For an nD array, it would be nice to be able to auto squeeze to remove singleton dimensions. Is there a way to do this that I don't know about? This would be especially useful for aggregate functions (e.g. sum, mean, etc) where you always expect a result with fewer dimensions.
Here's a simple example:
>> A = ones(3,3,3);
>> B = mean(A);
>> size(B)
ans =
1 3 3
>> squeeze(B)
ans =
1 1 1
1 1 1
1 1 1
It would be nice if Matlab/Octave would automatically do the squeezing for me. Or if there was a way to turn that option on (something similar to hold on for plots).
As far as I know, Matlab does not have that. And I don't think it would be a good idea. Consider a modified version of your example:
>> A = ones(3,1,1,3);
>> B = mean(A);
>> size(B)
ans =
1 1 1 3
What should "auto-squeeze" do here? Reduce B to size [1 1 3] or to [1 3]?
You could argue that it should remove the same dimension that mean has turned into a singleton. But then it would have to be done within the mean function, perhaps with an optional input argument. Once you get the function output, there is no information how it was obtained.
Or you could argue that it should remove all singleton dimensions, like squeeze (more or less) does. But then it would remove dimensions that were already singleton in the function input, which is probably unwanted.
If you ask me, having a second input in squeeze specifiyng which (singleton) dimensions to remove would be a nice addition (in the same vein as you can use mean(A, 1) to force the operation to be applied along the first dimension even if A happens to be a row vector).
I agree with Luis and Cris, but I would add the following.
Both Matlab and Octave do automatically squeeze extra dimensions, in a very particular scenario: any dimensions at the end that have been reduced to singletons, are automatically squeezed out.
E.g.
A = ones([1,2,3,4]);
B = mean(A, 4);
size(B)
% ans = 1 2 3
Note, how the answer is [1,2,3], and not [1,2,3,1]. This is in contrast to languages like python, for instance, where a size of (1,1) is very different to a size of (1,).
Therefore, with regard to your questions, one way to use this to your advantage could be to ensure that the dimension that is to be reduced is always found at the end, and thus automatically simplified.
This becomes even more useful when you realise that:
size(A(:)) % ans = 24 1 (i.e. 24)
size(A(:,:)) % ans = 1 24
size(A(:,:,:)) % ans = 1 2 12
size(A(:,:,:,:)) % ans = 1 2 3 4
Meaning, if you order your dimensions hierarchically you can ensure that any operations that need to take place over the higher dimensions, can a) be vectorised easily, and b) give a natural result, without the need to waste time squeezing or permuting the resulting dimensions.
Related
I know that bsxfun(which works fast!) and arrayfun(as far as I could understand, uses loops internally which is expected to be slow) are intended for different uses, at least, at the most basic level.
Having said this, I am trying
to sum up all the numbers in a given array, say y, before a certain index
add the number at the specific location(which is the number at the above index location) to the above sum.
I could perform this with the below piece of example code easily:
% index array
x = [ 1:6 ]; % value array
y = [ 3 3 4 4 1 1 ];
% arrayfun version
o2 = arrayfun(#(a) ...
sum(y(1:(a-1)))+...
y(a), ...
x)
But it seems to be slow on large inputs.
I was wondering what would be a good way to convert this to a version that works with bsxfun, if possible.
P.S. the numbers in y do not repeat as given above, this was just an example, it could also be [3 4 3 1 4 ...]
Is x always of the form 1 : n? Assuming the answer is yes, then you can get the same result with the much faster code:
o2 = cumsum(y);
Side note: you don't need the brackets in the definition of x.
if you have a supported GPU device, you can define your variables as gpuArray type since arrayfun, bsxfun and pagefun are compatible with GPUs.
GPU computing is supposed to be faster for large data.
I would like to compute the product of the next n adjacent elements of a matrix. The number n of elements to be multiplied should be given in function's input.
For example for this input I should compute the product of every 3 consecutive elements, starting from the first.
[p, ind] = max_product([1 2 2 1 3 1],3);
This gives [1*2*2, 2*2*1, 2*1*3, 1*3*1] = [4,4,6,3].
Is there any practical way to do it? Now I do this using:
for ii = 1:(length(v)-2)
p = prod(v(ii:ii+n-1));
end
where v is the input vector and n is the number of elements to be multiplied.
in this example n=3 but can take any positive integer value.
