I would like to create a column vector X by repeating a smaller column vector G of length h a number n of times. The final vector X will be of length h*n. For example
G = [1;2;3;4] #column vector of length h
X = [1;2;3;4;1;2;3;4;1;2;3;4] #ie X = [G;G;G;G] column vector of
length h*n
I can do this in a loop but is there an equivalent to the 'fill' function that can be used without the dimensions going wrong. When I try to use fill for this case, instead of getting one column vector of length h*n I get a column vector of length n where each row is another vector of length h. For example I get the following:
X = [[1,2,3,4];[1,2,3,4];[1,2,3,4];[1,2,3,4]]
This doesn't make sense to me as I know that the ; symbol is used to show elements in a row and the space is used to show elements in a column. Why is there the , symbol used here and what does it even mean? I can access the first row of the final output X by X[1] and then any element of this by X[1][1] for example.
Either I would like to use some 'fill' equivalent or some sort of 'flatten' function if it exists, to flatten all the elements of the X into one column vector with each entry being a single number.
I have also tried the reshape function on the output but I can't get this to work either.
Thanks Dan Getz for the answer:
repeat([1, 2, 3, 4], outer = 4)
Type ?repeat at the REPL to learn about this useful function.
In older versions of Julia, repmat was an alternative, but it has now been deprecated and absorbed into repeat
As #DanGetz has pointed out in a comment, repeat is the function you want. From the docs:
repeat(A, inner = Int[], outer = Int[])
Construct an array by repeating the entries of A. The i-th element of inner specifies the number of times that the individual entries of the i-th dimension of A should be repeated. The i-th element of outer specifies the number of times that a slice along the i-th dimension of A should be repeated.
So an example that does what you want is:
X = repeat(G; outer=[k])
where G is the array to be repeated, and k is the number of times to repeat it.
I will also attempt to answer your confusion about the result of fill. Julia (like most languages) makes a distinction between vectors containing numbers and numbers themselves. We know that fill(5, 5) produces [5, 5, 5, 5, 5], which is a one-dimensional array (a vector) where each element is 5.
Note that fill([5], 5), however, produces a one-dimensional array (a vector) where each element is [5], itself a vector. This prints as
5-element Array{Array{Int64,1},1}:
[5]
[5]
[5]
[5]
[5]
and we see from the type that this is indeed a vector of vectors. That of course is not the same thing as the concatenation of vectors. Note that [[5]; [5]; [5]; [5]; [5]] is syntax for concatenation, and will return [5, 5, 5, 5, 5] as you might expect. But although ; syntax (vcat) does concatenation, fill does not do concatenation.
Mathematically (under certain definitions), we may imagine R^(kn) to be distinct (though isomorphic to) from (R^k)^n, for instance, where R^k is the set of k-tuples of real numbers. fill constructs an object of the latter, whereas repeat constructs an object of the former.
As long as you are working with 1-dimensional arrays (Vectors)...
X=repmat(G,4) should do it.
--
On another note, Julia makes no distinction between row and column vector, they are both one-dimensional arrays.
[1,2,3]==[1;2;3] returns true as they are both 3-element Array{Int64,1} or vectors (Array{Int,1} == Vector{Int} returns true)
This is one of the differences between Matlab and Julia...
If, for some specific reason you want to do it, you can create 2-dimensional Arrays (or Matrices) with one of the dimensions equal to 1.
For example:
C = [1 2 3 4] will create a 1x4 Array{Int64,2} the 2 there indicates the dimensions of the Array.
D = [1 2 3 4]' will create a 4x1 Array{Int64,2}.
In this case, C == D returns false of course. But neither is a Vector for Julia, they are both Matrices (Array{Int,2} == Matrix{Int} returns true).
Related
New to julia, so this is probably very easy.
I have an n-by-m array and a vector of length n and want to repeat each row of the array the number of times in the corresponding element of the vector. For example:
mat = rand(3,6)
v = vec([2 3 1])
The result should be a 6-by-6 array. I tried the repeat function but
repeat(mat, inner = v)
yields a 6×18×1 Array{Float64,3}: array instead so it takes v to be the dimensions along which to repeat the elements. In matlab I would use repelem(mat, v, 1) and I hope julia offers something similar. My actual matrix is a lot bigger and I will have to call the function many times, so this operation needs to be as fast as possible.
