I am doing a matlab homework and I solved the next problem. and the grader say it is a correct answer. I used for in the program and we didn't take yet in the course. can someone suggest a program with out for or if.
Write a function called pitty that takes a matrix called ab as an input argument. The matrix ab has exactly two columns. The function should return a column vector c that contains positive values each of which satisfies the Pythagorean Theorem, a2 + b2 = c2, for the corresponding row of ab assuming that the two elements on each row of ab correspond to one pair, a and b, respectively, in the theorem. Note that the built-in MATLAB function sqrt computes the square root and you are allowed to use it.
my code
function c = pitty(ab)
[n , m] = size(ab)
for i = 1:n
c(i) = sqrt(ab(i,1)^2 + ab(i,2)^2)
end
c = c'
end
You can square each element of the matrix by using the .^2 operator. Then summing along each row sum(...,2) and finally taking the root.
ab = [1,2;3,4;5,6]
c = sqrt(sum(ab.^2,2));
No for needed for that.
MATLAB has a function for this called hypot short for hypotenuse. The main reason for existence of it is that it takes care of overflow (and underflow) problem. If the input values are too large (or small) the square of them (or sum of square of them) can be larger (smaller) than the largest (smallest) representable value in floating-point, while still the corresponding c value is representable. In your case you can use it like this:
c=hypot(ab(:,1), ab(:,2));
Cleve Moler, one of the founders of MathWorks and original author of MATLAB, tells the story behind hypotin this article.
I'd recommend hypot as in Mohsen's answer.
Just for some variety, here's another approach, using complex numbers. This approach avoids overflow and underflow, just like hypot does:
abs(ab*[1; 1j])
Examples (taken from Cleve Moler's post):
>> ab = [1e154 1e154]; %// LARGE VALUES: possible overflow
>> sqrt(sum(ab.^2,2))
ans =
Inf %// overflow
>> hypot(ab(:,1), ab(:,2))
ans =
1.414213562373095e+154 %// correct result
>> abs(ab*[1; 1j])
ans =
1.414213562373095e+154 %// correct result
>> ab = [3e-200 4e-200]; %// SMALL VALUES: possible underflow
>> sqrt(sum(ab.^2,2))
ans =
0 %// underflow
>> hypot(ab(:,1), ab(:,2))
ans =
5.000000000000000e-200 %// correct result
>> abs(ab*[1; 1j])
ans =
5.000000000000000e-200 %// correct result
Related
Suppose I have 4D matrix:
>> A=1:(3*4*5*6);
>> A=reshape(A,3,4,5,6);
And now I want to cut given number of rows and columns (or any given chunks at known dimensions).
If I would know it's 4D I would write:
>> A1=A(1:2,1:3,:,:);
But how to write universally for any given number of dimensions?
The following gives something different:
>> A2=A(1:2,1:3,:);
And the following gives an error:
>> A2=A;
>> A2(3:3,4:4)=[];
It is possible to generate a code with general number of dimension of A using the second form of indexing you used and reshape function.
Here there is an example:
Asize = [3,4,2,6,4]; %Initialization of A, not seen by the rest of the code
A = rand(Asize);
%% This part of the code can operate for any matrix A
I = 1:2;
J = 3:4;
A1 = A(I,J,:);
NewSize = size(A);
NewSize(1) = length(I);
NewSize(2) = length(J);
A2 = reshape(A1,NewSize);
A2 will be your cropped matrix. It works for any Asize you choose.
I recommend the solution Luis Mendo suggested for the general case, but there is also a very simple solution when you know a upper limit for your dimensions. Let's assume you have at most 6 dimensions. Use 6 dimensional indexing for all matrices:
A1=A(1:2,1:3,:,:,:,:);
Matlab will implicit assume singleton dimensions for all remaining dimension, returning the intended result also for matrices with less dimensions.
It sounds like you just want to use ndims.
num_dimensions = ndims(A)
if (num_dimensions == 3)
A1 = A(1:2, 1:3, :);
elseif (num_dimensions == 4)
A1 = A(1:2, 1:3, :, :);
end
If the range of possible matrix dimensions is small this kind of if-else block keeps it simple. It seems like you want some way to create an indexing tuple (e.g. (1:2,:,:,:) ) on the fly, which I don't know if there is a way to do. You must match the correct number of dimensions with your indexing...if you index in fewer dimensions than the matrix has, matlab returns a value with the unindexed dimensions collapsed into a single array (similar to what you get with
A1 = A(:);
So I have the following matrices:
A = [1 2 3; 4 5 6];
B = [0.5 2 3];
I'm writing a function in MATLAB that will allow me to multiply a vector and a matrix by element as long as the number of elements in the vector matches the number of columns. In A there are 3 columns:
1 2 3
4 5 6
B also has 3 elements so this should work. I'm trying to produce the following output based on A and B:
0.5 4 9
2 10 18
My code is below. Does anyone know what I'm doing wrong?