Depending whether n is odd or even or length(v) is odd or even, I get sometimes right answers but sometimes an error.
For example for arguments:
v = [1.35912281237829 -0.958120385352704 -0.553335935098461 1.44601450110386 1.43760259196739 0.0266423803393867 0.417039432979809 1.14033971399183 -0.418125096873537 -1.99362640306847 -0.589833539347417 -0.218969651537063 1.49863539349242 0.338844452879616 1.34169199365703 0.181185490389383 0.102817336496793 0.104835620599133 -2.70026800170358 1.46129128974515 0.64413523430416 0.921962619821458 0.568712984110933]
n = 7
I get the error:
Index exceeds matrix dimensions.
Error in max_product (line 6)
p = prod(v(ii:ii+n-1));
Is there any correct general way to do it?
Based on the solution in Fast numpy rolling_product, I'd like to suggest a MATLAB version of it, which leverages the movsum function introduced in R2016a.
The mathematical reasoning is that a product of numbers is equal to the exponent of the sum of their logarithms:
A possible MATLAB implementation of the above may look like this:
function P = movprod(vec,window_sz)
P = exp(movsum(log(vec),[0 window_sz-1],'Endpoints','discard'));
if isreal(vec) % Ensures correct outputs when the input contains negative and/or
P = real(P); % complex entries.
end
end
Several notes:
I haven't benchmarked this solution, and do not know how it compares in terms of performance to the other suggestions.
It should work correctly with vectors containing zero and/or negative and/or complex elements.
It can be easily expanded to accept a dimension to operate along (for array inputs), and any other customization afforded by movsum.
The 1st input is assumed to be either a double or a complex double row vector.
Outputs may require rounding.
Update
Inspired by the nicely thought answer of Dev-iL comes this handy solution, which does not require Matlab R2016a or above:
out = real( exp(conv(log(a),ones(1,n),'valid')) )
The basic idea is to transform the multiplication to a sum and a moving average can be used, which in turn can be realised by convolution.
Old answers
This is one way using gallery to get a circulant matrix and indexing the relevant part of the resulting matrix before multiplying the elements:
a = [1 2 2 1 3 1]
n = 3
%// circulant matrix
tmp = gallery('circul', a(:))
%// product of relevant parts of matrix
out = prod(tmp(end-n+1:-1:1, end-n+1:end), 2)
out =
4
4
6
3
More memory efficient alternative in case there are no zeros in the input:
a = [10 9 8 7 6 5 4 3 2 1]
n = 2
%// cumulative product
x = [1 cumprod(a)]
%// shifted by n and divided by itself
y = circshift( x,[0 -n] )./x
%// remove last elements
out = y(1:end-n)
out =
90 72 56 42 30 20 12 6 2
Your approach is correct. You should just change the for loop to for ii = 1:(length(v)-n+1) and then it will work fine.
If you are not going to deal with large inputs, another approach is using gallery as explained in #thewaywewalk's answer.
I think the problem may be based on your indexing. The line that states for ii = 1:(length(v)-2) does not provide the correct range of ii.
Try this:
function out = max_product(in,size)
size = size-1; % this is because we add size to i later
out = zeros(length(in),1) % assuming that this is a column vector
for i = 1:length(in)-size
out(i) = prod(in(i:i+size));
end
Your code works when restated like so:
for ii = 1:(length(v)-(n-1))
p = prod(v(ii:ii+(n-1)));
end
That should take care of the indexing problem.
using bsxfun you create a matrix each row of it contains consecutive 3 elements then take prod of 2nd dimension of the matrix. I think this is most efficient way:
max_product = #(v, n) prod(v(bsxfun(#plus, (1 : n), (0 : numel(v)-n)')), 2);
p = max_product([1 2 2 1 3 1],3)
Update:
some other solutions updated, and some such as #Dev-iL 's answer outperform others, I can suggest fftconv that in Octave outperforms conv
If you can upgrade to R2017a, you can use the new movprod function to compute a windowed product.
