Suppose A and B are two 1d arrays of different sizes such that pythonically B = A[C] where C is a particular list of indices meeting some specific but otherwise known condition. How to get C if A and B are both known? I tried C = np.where(np.close(A, B)) but I get the following error message:
File "/home/username/../my_script.py", line 897, in get_histogram
i0, i1 = np.where(np.isclose(hist_xs, vals1)), np.where(np.isclose(hist_ys, vals2))
File "<__array_function__ internals>", line 6, in isclose
File "/usr/local/anaconda3/lib/python3.7/site-packages/numpy/core/numeric.py", line 2260, in isclose
return within_tol(x, y, atol, rtol)
File "/usr/local/anaconda3/lib/python3.7/site-packages/numpy/core/numeric.py", line 2246, in within_tol
return less_equal(abs(x-y), atol + rtol * abs(y))
ValueError: operands could not be broadcast together with shapes (722,) (1536,)
In other words, I am trying to get only those elements of A whose right indices would correspond to the known B array.
Are you looking for this?:
sorti = np.argsort(A)
C_inv = sorti[np.searchsorted(A,B,sorter=sorti)]
sample code (it works on any sortable array, if your array elements do not have comparison operator, you could write your own. If (less/greater) comparison is not applicable, you would need a for loop to find elements which seems to easy to include here):
A: [89 28 86 73 29 71 37 46 15 52]
B: [86 52 15]
C: [2 9 8]
C_inv: [2 9 8]
Related
It would really help me reason about my MATLAB code if I didn't have to worry about accidental 2d operations. For instance, if I want to do element-wise multiplication of 1d arrays, but one is a row and another is a column, I end up with a 2d result.
>> a = 1:8;
>> a = a(:);
>> a .* cumsum(ones(8))
ans =
1 1 1 1 1 1 1 1
4 4 4 4 4 4 4 4
9 9 9 9 9 9 9 9
16 16 16 16 16 16 16 16
25 25 25 25 25 25 25 25
36 36 36 36 36 36 36 36
49 49 49 49 49 49 49 49
64 64 64 64 64 64 64 64
I'd like to prevent this type of thing, and likely other problems that I can't foresee, by keeping all my arrays 1d wherever I can. But every time I check the size() of vector, I get at least 2 elements back:
>> size(1:1:6)
ans =
1 6
>> size(linspace(0, 5, 10))
ans =
1 10
I've tried the suggestions at How to create single dimensional array in matlab? and some of the options here (PDF download), and I can't get a "truly" 1d array. How would you deal with this type of issue?
There is no such thing as 1D array. The documentation says (emphasis mine):
All MATLAB variables are multidimensional arrays, no matter what type of data. A matrix is a two-dimensional array often used for linear algebra.
You may use isvector, isrow and iscolumn to identify vectors, row vectors and column vectors respectively.
#Sardar has already said the last word. Another clue is ndims:
N = ndims(A) returns the number of dimensions in the array A. The
number of dimensions is always greater than or equal to 2. ...
But about your other question:
How would you deal with this type of issue?
There's not much you can do. Debug, find the mistake and fix it. If it's some one-time script, you are done. But if you are writing functions that may be used later, it's better to protect them from accepting arguments with unequal dimensions:
function myFunc(A, B)
if ndims(A)~=ndims(B) || any(size(A)~=size(B))
error('Matrix dimensions must agree.');
end
% ...
end
Or, if your function really needs them to be vectors:
function myFunc(A, B)
if ~isvector(A) || ~isvector(B) || any(size(A)~=size(B))
error('A and B must be vectors with same dimensions.');
end
% ...
end
You can also validate different attributes of arguments using validateattributes:
function myFunc(A, B)
validateattributes(A, {'numeric'},{'vector'}, 'myFunc', 'A')
validateattributes(B, {'numeric'},{'size', size(A)}, 'myFunc', 'B')
% ...
end
Edit:
Also, if the function only needs the inputs to be vectors and their orientation does not matter, you can modify them inside the function (thanks to #CrisLuengo for commenting).
function myFunc(A, B)
if ~isvector(A) || ~isvector(B) || length(A)~=length(B)
error('A and B must be vectors with the same length.');
end
A = A(:);
B = B(:);
% ...
end
However, this is not recommended when the output of the function is also a vector with the same size as the inputs. This is because the caller expects the output to be in the same orientation as the inputs, and if this is not the case, problems may arise.
I have an array of mxnx4 dimensions and a function with a 4-element vector input. I want to apply my function to give me an mxnx4 output, such that the function is applied mxn times (i.e. the 3rd dimension of my array becomes in the input of my function). Any ideas on how do do this would be massively appreciated. I have looked at arrayfun, but this seems not to address what I want to do. I also want to maintain the formatting of my function as it requires this formatting elsewhere in my code.
I.e.:
F=#(V)[V(1)+V(2);V(2)+V(3); V(3)+V(4); V(4)+V(1)]
to be applied to array M of dimensions 2x3x4 (m=2; n=3)
M = zeros(2,3,4);
M(:,:,1) = [1 2 3;4 5 6];
M(:,:,2) = [7 8 9;10 11 12];
M(:,:,3) = [13 14 15; 16 17 18];
M(:,:,4) = [19 20 21; 22 23 24];
to generate an array of dimensions 2x3x4
C(1,1,:) = F([1 7 13 19])
C(1,2,:) = F([2 8 14 20])
etc.
