I need to populate a vector with elements of another, smaller vector. So say the vector I need to populate is of length ten and is currently all zeros, i.e.
vector = [0,0,0,0,0,0,0,0,0,0]
Now suppose I have already define a vector
p = [1, 2, 3, 4, 5]
How could I populate "vector" with the array "p" so that the result is [1, 2, 3, 4, 5, 0, 0, 0, 0, 0]? Bear in mind, I want the other positions in "vector" to remain unchanged. I have already tried using repmat(p, length(p)) but that ends up giving me something of the form [1,2,3,4,5,1,2,3,4,5]. Thanks!
Try a combination of vector slicing and concatenation:
vector = cat(1, p, vector(5:))
This is faster:
vector(1:5) = p
More generally,
vector(1:numel(p)) = p
Related
Based on a 1D binary mask, for example, np.array([0,0,0,1,1,1,0,0,1,1,0]), I would like to extract the columns of another array, indicated by the 1's in the binary mask, as as sub-arrays/separate blocks, like [9, 3.5, 7]) and [2.8, 9.1] (I am just making up the numbers to illustrate the point).
So far what I have (again just as a demo to illustrate what my goal is, not the data where this operation will be performed):
arr = torch.from_numpy(np.array([0,0,0,1,1,1,0,0,1,1,0]))
split_idx = torch.where(torch.diff(arr) == 1)[0]+1
torch.tensor_split(arr, split_idx.tolist())
The output is:
(tensor([0, 0, 0]),
tensor([1, 1, 1]),
tensor([0, 0]),
tensor([1, 1]),
tensor([0]))
What I would like to have in the end is:
(tensor([1, 1, 1]),
tensor([1, 1]))
Do you know how to implement it, preferably in pytorch, but numpy functions are also fine. A million thanks in advance!!
You can construct your tensor of slice indices with your approach. Only thing is you were missing the indices for the position of the end of each slice. You can do something like:
>>> slices = arr.diff().abs().nonzero().flatten()+1
tensor([ 3, 6, 8, 10])
Then apply tensor_split and slice to only keep every other element:
>>> torch.tensor_split(arr, slices)[1::2]
(tensor([1, 1, 1]), tensor([1, 1]))
Given array=[1, 2, 3, 4, 5, 6]
I want to choose the 0-th 2-nd, 4-th index value to build a new array
array1=[1, 3, 5]
Could someone show me how to do using python? Thanks~
If it is just 0, 2, and 4, you can use operator.itemgetter():
from operator import itemgetter
array1 = itemgetter(0, 2, 4)(array)
That will be a tuple. If it must be a list, convert it:
array1 = list(itemgetter(0, 2, 4)(array))
If the point is to get the even numbered indices, use slicing:
array1 = array[::2]
Whichever you are looking for, you could use a list comprehension:
array1 = [array[i] for i in (0, 2, 4)]
or
array1 = [array[i] for i in xrange(0, len(array), 2)]
You can try something like this. In python the nth term of a list has index (n-1). Suppose the first element you want is 2, which happens to be the element 1 of array. Just save the first element index in a variable. Append it to the new list array1 and increase the index by 2. Continue doing this until the list array is exhausted.
from numpy import*
array=[1,2,3,4,5,6]
array1=[]
term=1
while term<len(array): # if the array length is 6 then it will have upto element 5.
array1.append(array[term])
term=term+2 # 2 is the gap between elements. You can replace it with your required step size.
Suppose I have an empty m-by-n-by-p dimensional cell called "cellPoints", and I also have a D-by-3 dimensional array called "cellIdx" where each row i contains the subscripts in "cellPoints". Now I want to compute "cellPoints" so that cellPoints{x, y, z} contains an array of row numbers in "cellIdx".
A naive implementation could be
for i = 1:size(cellIdx, 1)
cellPoints{cellIdx(i, 1), cellIdx(i, 2), cellIdx(i, 3)} = ...
[cellPoints{cellIdx(i, 1), cellIdx(i, 2), cellIdx(i, 3)};i];
end
As an example, suppose
cellPoints = cell(10, 10, 10);% user defined, cannot change
cellIdx = [1, 3, 2;
3, 2, 1;
1, 3, 2;
1, 4, 2]
Then
cellPoints{1, 3, 2} = [1;3];
cellPoints{3, 2, 1} = [2];
cellPoints{1, 4, 2} = [4];
and other indices of cellPoints should be empty
Since cellIdx is a large matrix and this is clearly inefficient, are there any other better implementations?
I've tried using unique(cellIdx, 'rows') to find unique rows in cellIdx, and then writing a for-loop to compute cellPoints, but it's even slower than above.