Using timing tests, I found that it's much more performant to grow Vector{Array{Float64}} objects using push! than it is to simply use an Array{Float64} object and either hcat or vcat. However, after the computation is completed, I need to change the resulting object to an Array{Float64} for further analysis. Is there a way that works regardless of the dimensions? For example, if I generate the Vector of Arrays via
u = [1 2 3 4
1 3 3 4
1 5 6 3
5 2 3 1]
uFull = Vector{Array{Int}}(0)
push!(uFull,u)
for i = 1:10000
push!(uFull,u)
end
I can do the conversion like this:
fill = Array{Int}(size(uFull)...,size(u)...)
for i in eachindex(uFull)
fill[i,:,:] = uFull[i]
end
but notice this requires that I know the arrays are matrices (2-dimensional). If it's 3-dimensional, I would need another :, and so this doesn't work for arbitrary dimensions.
Note that I also need a form of the "inverse transform" (except first indexed by the last index of the full array) in arbitrary dimensions, and I currently have
filla = Vector{Array{Int}}(size(fill)[end])
for i in 1:size(fill)[end]
filla[i] = fill[:,:,i]'
end
I assume the method for the first conversion will likely solve the second as well.
This is the sort of thing that Julia's custom array infrastructure excels at. I think the simplest solution here is to actually make a special array type that does this transformation for you:
immutable StackedArray{T,N,A} <: AbstractArray{T,N}
data::A # A <: AbstractVector{<:AbstractArray{T,N-1}}
dims::NTuple{N,Int}
end
function StackedArray(vec::AbstractVector)
#assert all(size(vec[1]) == size(v) for v in vec)
StackedArray(vec, (length(vec), size(vec[1])...))
end
StackedArray{T, N}(vec::AbstractVector{T}, dims::NTuple{N}) = StackedArray{eltype(T),N,typeof(vec)}(vec, dims)
Base.size(S::StackedArray) = S.dims
#inline function Base.getindex{T,N}(S::StackedArray{T,N}, I::Vararg{Int,N})
#boundscheck checkbounds(S, I...)
S.data[I[1]][Base.tail(I)...]
end
Now just wrap your vector in a StackedArray and it'll behave like an N+1 dimensional array. This could be expanded and made more featureful (it could similarly support setindex! or even push!ing arrays to concatenate natively), but I think that it's sufficient to solve your problem. By simply wrapping uFull in a StackedArray you get an object that acts like an Array{T, N+1}. Make a copy, and you get exactly a dense Array{T, N+1} without ever needing to write a for loop yourself.
julia> S = StackedArray(uFull)
10001x4x4 StackedArray{Int64,3,Array{Array{Int64,2},1}}:
[:, :, 1] =
1 1 1 5
1 1 1 5
1 1 1 5
…
julia> squeeze(S[1:1, :, :], 1) == u
true
julia> copy(S) # returns a dense Array{T,N}
10001x4x4 Array{Int64,3}:
[:, :, 1] =
1 1 1 5
1 1 1 5
…
Finally, I'll just note that there's another solution here: you could introduce the custom array type sooner, and make a GrowableArray that internally stores its elements as a linear Vector{T}, but allows pushing entire columns or arrays directly.
Matt B.'s answer is great, because it "simulates" an array without actually having to create or store it. When you can use this solution, it's likely to be your best choice.
However, there might be circumstances where you need to create a concatenated array (e.g., if you're passing this to some C code which requires contiguous memory). In that case you can just call cat, which is generic (it can handle arbitrary dimensions).
For example:
u = [1 2 3 4
1 3 3 4
1 5 6 3
5 2 3 1]
uFull = Vector{typeof(u)}(0)
push!(uFull,u)
for i = 1:10000
push!(uFull,u)
end
ucat = cat(ndims(eltype(uFull))+1, uFull)
I took the liberty of making one important change to your code: uFull = Vector{typeof(u)}(0) because it ensures that the objects stored in the Vector container have concrete type. Array{Int} is actually an abstract type, because you'd need to specify the dimensionality too (Array{Int,2}).
(* I don't know programming in Matlab. It is just a general question about Matlab language. *)
In Excel, we can store a formula in a cell. For instance, if A2 contains a formula = A1+10, the re-evaluation of A2 returns 30 when the value of A1 is 20.
My question is, is there a similar mechanism in matlab? That said, can we specify a formula in a element of an array in Matlab, so that we can re-evaluate the array later?
Edit 1:
Following the comment of #rayryeng I try to make an example to illustrate the concept... Actually, this is exactly what spreadsheet languages such as Excel can do.
