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Is there a way to write a python function that will create 'N' arrays? (see body)

I have an numpy array that is shape 20, 3. (So 20 3 by 1 arrays. Correct me if I'm wrong, I am still pretty new to python)
I need to separate it into 3 arrays of shape 20,1 where the first array is 20 elements that are the 0th element of each 3 by 1 array. Second array is also 20 elements that are the 1st element of each 3 by 1 array, etc.
I am not sure if I need to write a function for this. Here is what I have tried:
Essentially I'm trying to create an array of 3 20 by 1 arrays that I can later index to get the separate 20 by 1 arrays.
a = np.load() #loads file
num=20 #the num is if I need to change array size
num_2=3
for j in range(0,num):
for l in range(0,num_2):
array_elements = np.zeros(3)
array_elements[l] = a[j:][l]
This gives the following error:
'''
ValueError: setting an array element with a sequence
'''
I have also tried making it a dictionary and making the dictionary values lists that are appended, but it only gives the first or last value of the 20 that I need.

Your array has shape (20, 3), this means it's a 2-dimensional array with 20 rows and 3 columns in each row.
You can access data in this array by indexing using numbers or ':' to indicate ranges. You want to split this in to 3 arrays of shape (20, 1), so one array per column. To do this you can pick the column with numbers and use ':' to mean 'all of the rows'. So, to access the three different columns: a[:, 0], a[:, 1] and a[:, 2].
You can then assign these to separate variables if you wish e.g. arr = a[:, 0] but this is just a reference to the original data in array a. This means any changes in arr will also be made to the corresponding data in a.
If you want to create a new array so this doesn't happen, you can easily use the .copy() function. Now if you set arr = a[:, 0].copy(), arr is completely separate to a and changes made to one will not affect the other.

Essentially you want to group your arrays by their index. There are plenty of ways of doing this. Since numpy does not have a group by method, you have to horizontally split the arrays into a new array and reshape it.
old_length = 3
new_length = 20
a = np.array(np.hsplit(a, old_length)).reshape(old_length, new_length)
Edit: It appears you can achieve the same effect by rotating the array -90 degrees. You can do this by using rot90 and setting k=-1 or k=3 telling numpy to rotate by 90 k times.
a = np.rot90(a, k=-1)

How to transform 2D array using values as another array's indices?

I have a 2D array with indices refering to another array:
indexarray = np.array([[0,0,1,1],
[1,2,3,0]])
The array which these indices refer to is:
valuearray = np.array([8,7,6,5])
I would like to get an array with the numbers from valuearray in the shape of indexarray, each item in this array corresponds to the value in valuearray with the index on the same location in indexarray, ie:
targetarray = np.array([[8,8,7,7],
[7,6,5,8]])
How can I do this without iteration?
What I do now to achieve this is:
np.apply_along_axis(func1d = lambda row: valuearray[row],axis=0,arr = indexarray)
If there is a simpler way, I am interested.

One way is to flatten the index array and get the values and reshape it back as follows.
targetarray = valuearray[indexarray.flatten()].reshape(indexarray.shape)

Index multidimensional torch tensor by another multidimensional tensor

I have a tensor x in pytorch let's say of shape (5,3,2,6) and another tensor idx of shape (5,3,2,1) which contain indices for every element in first tensor. I want a slicing of the first tensor with the indices of the second tensor. I tried x= x[idx] but I get a weird dimensionality when I really want it to be of shape (5,3,2) or (5,3,2,1).
I'll try to give an easier example:
Let's say
x=torch.Tensor([[10,20,30],
[8,4,43]])
idx = torch.Tensor([[0],
[2]])
I want something like
y = x[idx]
such that 'y' outputs [[10],[43]] or something like.
The indices represent the position of the wanted elements the last dimension. for the example above where x.shape = (2,3) the last dimension are the columns, then the indices in 'idx' is the column. I want this but for more than 2 dimensions

From what I understand from the comments, you need idx to be index in the last dimension and each index in idx corresponds to similar index in x (except for the last dimension). In that case (this is the numpy version, you can convert it to torch):
ind = np.indices(idx.shape)
ind[-1] = idx
x[tuple(ind)]
output:
[[10]
[43]]

You can use range; and squeeze to get proper idx dimension like
x[range(x.size(0)), idx.squeeze()]
tensor([10., 43.])
# or
x[range(x.size(0)), idx.squeeze()].unsqueeze(1)
tensor([[10.],
[43.]])

Here's the one that works in PyTorch using gather. The idx needs to be in torch.int64 format which the following line will ensure (note the lowercase of 't' in tensor).
idx = torch.tensor([[0],
[2]])
torch.gather(x, 1, idx) # 1 is the axis to index here
tensor([[10.],
[43.]])

Set a column in numpy array to zero

I want to set a column in numpy array to zero at different times, in other words, I have numpy array M with size 5000x500. When I enter shape command the result is (5000,500), I think 5000 are rows and 500 are columns
shape(M)
(5000,500)
But the problem when I want to access one column like first column
Mcol=M[:][0]
Then I check by shape again with new matrix Mcol
shape(Mcol)
(500,)
I expected the results will be (5000,) as the first has 5000 rows. Even when changed the operation the result was the same
shape(M)
(5000,500)
Mcol=M[0][:]
shape(Mcol)
(500,)
Any help please in explaining what happens in my code and if the following operation is right to set one column to zero
M[:][0]=0

You're doing this:
M[:][0] = 0
But you should be doing this:
M[:,0] = 0
The first one is wrong because M[:] just gives you the entire array, like M. Then [0] gives you the first row.
Similarly, M[0][:] gives you the first row as well, because again [:] has no effect.

Replace row with another row in 3D numpy array

I am trying to replace a specific row of NaN's in a 3-D array (filled with NaN's) with rows of known integer values from a specific column in a text file (ex: 24 rows of column 8). Is there a method to perform this replacement that I have missed in my search for help?
My most recent trial code (of many) is as follows:
import numpy as np
tfile = "C:\...\Lee_Gilmer_MEM_GA_01_02_2015.txt"
data = np.genfromtxt(tfile, dtype=None)
#creation of empty 24 hour global matrix
s_array = np.empty((24,361,720))
s_array[:] = np.NAN
#Get values from column 8
c_data = data[:,7]
#Replace all 24 NaN's slices of row 1 column 1 with corresponding 24 row values from column 8
s_array[:,0:1,0:1] = c_data
print s_array
This produces a result of:
ValueError: could not broadcast input array from shape (24) into shape (24,1,1)
When I print out the shape of c_data, I get:
(24L,)
Is this at all possible to do without having to use a loop and replacing each one individually?

The error message tells you pretty much everything you need to know: the array slice on the left-hand side of the assignment has a shape of (24,1,1), whereas the right-hand side has shape (24,). Since these shapes don't match, numpy raises a ValueError.
There are two ways to solve this:
Make the shape of the LHS (24,) rather than (24, 1, 1). A nice way to do this would be to index with an integer rather than a slice for the last two dimensions:
s_array[:, 0, 0] = c_data
Reshape c_data to match the shape of the LHS:
s_array[:, 0:1, 0:1] = c_data.reshape(24, 1, 1)
I think option 1 is a lot more readable.