We Keep Coding

Append indices of element to each element

So basically I want to create a new array for each element and append the coordinates of the element to the original value (so adding the x and y position to the original element):
[ [7,2,4],[1,5,3] ]
then becomes
[ [[0,0,7][0,1,2][0,2,4]],
[[1,0,1][1,1,5][1,2,3]] ]
I've been looking for different ways to make this work with the axis system in NumPy but I'm probably overseeing some more obvious way.

You can try np.meshgrid to create a grid and then np.stack to combine it with input array:
import numpy as np
a = np.asarray([[7,2,4],[1,5,3]])
result = np.stack(np.meshgrid(range(a.shape[1]), range(a.shape[0]))[::-1] + [a], axis=-1)
Output:
array([[[0, 0, 7],
[0, 1, 2],
[0, 2, 4]],
[[1, 0, 1],
[1, 1, 5],
[1, 2, 3]]])
Let me know if it helps.

Without numpy you could use list comprehension:
old_list = [ [7,2,4],[1,5,3] ]
new_list = [ [[i,j,old_list[i][j]] for j in range(len(old_list[i]))] for i in range(old_list) ]
I'd assume that numpy is faster but the sublists are not required to have equal length in this solution.

Another approach using enumerate
In [38]: merge = list()
...: for i,j in enumerate(val):
...: merge.append([[i, m, n] for m, n in enumerate(j)])
...:
In [39]: merge
Out[39]: [[[0, 0, 7], [0, 1, 2], [0, 2, 4]], [[1, 0, 1], [1, 1, 5], [1, 2, 3]]]

Hope it useful
a = np.array([[7,2,4], [1,5,3]])
idx = np.argwhere(a)
idx = idx.reshape((*(a.shape), -1))
a = np.expand_dims(a, axis=-1)
a = np.concatenate((idx, a), axis=-1)

How to iterate over a numpy array at a given axis?

I have an ndarray of shape, say
my_array = np.zeros((Na, Nb, Nc))
and I have Nb bunches of data of shape (Na, Nc)
I would like to fill "my_array" with my bunches of data as follows:
for b, bit in zip(range(Nb), my_bits):
my_array[:, b, :] = bit
but I would like to do this for any arbitray axis for an arbitrary shape data set. How would I do this?
** EDIT **
Someone pointed out an ambiguity in my question. Below is an example code of what I know how to do, leading to what I don't know.
What I know:
my_bits = [np.ones((Na, Nc)) for j in range(Nb)]
my_array = np.zeros((Na, Nb, Nc))
for b, bit in zip(range(Nb), my_bits):
my_array[:, b, :] = bit
Generically we could consider
my_array = np.zeros((N1, N2, ..., Nj, ..., Nmax))
bit = np.ones((N1, N2, ..., Nj-1, Nj+1, ..., Nmax))
my_bits = [bit]*Nj
From there I would like to fill my_array with my_bits along axis Nj. How would I do this? I am looking at the nditer function, but I am not sure what to do with it.

There is a problem with dimensions here.
If I understood it correctly, my_array[:, 0, :] is same dimension as my_bits. So the assignment should be done like this:
import numpy as np
Na = 2
Nb = 2
Nc = 2
my_array = np.zeros((Na, Nb, Nc))
my_bits = np.ones((Na, Nc))
print('my_array before', my_array)
for b in range(Nb):
my_array[:, b, :] = my_bits
print('my_array after', my_array)

I think you might want numpy.transpose
list_of_2d_arrays = [ make_2d_array() for i in range(Nb) ] # each has shape=(Na,Nc)
tensor3d = np.array(list_of_2d_arrays) # shape = (Nb,Na,Nc)
# now shift the dimensions around
X = np.transpose(tensor3d,(1,0,2)) # has shape (Na,Nb,Nc)

Using transpose to move the iteration axis to a known position, front or last, is done in some numpy functions.
Another approach is to construct a slice object, and iterate that.
In [46]: arr = np.zeros((2,3,4),int)
In [47]: idx = np.full((arr.ndim,), slice(None))
In [48]: n=2
In [49]: for i in range(arr.shape[n]):
...: idx[n] = i
...: arr[tuple(idx)] = i+1
...:
In [50]: arr
Out[50]:
array([[[1, 2, 3, 4],
[1, 2, 3, 4],
[1, 2, 3, 4]],
[[1, 2, 3, 4],
[1, 2, 3, 4],
[1, 2, 3, 4]]])
for another axis
In [56]: arr = np.zeros((2,3,4),int)
In [57]: idx = np.full((arr.ndim,), slice(None))
In [58]: n=1
In [59]: for i in range(arr.shape[n]):
...: idx[n] = i
...: arr[tuple(idx)] = i+1
...:
In [60]: arr
Out[60]:
array([[[1, 1, 1, 1],
[2, 2, 2, 2],
[3, 3, 3, 3]],
[[1, 1, 1, 1],
[2, 2, 2, 2],
[3, 3, 3, 3]]])
In this case idx varies as:
[slice(None, None, None) 0 slice(None, None, None)]
[slice(None, None, None) 1 slice(None, None, None)]
[slice(None, None, None) 2 slice(None, None, None)]
The key is that an indexing expression has a tuple equivalent, which can be constructed programatically:
In [61]: np.s_[:,3,:]
Out[61]: (slice(None, None, None), 3, slice(None, None, None))

Put numpy arrays split with np.split() back together

I have split a numpy array like so:
x = np.random.randn(10,3)
x_split = np.split(x,5)
which splits x equally into five numpy arrays each with shape (2,3) and puts them in a list. What is the best way to combine a subset of these back together (e.g. x_split[:k] and x_split[k+1:]) so that the resulting shape is similar to the original x i.e. (something,3)?
I found that for k > 0 this is possible with you do:
np.vstack((np.vstack(x_split[:k]),np.vstack(x_split[k+1:])))
but this does not work when k = 0 as x_split[:0] = [] so there must be a better and cleaner way. The error message I get when k = 0 is:
ValueError: need at least one array to concatenate