So my question is, is there a mechanism that permits the following in Matlab? (Note that the following syntax is just symbolic)
>> B = [1 2; B{1,1}+2 4] // store some values and a formula in the array
B =
1 2
3 4
>> B{1,1} = 10 // change the value of one cell
B =
10 2
3 4
>> EVAL(B) // there is a re-evaluation command to re-calculate all the cells
ans =
10 2
13 4
Hopefully I'm understanding what you want correctly, but the answer is indeed yes. You can store "formulas" in a cell array where each element is a handle or an anonymous function.
Perhaps you mean something like this:
formulae = {#(x) x+10, #sin, #cos, #(x) x / 3};
The syntax # denotes a function handle and the (x) denote that this is an anonymous function with the input variable x. The first cell element provides a function that adds 10 to every value that goes into it, the second and third parameters are handles to sin and cos, so these act like those trigonometric functions. The last handle divides every value that goes into it by 3.
To demonstrate, let's create a small array, then go through each formula and apply each of them to the small array:
>> formulae = {#(x) x+10, #sin, #cos, #(x) x / 3};
>> A = [1 2; 3 4]
A =
1 2
3 4
>> formulae{1}(A)
ans =
11 12
13 14
>> formulae{2}(A)
ans =
0.8415 0.9093
0.1411 -0.7568
>> formulae{3}(A)
ans =
0.5403 -0.4161
-0.9900 -0.6536
>> formulae{4}(A)
ans =
0.3333 0.6667
1.0000 1.3333
We first create the formulae, then create a small 2 x 2 matrix of [1 2; 3 4]. After, we access each formula's cell, then put in the input A into the function and we get what you see.
However, when you're starting out, start with actually declaring functions in function scripts.... don't use this kind of style of programming for practical applications. It makes your code less readable. For example, doing sin(A) is much more readable than formula{2}(A). People reading your code have to remember what position in the array corresponds to what formula you are applying to each element in the input.
Because I am trying to let a GUI element slice my array, there will be a : (colon) sign in the variables. This returns me an error:
Error in gui_mainfcn (line 96)
feval(varargin{:});
line 96 refers to this code:
image(handles.data(1:handles.rows,1:handles.cols, temp))
Temp looks like this
temp =
1 1 1 1 2 1 1 1 1
And both handles.rows and cols are the value 64. So the problem seems to be that I use colons in the gui function. However, to slice I need to use colons. My question now is: Any idea how to work around this?
To clarify as requested below
The above code works when I manually enter it in the console. Also when I use handles.data(:,:,1,1,1,1,2,1,1,1,1), handles.data(1:end,1:end,1,1,1,1,2,1,1,1,1), handles.data(1:64,1:64,1,1,1,1,2,1,1,1,1), etc I get the same error from the gui. Manually they all work and return a 64 by 64 array of doubles which I can plot with image().
Might be related to these questions, however those deal with parfor difficulties and dont seem to answer my question:
matlab-parfor-slicing-issue
index-inside-parfor-slicing
I am now also reading the advanced topics for slicing variables. Still dont see what I am doing wrong though, so any help or explanation would still be greatly apprectiated. Thanks!
Explanation
By putting the vector temp as the third index into your data, you are not indexing the higher dimensions - you are repeatedly indexing the third. In other words, you get handles.data(:,:,[1 1 1 1 2 1 1 1 1]) instead of handles.data(:,:,1,1,1,1,2,1,1,1,1).
Solution
Here's a solution that doesn't require squeeze or eval. It exploits the comma-separated lists output of the {:} syntax with cell arrays, and the ability to apply linear indexing on the last subscripted dimension.
ctemp = num2cell(temp); % put each index into a cell
sz = size(handles.data); % i.e. sz = [256 256 1 1 2 1 2]
sliceind = sub2ind(sz(3:end),ctemp{:}); % compute high dim. linear index (scalar)
image(handles.data(:,:,sliceind));
This performs subscripting of a >3D array with only 3 subscripts by computing the last subscript as a linear index. It's weird, but convenient sometimes.
A heads up for people with the same problem, this error can not only result from not knowing how to slice, it could also result from not having defined your variables correctly: http://www.mathworks.nl/matlabcentral/answers/87417-how-to-slice-inside-gui-without-error-feval-varargin