It has been discussed to add a similar thing to Julia Base, but currently it is not implemented yet AFAIK. You can achieve what you want using the inverse_rle function from StatsBase.jl:
julia> row_idx = inverse_rle(axes(v, 1), v)
6-element Array{Int64,1}:
1
1
2
2
2
3
and now you can write:
mat[row_idx, :]
or
#view mat[row_idx, :]
(the second option creates a view which might be relevant in your use case if you say that your mat is large and you need to do such indexing many times - which option is faster will depend on your exact use case).
I'm wondering if there's an indexable way of doing the following code on Octave, as it's iterative and thus really slow compared to working with indexation.
for i = [1:size(A, 1)]
for j = [1:size(A, 2)]
if (max(A(i, j, :)) == 0)
A(i, j, :) = B(i, j, :);
endif
endfor
endfor
A and B are two RGB images that overlaps and I want A(i,j) to have B(i,j) value if A(i,j) is 0 on all of the three channels. It is very slow in this form but I'm not experimented enough with this language to vectorize it.
Your code can be vectorized as follows:
I = max(A,[],3) == 0;
I = repmat(I,1,1,3);
A(I) = B(I);
The first line is a direct copy of your max conditional statement within the loop, but vectorized across all of A. This returns a 2D array, which we cannot directly use to index into the 3D arrays A and B, so we apply repmat to replicate it along the 3rd dimension (the 3 here is the number of repetitions, we're assuming A and B are RGB images with 3 elements along the 3rd dimension). Finally, an indexed assignment copies the relevant values over from B to A.
To generalize this to any array size, replace the "3" in the repmat statement with size(A,3).
Not adding much here, but perhaps this will give you a better understanding so worth adding another solution.
% example data
A = randi( 255, [2,4,3] ); A(2,2,:) = [0,0,0];
B = randi( 255, [2,4,3] );
% Logical array with size [Dim1, Dim2], such that Dim3 is 'squashed' into a
% single logical value at each position, indicating whether the third dimension
% at that position does 'not' have 'any' true (i.e. nonzero) values.
I = ~any(A, 3);
% Use this to index A and B for assignment.
A([I,I,I]) = B([I,I,I])
This approach may be more efficient than the repmat one, which is a slightly more expensive operation, but may be slightly less obvious to understand why it works. But. Understanding how this works teaches you something about matlab/octave, so it's a nice learning point.
Matlab and Octave store arrays in column major order (as opposed to, say, Python). This is also the reason that doing A(:) will return A as a vector, constructed in a column-by-column basis. It is also the reason that you can index a 3-dimensional array using a single index (called a "linear index"), which will correspond to the element you reach when you count that number of elements going down columns.
When performing logical indexing, matlab/octave effectively takes a logical vector, matches each linear index of that vector to the equivalent linear index of A and decides whether to return it or not, based on whether the boolean value of the logical index at that linear index is true or false. If you provide a logical array I that is of a smaller size than A, the indexing will simply stop at the last linear index of I. Specifically, note that the shape of I is irrelevant, since it will be interpreted in a linear indexing manner anyway.
In other words, logical indexing with I is the same as logical indexing with I(:), and logical indexing with [I,I,I] is the same as logical indexing with [ I(:); I(:); I(:) ].
And if I is of size A(:,:,1) then [I,I,I] is of size A(:,:,:), such that in a linear indexing sense it can be used as a valid logical index matching each linear index of I to the equivalent linear index of A.
The max() function can take a single matrix and return the maximum value along a dimension
There's also the all() function that tells you if all values along a dimension are nonzero, and the any() function that tells you if any of the values along a dimension are nonzero
A = reshape(1:75, 5, 5, 3)
A(2, 3, :) = 0;
B = ones(size(A)) * 1000
use_pixel_from_A = any(A, 3)
use_pixel_from_B = ~use_pixel_from_A
Now for each element of the 3rd axis, you know which pixels to take from A and which to take from B. Since our use_pixel... matrices contain 0 and 1, we can element-wise multiply them to A and B to filter out elements of A and B as required.
C = zeros(size(A));
for kk = 1:size(A, 3)
C(:, :, kk) = A(:, :, kk) .* use_pixel_from_A + B(:, :, kk) .* use_pixel_from_B
end
Say if I have a 5D Array with size 1024x1024x1x1x100. How can I make a new array that is 1024x1024x100?