The comment by Paul Panzer is right on target, but since NumPy now gently discourages vstack, here is the concatenate version:
x = np.random.randn(10, 3)
x_split = np.split(x, 5, axis=0)
k = 0
np.concatenate(x_split[:k] + x_split[k+1:], axis=0)
Note the explicit axis argument passed both times (it has to be the same); this makes it easy to adapt the code to work for other axes if needed. E.g.,
x_split = np.split(x, 3, axis=1)
k = 0
np.concatenate(x_split[:k] + x_split[k+1:], axis=1)

np.r_ can turn several slices into a list of indices.
In [20]: np.r_[0:3, 4:5]
Out[20]: array([0, 1, 2, 4])
In [21]: np.vstack([xsp[i] for i in _])
Out[21]:
array([[9, 7, 5],
[6, 4, 3],
[9, 8, 0],
[1, 2, 2],
[3, 3, 0],
[8, 1, 4],
[2, 2, 5],
[4, 4, 5]])
In [22]: np.r_[0:0, 1:5]
Out[22]: array([1, 2, 3, 4])
In [23]: np.vstack([xsp[i] for i in _])
Out[23]:
array([[9, 8, 0],
[1, 2, 2],
[3, 3, 0],
[8, 1, 4],
[3, 2, 0],
[0, 3, 8],
[2, 2, 5],
[4, 4, 5]])
Internally np.r_ has a lot of ifs and loops to handle the slices and their boundaries, but it hides it all from us.
If the xsp (your x_split) was an array, we could do xsp[np.r_[...]], but since it is a list we have to iterate. Well we could also hide that iteration with an operator.itemgetter object.
In [26]: operator.itemgetter(*Out[22])
Out[26]: operator.itemgetter(1, 2, 3, 4)
In [27]: np.vstack(operator.itemgetter(*Out[22])(xsp))

I have a numpy array, and an array of indexs, how can I access to these positions at the same time

for example, I have the numpy arrays like this
a =
array([[1, 2, 3],
[4, 3, 2]])
and index like this to select the max values
max_idx =
array([[0, 2],
[1, 0]])
how can I access there positions at the same time, to modify them.
like "a[max_idx] = 0" getting the following
array([[1, 2, 0],
[0, 3, 2]])

Simply use subscripted-indexing -
a[max_idx[:,0],max_idx[:,1]] = 0
If you are working with higher dimensional arrays and don't want to type out slices of max_idx for each axis, you can use linear-indexing to assign zeros, like so -
a.ravel()[np.ravel_multi_index(max_idx.T,a.shape)] = 0
Sample run -
In [28]: a
Out[28]:
array([[1, 2, 3],
[4, 3, 2]])
In [29]: max_idx
Out[29]:
array([[0, 2],
[1, 0]])
In [30]: a[max_idx[:,0],max_idx[:,1]] = 0
In [31]: a
Out[31]:
array([[1, 2, 0],
[0, 3, 2]])

Numpy support advanced slicing like this:
a[b[:, 0], b[:, 1]] = 0
Code above would fit your requirement.
If b is more than 2-D. A better way should be like this:
a[np.split(b, 2, axis=1)]
The np.split will split ndarray into columns.

How to append last row from 2D array in Python

How can I append the last row of an array to itself ?
something like:
x= np.array([(1,2,3,4,5)])
x= np.append(x, x[0], 1)
Also, Could you explain why this way of working with vectors yields an error?
for i in range(3):
x.append(0)
x
[0, 0, 0]
x= np.append(x, x[0],0)
Which way of working with vectors would be best ? I am trying to work with 2D vectors as being a matrix, keeping in mind i would like to do some future matrix calculations like multiplication etc.

In [3]: x=np.array([(1,2,3,4,5)])
In [4]: x
Out[4]: array([[1, 2, 3, 4, 5]])
In [5]: x=np.append(x,x[0],1)
...
ValueError: all the input arrays must have same number of dimensions
x is (1,5), x[0] is (5,) - one is 2d, the other 1d.
In [11]: x=np.vstack([x,x[0]])
In [12]: x
Out[12]:
array([[1, 2, 3, 4, 5],
[1, 2, 3, 4, 5]])
this works because vstack changes the x[0] to 2d, e.g. (1,5), so it can concatenate it with x.
In [16]: x=np.concatenate([x, np.atleast_2d(x[-1,:])])
In [17]: x.shape
Out[17]: (3, 5)
We can use concatenate (or append) by first expanding x[-1,:] to 2d.
But in general repeated concatenation is a slow way of building an array.
For a list, repeated append like this works. But it does not work for arrays. For one thing, an array does not have an append method. And np.append function returns a new array. It does not change x in place.
In [19]: z=[]
In [20]: for i in range(3):
...: z.append(0)
...:
In [21]: z
Out[21]: [0, 0, 0]
Repeated append to a list is fine. Repeated append to an array is slow.
In [25]: z=[]
In [26]: for i in range(3):
...: z.append(list(range(i,i+4)))
In [27]: z
Out[27]: [[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5]]
In [28]: np.array(z)
Out[28]:
array([[0, 1, 2, 3],
[1, 2, 3, 4],
[2, 3, 4, 5]])

>>> np.append(x,x[-1:],0)
array([[1, 2, 3, 4, 5],
[1, 2, 3, 4, 5]])

How about this:
np.append(arr=x, values=x[-1,None], axis=0)
#array([[1, 2, 3, 4, 5],
# [1, 2, 3, 4, 5]])