I can see a for loop would work, applying the following across the mxn matrix
C(m,n,:)=F(M(m,n,:))
However, I need to run this millions of times so would like a faster approach
I am new to using Julia and have little experience with the language. I am trying to understand how multi-dimensional arrays work in it and how to access the array at the different dimensions. The documentation confuses me, so maybe someone here can explain it better.
I created an array (m = Array{Int64}(6,3)) and am trying to access the different parts of that array. Clearly I am understanding it wrong so any help in general about Arrays/Multi-Dimensional Arrays would help.
Thanks
Edit I am trying to read a file in that has the contents
58 129 10
58 129 7
25 56 10
24 125 25
24 125 15
13 41 10
0
The purpose of the project is to take these fractions (58/129) and round the fractions using farey sequence. The last number in the row is what both numbers need to be below. Currently, I am not looking for help on how to do the problem, just how to create a multidimensional array with all the numbers except the last row (0). My trouble is how to put the numbers into the array after I have created it.
So I want m[0][0] = 58, so on. I'm not sure how syntax works for this and the manual is confusing. Hopefully this is enough information.
Julia's arrays are not lists-of-lists or arrays of pointers. They are a single container, with elements arranged in a rectangular shape. As such, you do not access successive dimensions with repeated indexing calls like m[j][i] — instead you use one indexing call with multiple indices: m[i, j].
If you trim off that last 0 in your file, you can just use the built-in readdlm to load that file into a matrix. I've copied those first six rows into my clipboard to make it a bit easier to follow here:
julia> str = clipboard()
"58 129 10\n58 129 7\n25 56 10\n24 125 25\n24 125 15\n13 41 10"
julia> readdlm(IOBuffer(str), Int) # or readdlm("path/to/trimmed/file", Int)
6×3 Array{Int64,2}:
58 129 10
58 129 7
25 56 10
24 125 25
24 125 15
13 41 10
That's not very helpful in teaching you how Julia's arrays work, though. Constructing an array like m = Array{Int64}(6,3) creates an uninitialized matrix with 18 elements arranged in 6 rows and 3 columns. It's a bit easier to see how things work if we fill it with a sensible pattern:
julia> m .= [10,20,30,40,50,60] .+ [1 2 3]
6×3 Array{Int64,2}:
11 12 13
21 22 23
31 32 33
41 42 43
51 52 53
61 62 63
This has set up the values of the array to have the row number in their tens place and the column number in the ones place. Accessing m[r,c] returns the value in m at row r and column c.
julia> m[2,3] # second row, third column
23
Now, r and c don't have to be integers — they can also be vectors of integers to select multiple rows or columns:
julia> m[[2,3,4],[1,2]] # Selects rows 2, 3, and 4 across columns 1 and 2
3×2 Array{Int64,2}:
21 22
31 32
41 42
Of course ranges like 2:4 are just vectors themselves, so you can more easily and efficiently write that example as m[2:4, 1:2]. A : by itself is a shorthand for a vector of all the indices within the dimension it indexes into:
julia> m[1, :] # the first row of all columns
3-element Array{Int64,1}:
11
12
13
julia> m[:, 1] # all rows of the first column
6-element Array{Int64,1}:
11
21
31
41
51
61
Finally, note that Julia's Array is column-major and arranged contiguously in memory. This means that if you just use one index, like m[2], you're just going to walk down that first column. As a special extension, we support what's commonly referred to as "linear indexing", where we allow that single index to span into the higher dimensions. So m[7] accesses the 7th contiguous element, wrapping around into the first row of the second column:
julia> m[5],m[6],m[7],m[8]
(51, 61, 12, 22)
I have an array of numbers A, and I want to create a new array B representing the indicies of A after A is reordered from smallest to largest.
For example:
A = [50 10 60 90 30];
The reordered A would then be:
A = [10 30 50 60 90];
And I want to get the output:
B = [2 5 1 3 4];
These indices in B therefore correspond to the original indices in A, but are written in the order of the reordered A.
How can I do this?
Use the second output of sort:
[A_sorted, B] = sort(A);
I have a cell array like this: [...
0
129
8...2...3...4
6...4
0
I just want to find and replace specific values, but I can't use the ordinary function because the cells are different lengths. I need to replace many specific values at the same time and there is no general function about how values are replaced. However, sometimes several input values should be replaced by the same output.
so I want to say
for values 1:129
'if 0, then 9'
'elseif 1 then 50'
'elseif 2 or 3 or 4 then 61'
etc...up to 129
where these rules are applied to the entire array.
I've tried to work it out myself, but still getting nowhere. Please help!
Since your values appear to span the range 0 to 129, one solution is to add one to these values (so they span the range 1 to 130) and use them as indices into a vector of replacement values. Then you can apply this operation to each cell using the function CELLFUN. For example:
>> C = {0, 129, [8 2 3 4], [6 4], 0}; %# The sample cell array you give above
>> replacement = [9 50 61 61 61 100.*ones(1,125)]; %# A 1-by-130 array of
%# replacement values (I
%# added 125 dummy values)
>> C = cellfun(#(v) {replacement(v+1)},C); %# Perform the replacement
>> C{:} %# Display the contents of C
ans =
9
ans =
100
ans =
100 61 61 61
ans =
100 61
ans =
9