The following works if you know which dimensions you want to keep ahead of time:
arr = arr[:, :, 1, 1, :]
But I don't know which dimensions are what size ahead of time and I would like to only keep dimensions given a boolean mask; something like this...
arr2 = arr[(size(arr) .> 1)]
The squeeze function was defined specifically for the purpose of removing dimensions of length 1. From the manual:
Base.squeeze — Function.
squeeze(A, dims)
Remove the dimensions
specified by dims from array A. Elements of dims must be unique and
within the range 1:ndims(A). size(A,i) must equal 1 for all i in dims.
To "squeeze" all the dimensions of size 1 (when they are unknown in advance), we need to find them and make them into a tuple. This is accomplished by ((size(arr).==1)...). So the result is:
squeeze(a,(find(size(a).==1)...))
I have an n-dimensional array A and want to slice it dynamically, i.e., given a list of array dimensions, like [2 4], and a list of values, like [6 8], I want
B = A(:,6,:,8,:,:,:,:,...)
List lengths are unknown. Using eval would work but is not an option. This question is a generalization of a previous post to multiple indices and dimensions without a for-loop.
You can still use the previous post I linked to (which I originally flagged as a duplicate) to answer your question. This original post only slices in one dimension. I originally flagged it as a duplicate and closed it because all you need to do is replace one line of code in the original post's accepted answer to achieve what you want. However, because it isn't that obvious, I have decided to reopen the question and answer the question for you.
Referring to the previous post, this is what Andrew Janke (the person with the accepted answer on the linked post) did (very clever I might add):
function out = slice(A, ix, dim)
subses = repmat({':'}, [1 ndims(A)]);
subses{dim} = ix;
out = A(subses{:});
Given a matrix A, an index number ix and the dimension you want to access dim, the above function would equivalently perform:
out = A(:, :, ..., ix, :, :,...:);
^ ^ ^ ^
dimensions --> 1 2 dim dim+1
You would access your desired dimension in dim, and place what value you want to use to slice into that dimension. As such, you'd call it like this:
out = slice(A, ix, dim);
How the function works is that subses would generate a cell array of ':' strings (that will eventually be converted into ':' operators) that is as long as the total number of dimensions of A. Next, you would access the element at dim, which corresponds to the dimension you want and you would replace this with ix. You would then unroll this cell array so that we would access A in the manner that you see in the above equivalent statement.
Who would have thought that you can use strings to index into an array!?
Now, to generalize this, all you have to do is make one small but very crucial change. ix would now be a vector of indices, and dim would be a vector of dimensions you want to access. As such, it would look something like this:
function out = slice(A, ix, dim)
subses = repmat({':'}, [1 ndims(A)]);
subses(dim) = num2cell(ix);
out = A(subses{:});
The only difference we see here is the second line of the code. We have to use num2cell so that you can convert each element into a cell array, and we slice into this cell array to replace the : operators with your desired dimensions. Note that we are using () braces and not {} braces. () braces are used to slice through cell arrays while {} are used to access cell array contents. Because we are going to assign multiple cells to subses, () is needed. We then perform our slicing in A accordingly.
As such, given your problem and with the above modifications, you would do:
out = slice(A, [6 8], [2 4]);
Be advised that ix and dim must contain the same number of elements and they must be 1D. Also, ix and dim should be sensible inputs (i.e. not floating point and negative). I don't do this error checking because I'm assuming you know what you're doing and you're smart enough to know how to use this properly.
Good luck!
I am not 100% what the role of the 1: is here. At which index to start the copy? But then why not two such parameters for the rank 2 array?
To be a little more explicit:
In Fortran90 and up you access a single array value by giving a single index, and access a subarray (a window of the array) by giving a range of indices separated by a colon.
a(1) = 0.
a(2:5) = (/3.14,2.71,1.62,0./)
You can also give a step size.
a(1:5:2) = (/0,2.71,0./)
And finally, you can leave out values and a default will be inserted. So if a runs from index 1 to 5 then I could write the above as
a(::2) = (/0,2.71,0./)
and the 1 and 5 are implied. Obviously, you shouldn't leave these out if it makes the code unclear.
With a multidimensional array, you can mix and match these on each dimension, as in your example.
You're taking a slice of array2, namely the elements in the D'th column from row 1 to C and putting them in the slice of array1 which is elements 1 through A
So both slices are 1-dimensional arrays
Slice may not be the correct term in Fortran