function C = lab11(mat, vec)
C = zeros(2,3);
[a, b] = size(mat);
[c, d] = size(vec);
for i = 1:a
for k = 1:b
for j = 1
C(i,k) = C(i,k) + A(i,j) * B(j,k);
end
end
end
end
MATLAB already has functionality to do this in the bsxfun function. bsxfun will take two matrices and duplicate singleton dimensions until the matrices are the same size, then perform a binary operation on the two matrices. So, for your example, you would simply do the following:
C = bsxfun(#times,mat,vec);
Referencing MrAzzaman, bsxfun is the way to go with this. However, judging from your function name, this looks like it's homework, and so let's stick with what you have originally. As such, you need to only write two for loops. You would use the second for loop to index into both the vector and the columns of the matrix at the same time. The outer most for loop would access the rows of the matrix. In addition, you are referencing A and B, which are variables that don't exist in your code. You are also initializing the output matrix C to be 2 x 3 always. You want this to be the same size as mat. I also removed your checking of the length of the vector because you weren't doing anything with the result.
As such:
function C = lab11(mat, vec)
[a, b] = size(mat);
C = zeros(a,b);
for i = 1:a
for k = 1:b
C(i,k) = mat(i,k) * vec(k);
end
end
end
Take special note at what I did. The outer-most for loop accesses the rows of mat, while the inner-most loop accesses the columns of mat as well as the elements of vec. Bear in mind that the number of columns of mat need to be the same as the number of elements in vec. You should probably check for this in your code.
If you don't like using the bsxfun approach, one alternative is to take the vector vec and make a matrix out of this that is the same size as mat by stacking the vector vec on top of itself for as many times as we have rows in mat. After this, you can do element-by-element multiplication. You can do this stacking by using repmat which repeats a vector or matrices a given number of times in any dimension(s) you want. As such, your function would be simplified to:
function C = lab11(mat, vec)
rows = size(mat, 1);
vec_mat = repmat(vec, rows, 1);
C = mat .* vec_mat;
end
However, I would personally go with the bsxfun route. bsxfun basically does what the repmat paradigm does under the hood. Internally, it ensures that both of your inputs have the same size. If it doesn't, it replicates the smaller array / matrix until it is the same size as the larger array / matrix, then applies an element-by-element operation to the corresponding elements in both variables. bsxfun stands for Binary Singleton EXpansion FUNction, which is a fancy way of saying exactly what I just talked about.
Therefore, your function is further simplified to:
function C = lab11(mat, vec)
C = bsxfun(#times, mat, vec);
end
Good luck!
I've been searching the net for a couple of mornings and found nothing, hope you can help.
I have an anonymous function like this
f = #(x,y) [sin(2*pi*x).*cos(2*pi*y), cos(2*pi*x).*sin(2*pi*y)];
that needs to be evaluated on an array of points, something like
x = 0:0.1:1;
y = 0:0.1:1;
w = f(x',y');
Now, in the above example everything works fine, the result w is a 11x2 matrix with in each row the correct value f(x(i), y(i)).
The problem comes when I change my function to have constant values:
f = #(x,y) [0, 1];
Now, even with array inputs like before, I only get out a 1x2 array like w = [0,1];
while of course I want to have the same structure as before, i.e. a 11x2 matrix.
I have no idea why Matlab is doing this...
EDIT 1
Sorry, I thought it was pretty clear from what I wrote in the original question, but I see some of you asking, so here is a clarification: what I want is to have again a 11x2 matrix, since I am feeding the function with arrays with 11 elements.
This means I expect to have an output exactly like in the first example, just with changed values in it: a matrix with 11 rows and 2 columns, with only values 0 in the first column and only values 1 in the second, since for all x(i) and y(i) the answer should be the vector [0,1].
It means I expect to have:
w = [0 1
0 1
0 1
...
0 1]
seems pretty natural to me...
You are defining a function f = #(x,y) [0, 1]; which has the input parameters x,y and the output [0,1]. What else do you expect to happen?
Update:
This should match your description:
g=#(x,y)[zeros(size(x)),ones(size(y))]
g(x',y')
Defining an anonymous function f as
f = #(x,y) [0,1];
naturally returns [0,1] for any inputs x and y regardless of the length of those vectors.
