1 2 3 4 1 2 3 4 ], [2 3] )
MATLAB function isnumeric() returns false for booleans. This feels just ... false :-)
About which further unexpected/unusual MATLAB features are you aware?
Image coordinates vs plot coordinates Used to get me every time.
%# create an image with one white pixel
img = zeros(100);
img(25,65) = 1;
%# show the image
figure
imshow(img);
%# now circle the pixel. To be sure of the coordinate, let's run find
[x,y] = find(img);
hold on
%# plot a red circle...
plot(x,y,'or')
%# ... and it's not in the right place
%# plot a green circle with x,y switched, and it works
plot(y,x,'og')
Edit 1
Array dimensions
Variables have at least two dimensions. Scalars are size [1,1], vectors are size [1,n] or [n,1]. Thus, ndims returns 2 for any of them (in fact, ndims([]) is 2 as well, since size([]) is [0,0]). This makes it a bit cumbersome to test for the dimensionality of your input. To check for 1D arrays, you have to use isvector, 0D arrays need isscalar.
Edit 2
Array assignments
Normally, Matlab is strict with array assignments. For example
m = magic(3);
m(1:2,1:3) = zeros(3,2);
throws a
??? Subscripted assignment dimension mismatch.
However, these work:
m(1:2,1:2) = 1; %# scalar to vector
m(2,:) = ones(3,1); %# vector n-by-1 to vector 1-by-n (for newer Matlab versions)
m(:) = 1:9; %# vector to 'linearized array'
Edit 3
Logical indexing with wrongly sized arrays Good luck debugging this!
Logical indexing seems to make a call to find, since your logical array doesn't need the same amount of elements as there are indices!
>> m = magic(4); %# a 4-by-4 array
>> id = logical([1 1 0 1 0])
id =
1 1 0 1 0
>> m(id,:) %# id has five elements, m only four rows
ans =
16 2 3 13
5 11 10 8
4 14 15 1
%# this wouldn't work if the last element of id was 1, btw
>> id = logical([1 1 0])
id =
1 1 0
>> m(id,:) %# id has three elements, m has four rows
ans =
16 2 3 13
5 11 10 8
Instead of listing examples of weird MATLAB syntax, I'll address some of the examples in the question that I think make sense or are expected/documented/desired behavior.
How ANY and ALL handle empty arguments:
The result of any([]) makes sense: there are no non-zero elements in the input vector (since it's empty), so it returns false.
The result of all([]) can be better understood by thinking about how you might implement your own version of this function:
function allAreTrue = my_all(inArray)
allAreTrue = true;
N = numel(inArray);
index = 1;
while allAreTrue && (index <= N)
allAreTrue = (inArray(index) ~= 0);
index = index + 1;
end
end
This function loops over the elements of inArray until it encounters one that is zero. If inArray is empty, the loop is never entered and the default value of allAreTrue is returned.
Concatenating unlike classes:
When concatenating different types into one array, MATLAB follows a preset precedence of classes and converts values accordingly. The general precedence order (from highest to lowest) is: char, integer (of any sign or number of bits), single, double, and logical. This is why [double(1.8), uint8(123)] gives you a result of type uint8. When combining unlike integer types (uint8, int32, etc.), the left-most matrix element determines the type of the result.
Multiple lines without using the line continuation operator (...):
When constructing a matrix with multiple rows, you can simply hit return after entering one row and enter the next row on the next line, without having to use a semicolon to define a new row or ... to continue the line. The following declarations are therefore equivalent:
a = {'aa', 'bb'
'cc', 'dd'};
a = {'aa', 'bb'; ...
'cc', 'dd'};
a = {'aa', 'bb'; 'cc', 'dd'};
Why would you want MATLAB to behave like this? One reason I've noticed is that it makes it easy to cut and paste data from, for example, an Excel document into a variable in the MATLAB command window. Try the following:
Select a region in an Excel file and copy it.
Type a = [ into MATLAB without hitting return.
Right-click on the MATLAB command window and select "Paste".
Type ]; and hit return. Now you have a variable a that contains the data from the rows and columns you selected in the Excel file, and which maintains the "shape" of the data.
Arrays vs. cells
Let's look at some basic syntax to start with. To create an array with elements a, b, c you write [a b c]. To create a cell with arrays A, B, C you write {A B C}. So far so good.
Accessing array elements is done like this: arr(i). For Cells, it's cell{i}. Still good.
Now let's try to delete an element. For arrays: arr(i) = []. Extrapolating from examples above, you might try cell{i} = {} for cells, but this is a syntax error. The correct syntax to delete an element of a cell is, in fact, the very same syntax you use for arrays: cell(i) = [].
So, most of the time you access cells using special syntax, but when deleting items you use the array syntax.
If you dig deeper you'll find that actually a cell is an array where each value is of a certain type. So you can still write cell(i), you'll just get {A} (a one-valued cell!) back. cell{i} is a shorthand to retrieve A directly.
All this is not very pretty IMO.