See if this is faster:
cellPoints = cell(10,10,10); %// initiallize to proper size
[~, jj, kk] = unique(cellIdx, 'rows', 'stable')
sz = size(cellPoints);
sz = [1 sz(1:end-1)];
csz = cumprod(sz).'; %'// will be used to build linear index
ind = 1+(cellIdx(jj,:)-1)*csz; %// linear index to fill cellPoints
cellPoints(ind) = accumarray(kk, 1:numel(kk), [], #(x) {sort(x)});
Or remove sort from the last line if order within each cell is not important.
I have a set of around 500 (x,y,z) real values. Since I will need to bin the values based on their (x,y) coordinates, I stripped the z values and stored in on a seperate list. I am left with only the x,y values; I rescaled and rounded them to index pairs in the range of, 1..100 range.
Now I want to populate an array with the z values in a 100x100 matrix at the particular (x,y) coordinates.
More precisely,
I have a set of values for example : data = {{2.62399, 0.338057, 2.09629}, {1.8424, 0.135817, 3.21925}, {0.702257, 1.14502, 3.9335}...
I stripped it of its zvalues and store it in zvalues list:
zvalues = {2.09629, 3.21925, 3.9335....
I rounded, rescaled and created a new array of indices
indices = {{53, 7}, {37, 3}, {14, 23}...
I want to create a new 100x100 matrix and place the zvalues on the coordinates corresponding to the indices matrix
For example, in pseudocode
For (int i = 1, i < 101, i++){
NewArray(indices[i]) = zvalues[i];
}
The first time the loop will run, it should do NewArray(53,7) = 2.09629.
I want to know the syntax to loop through the indices array and populate the 2 dimensional 100x100 NewArray with zvalues
to follow your basic approach you need to initialize the array:
newArray=Table[,{100},{100}]
then in the loop the syntax is:
newArray[[indices[[i,1]],indices[[i,2]]]]=zdata[[i]]
note the double square brackets for referencing parts of arrays (or lists in Mathematica terminology)
A better approach would be to create a SparseArray, which for one thing would not require pre-initialization, or even knowing the dimensions in advance.
Finally in mathematica you can usually use an object oriented approach, avioding the "do" loop all together:
data = {{1.5, 1.1, 1.1}, {2.2, 2.2, 2.2}, {1.01, 2.3, 1.2}};
m1 = Table[, {2}, {2}];
(m1[[Floor[#[[1]]], Floor[#[[2]]]]] = #[[3]]) & /# data;
m1
m2 = SparseArray[ Floor[#[[1 ;; 2]]] -> #[[3]] & /# data , Automatic,];
Normal[m2]
{{1.1, 1.2}, {Null, 2.2}}
{{1.1, 1.2}, {Null, 2.2}}
While I don't understand why you want to create a new way of indexing your array, this will do what you want :
data = {{2.62399, 0.338057, 2.09629}, {1.8424, 0.135817, 3.21925}, {0.702257, 1.14502, 3.9335}};
zvalues = {2.09629, 3.21925, 3.9335};
indices = {{53, 7}, {37, 3}, {14, 23}};
newArray[xIndex_, yIndex_]:=Take[data, Position[indices, {xIndex, yIndex}][[1, 1]]][[1, 3]]
newArray[53, 7]
(* 2.09629 *)
Hey, having a wee bit of trouble. Trying to assign a variable length 1d array to different values of an array, e.g.
a(1) = [1, 0.13,0.52,0.3];
a(2) = [1, 0, .268];
However, I get the error:
??? In an assignment A(I) = B, the number of elements in B and
I must be the same.
Error in ==> lab2 at 15
a(1) = [1, 0.13,0.52,0.3];
I presume this means that it's expecting a scalar value instead of an array. Does anybody know how to assign the array to this value?
I'd rather not define it directly as a 2d array as it is for are doing solutions to different problems in a loop
Edit: Got it!
a(1,1:4) = [1, 0.13,0.52,0.3];
a(2,1:3) = [1, 0, .268];
What you probably wanted to write was
a(1,:) = [1, 0.13,0.52,0.3];
a(2,:) = [1, 0, .268];
i.e the the first row is [1, 0.13,0.52,0.3] and the second row is [1, 0, .268]. This is not possible, because what would be the value of a(2,4) ?
There are two ways to fix the problem.
(1) Use cell arrays
a{1} = [1, 0.13,0.52,0.3];
a{2} = [1, 0, .268];
(2) If you know the maximum possible number of columns your solutions will have, you can preallocate your array, and write in the results like so (if you don't preallocate, you'll
get zero-padding. You also risk slowing down your loop a lot, if there are many iterations, because the array will have to be recreated at every iteration.
a = NaN(nIterations,maxNumCols); %# this fills the array with not-a-numbers
tmp = [1, 0.13,0.52,0.3];
a(1,1:length(tmp)) = tmp;
tmp = [1, 0, .268];
a(2,1:length(tmp)) = tmp;