This behavior puzzled me also until I realized that I expected f(a,b) to loop over a and b as if I had written
for inc = 1:length(a)
f(a(inc), b(inc))
end
However, f(a,b) does not loop over the length of its inputs, so it merely returns [0,1] regardless of the length of a and b.
The desired behavior can be obtained by defining f as
g=#(x,y)[zeros(size(x)),ones(size(y))]
as Daniel stated in his answer.
I'm trying to use MatLab code as a way to learn math as a programmer.
So reading I'm this post about subspaces and trying to build some simple matlab functions that do it for me.
Here is how far I got:
function performSubspaceTest(subset, numArgs)
% Just a quick and dirty function to perform subspace test on a vector(subset)
%
% INPUT
% subset is the anonymous function that defines the vector
% numArgs is the the number of argument that subset takes
% Author: Lasse Nørfeldt (Norfeldt)
% Date: 2012-05-30
% License: http://creativecommons.org/licenses/by-sa/3.0/
if numArgs == 1
subspaceTest = #(subset) single(rref(subset(rand)+subset(rand))) ...
== single(rref(rand*subset(rand)));
elseif numArgs == 2
subspaceTest = #(subset) single(rref(subset(rand,rand)+subset(rand,rand))) ...
== single(rref(rand*subset(rand,rand)));
end
% rand just gives a random number. Converting to single avoids round off
% errors.
% Know that the code can crash if numArgs isn't given or bigger than 2.
outcome = subspaceTest(subset);
if outcome == true
display(['subset IS a subspace of R^' num2str(size(outcome,2))])
else
display(['subset is NOT a subspace of R^' num2str(size(outcome,2))])
end
And these are the subset that I'm testing
%% Checking for subspaces
V = #(x) [x, 3*x]
performSubspaceTest(V, 1)
A = #(x) [x, 3*x+1]
performSubspaceTest(A, 1)
B = #(x) [x, x^2, x^3]
performSubspaceTest(B, 1)
C = #(x1, x3) [x1, 0, x3, -5*x1]
performSubspaceTest(C, 2)
running the code gives me this
V =
#(x)[x,3*x]
subset IS a subspace of R^2
A =
#(x)[x,3*x+1]
subset is NOT a subspace of R^2
B =
#(x)[x,x^2,x^3]
subset is NOT a subspace of R^3
C =
#(x1,x3)[x1,0,x3,-5*x1]
subset is NOT a subspace of R^4
The C is not working (only works if it only accepts one arg).
I know that my solution for numArgs is not optimal - but it was what I could come up with at the current moment..
Are there any way to optimize this code so C will work properly and perhaps avoid the elseif statments for more than 2 args..?
PS: I couldn't seem to find a build-in matlab function that does the hole thing for me..
Here's one approach. It tests if a given function represents a linear subspace or not. Technically it is only a probabilistic test, but the chance of it failing is vanishingly small.
First, we define a nice abstraction. This higher order function takes a function as its first argument, and applies the function to every row of the matrix x. This allows us to test many arguments to func at the same time.
function y = apply(func,x)
for k = 1:size(x,1)
y(k,:) = func(x(k,:));
end
Now we write the core function. Here func is a function of one argument (presumed to be a vector in R^m) which returns a vector in R^n. We apply func to 100 randomly selected vectors in R^m to get an output matrix. If func represents a linear subspace, then the rank of the output will be less than or equal to m.
function result = isSubspace(func,m)
inputs = rand(100,m);
outputs = apply(func,inputs);
result = rank(outputs) <= m;
Here it is in action. Note that the functions take only a single argument - where you wrote c(x1,x2)=[x1,0,x2] I write c(x) = [x(1),0,x(2)], which is slightly more verbose, but has the advantage that we don't have to mess around with if statements to decide how many arguments our function has - and this works for functions that take input in R^m for any m, not just 1 or 2.
>> v = #(x) [x,3*x]
>> isSubspace(v,1)
ans =
1
>> a = #(x) [x(1),3*x(1)+1]
>> isSubspace(a,1)
ans =
0
>> c = #(x) [x(1),0,x(2),-5*x(1)]
>> isSubspace(c,2)
ans =
1
The solution of not being optimal barely scratches the surface of the problem.
I think you're doing too much at once: rref should not be used and is complicating everything. especially for numArgs greater then 1.
Think it through: [1 0 3 -5] and [3 0 3 -5] are both members of C, but their sum [4 0 6 -10] (which belongs in C) is not linear product of the multiplication of one of the previous vectors (e.g. [2 0 6 -10] ). So all the rref in the world can't fix your problem.
So what can you do instead?
you need to check if
(randn*subset(randn,randn)+randn*subset(randn,randn)))
is a member of C, which, unless I'm mistaken is a difficult problem: Conceptually you need to iterate through every element of the vector and make sure it matches the predetermined condition. Alternatively, you can try to find a set such that C(x1,x2) gives you the right answer. In this case, you can use fminsearch to solve this problem numerically and verify the returned value is within a defined tolerance:
[s,error] = fminsearch(#(x) norm(C(x(1),x(2)) - [2 0 6 -10]),[1 1])
s =
1.999996976386119 6.000035034493023
error =
3.827680714104862e-05
Edit: you need to make sure you can use negative numbers in your multiplication, so don't use rand, but use something else. I changed it to randn.
You can apply a function to every item in a vector by saying, for example, v + 1, or you can use the function arrayfun. How can I do it for every row/column of a matrix without using a for loop?
Many built-in operations like sum and prod are already able to operate across rows or columns, so you may be able to refactor the function you are applying to take advantage of this.
If that's not a viable option, one way to do it is to collect the rows or columns into cells using mat2cell or num2cell, then use cellfun to operate on the resulting cell array.
As an example, let's say you want to sum the columns of a matrix M. You can do this simply using sum:
M = magic(10); %# A 10-by-10 matrix
columnSums = sum(M, 1); %# A 1-by-10 vector of sums for each column
And here is how you would do this using the more complicated num2cell/cellfun option:
M = magic(10); %# A 10-by-10 matrix
C = num2cell(M, 1); %# Collect the columns into cells
columnSums = cellfun(#sum, C); %# A 1-by-10 vector of sums for each cell
You may want the more obscure Matlab function bsxfun. From the Matlab documentation, bsxfun "applies the element-by-element binary operation specified by the function handle fun to arrays A and B, with singleton expansion enabled."
#gnovice stated above that sum and other basic functions already operate on the first non-singleton dimension (i.e., rows if there's more than one row, columns if there's only one row, or higher dimensions if the lower dimensions all have size==1). However, bsxfun works for any function, including (and especially) user-defined functions.
For example, let's say you have a matrix A and a row vector B. E.g., let's say:
A = [1 2 3;
4 5 6;
7 8 9]
B = [0 1 2]
You want a function power_by_col which returns in a vector C all the elements in A to the power of the corresponding column of B.
From the above example, C is a 3x3 matrix:
C = [1^0 2^1 3^2;
4^0 5^1 6^2;
7^0 8^1 9^2]
i.e.,
C = [1 2 9;
1 5 36;
1 8 81]
You could do this the brute force way using repmat:
C = A.^repmat(B, size(A, 1), 1)
Or you could do this the classy way using bsxfun, which internally takes care of the repmat step:
C = bsxfun(#(x,y) x.^y, A, B)
So bsxfun saves you some steps (you don't need to explicitly calculate the dimensions of A). However, in some informal tests of mine, it turns out that repmat is roughly twice as fast if the function to be applied (like my power function, above) is simple. So you'll need to choose whether you want simplicity or speed.
I can't comment on how efficient this is, but here's a solution:
applyToGivenRow = #(func, matrix) #(row) func(matrix(row, :))
applyToRows = #(func, matrix) arrayfun(applyToGivenRow(func, matrix), 1:size(matrix,1))'
% Example
myMx = [1 2 3; 4 5 6; 7 8 9];
myFunc = #sum;
applyToRows(myFunc, myMx)
Building on Alex's answer, here is a more generic function:
applyToGivenRow = #(func, matrix) #(row) func(matrix(row, :));
newApplyToRows = #(func, matrix) arrayfun(applyToGivenRow(func, matrix), 1:size(matrix,1), 'UniformOutput', false)';
takeAll = #(x) reshape([x{:}], size(x{1},2), size(x,1))';
genericApplyToRows = #(func, matrix) takeAll(newApplyToRows(func, matrix));
Here is a comparison between the two functions:
>> % Example
myMx = [1 2 3; 4 5 6; 7 8 9];
myFunc = #(x) [mean(x), std(x), sum(x), length(x)];
>> genericApplyToRows(myFunc, myMx)
ans =
2 1 6 3
5 1 15 3
8 1 24 3
>> applyToRows(myFunc, myMx)
??? Error using ==> arrayfun
Non-scalar in Uniform output, at index 1, output 1.
Set 'UniformOutput' to false.
Error in ==> #(func,matrix)arrayfun(applyToGivenRow(func,matrix),1:size(matrix,1))'
For completeness/interest I'd like to add that matlab does have a function that allows you to operate on data per-row rather than per-element. It is called rowfun (http://www.mathworks.se/help/matlab/ref/rowfun.html), but the only "problem" is that it operates on tables (http://www.mathworks.se/help/matlab/ref/table.html) rather than matrices.
Adding to the evolving nature of the answer to this question, starting with r2016b, MATLAB will implicitly expand singleton dimensions, removing the need for bsxfun in many cases.
From the r2016b release notes:
Implicit Expansion: Apply element-wise operations and functions to arrays with automatic expansion of dimensions of length 1
Implicit expansion is a generalization of scalar expansion. With
scalar expansion, a scalar expands to be the same size as another
array to facilitate element-wise operations. With implicit expansion,
the element-wise operators and functions listed here can implicitly
expand their inputs to be the same size, as long as the arrays have
compatible sizes. Two arrays have compatible sizes if, for every
dimension, the dimension sizes of the inputs are either the same or
one of them is 1. See Compatible Array Sizes for Basic Operations and
Array vs. Matrix Operations for more information.
Element-wise arithmetic operators — +, -, .*, .^, ./, .\
Relational operators — <, <=, >, >=, ==, ~=
Logical operators — &, |, xor
Bit-wise functions — bitand, bitor, bitxor
Elementary math functions — max, min, mod, rem, hypot, atan2, atan2d
For example, you can calculate the mean of each column in a matrix A,
and then subtract the vector of mean values from each column with A -
mean(A).
Previously, this functionality was available via the bsxfun function.
It is now recommended that you replace most uses of bsxfun with direct
calls to the functions and operators that support implicit expansion.
Compared to using bsxfun, implicit expansion offers faster speed,
better memory usage, and improved readability of code.
None of the above answers worked "out of the box" for me, however, the following function, obtained by copying the ideas of the other answers works:
apply_func_2_cols = #(f,M) cell2mat(cellfun(f,num2cell(M,1), 'UniformOutput',0));
It takes a function f and applies it to every column of the matrix M.
So for example:
f = #(v) [0 1;1 0]*v + [0 0.1]';
apply_func_2_cols(f,[0 0 1 1;0 1 0 1])
ans =
0.00000 1.00000 0.00000 1.00000
0.10000 0.10000 1.10000 1.10000
With recent versions of Matlab, you can use the Table data structure to your advantage. There's even a 'rowfun' operation but I found it easier just to do this:
a = magic(6);
incrementRow = cell2mat(cellfun(#(x) x+1,table2cell(table(a)),'UniformOutput',0))
or here's an older one I had that doesn't require tables, for older Matlab versions.
dataBinner = cell2mat(arrayfun(#(x) Binner(a(x,:),2)',1:size(a,1),'UniformOutput',0)')
The accepted answer seems to be to convert to cells first and then use cellfun to operate over all of the cells. I do not know the specific application, but in general I would think using bsxfun to operate over the matrix would be more efficient. Basically bsxfun applies an operation element-by-element across two arrays. So if you wanted to multiply each item in an n x 1 vector by each item in an m x 1 vector to get an n x m array, you could use:
vec1 = [ stuff ]; % n x 1 vector
vec2 = [ stuff ]; % m x 1 vector
result = bsxfun('times', vec1.', vec2);
This will give you matrix called result wherein the (i, j) entry will be the ith element of vec1 multiplied by the jth element of vec2.
You can use bsxfun for all sorts of built-in functions, and you can declare your own. The documentation has a list of many built-in functions, but basically you can name any function that accepts two arrays (vector or matrix) as arguments and get it to work.
I like splitapply, which allows a function to be applied to the columns of A using splitapply(fun,A,1:size(A,2)).
For example
A = magic(5);
B = splitapply(#(x) x+1, A, 1:size(A,2));
C = splitapply(#std, A, 1:size(A,2));
To apply the function to the rows, you could use
splitapply(fun, A', 1:size(A,1))';
(My source for this solution is here.)
Stumbled upon this question/answer while seeking how to compute the row sums of a matrix.
I would just like to add that Matlab's SUM function actually has support for summing for a given dimension, i.e a standard matrix with two dimensions.
So to calculate the column sums do:
colsum = sum(M) % or sum(M, 1)
and for the row sums, simply do
rowsum = sum(M, 2)
My bet is that this is faster than both programming a for loop and converting to cells :)
All this can be found in the matlab help for SUM.
if you know the length of your rows you can make something like this:
a=rand(9,3);
b=rand(9,3);
arrayfun(#(x1,x2,y1,y2,z1,z2) line([x1,x2],[y1,y2],[z1,z2]) , a(:,1),b(:,1),a(:,2),b(:,2),a(:,3),b(:,3) )