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Numpy array reshape customization python

I have an np array n and the shape is (250,250). after that I converted (final array w)it into (250,250,3) because I need to do some operation on it.
is it possible to convert the shape of w to (250,250)?
thanks in advance.
I tried some reshaping operation of Numpy but it does not work!
Numpy reshaping array
Comparing two NumPy arrays for equality, element-wise
Numpy reshaping array

numpy.reshape
Gives a new shape to an array without changing its data.
so this is not right to convert array with shape of (250,250,3) into array with shape of (250,250) as 1st does have 187500 cells and 2nd does have 62500 cells.
You probably should use slicing, consider following example
import numpy as np
arr = np.array([[[0,1],[2,3]],[[4,5],[6,7]]]) # has shape (2,2,2)
arr2 = arr[:,:,0] # get certain cross-section, check what will happend if you use 1 inplace of 0 and 2 inplace of 0
print("arr2")
print(arr2)
print("arr2.shape",arr2.shape)
output
arr2
[[0 2]
[4 6]]
arr2.shape (2, 2)

2D numpy array showing as 1D

I have a numpy ndarray train_data of length 200, where every row is another ndarray of length 10304.
However when I print np.shape(train_data), I get (200, 1), and when I print np.shape(train_data[0]) I get (1, ), and when I print np.shape(train_data[0][0]) I get (10304, ).
I am quite confused with this behavior as I supposed the first np.shape(train_data) should return (200, 10304).
Can someone explains to me why this is happening, and how could I get the array to be in shape of (200, 10304)?

This is because the arrays are constructed to be arrays of objects. Basically each element in the array is pointing to another array of size (1, ) which points to another array of size (10304, ). This is not equivalent to a normal ndarray in numpy so the shape is not recognized correctly. You can check this by looking at the dtypes.
To replicate what you see:
import numpy as np
arr = np.empty(200, dtype='object')
for i in range(200):
temp_arr = np.empty(1, dtype='object')
temp_arr[0] = np.zeros(10304)
arr[i] = temp_arr
print(arr.shape)
print(arr[0].shape)
print(arr[0][0].shape)
(200,)
(1,)
(10304,)
To get the (200, 10304) array back you need to "unpack" them:
new_arr = np.array([x[0] for x in arr])
#(200, 10304)

I'm not sure why that's happening, try reshaping the array:
B = np.reshape(A, (-1, 2))

How To ReShape a Numpy Array in Python

I have a numpy array of images with the shape of (5879,). Inside every index of the numpy array, I have the Pixels of the image with a shape of (640,640,3).
I want to reshape the complete array in such a way that the shape of the numpy array becomes (5879,640,640,3).

please check, whether below code works for you or not
import numpy as np
b = np.array([5879])
b.shape
output (1,)
a = np.array([[640],[640],[3]])
a = a.reshape((a.shape[0], 1))
a.shape
output (3, 1)
c = np.concatenate((a,b[:,None]),axis=0)
c.shape
Output:
(4, 1)
np.concatenate((a,b[:,None]),axis=0)
output
array([[ 640],
[ 640],
[ 3],
[5879]])

You want to stack your images along the first axis, into a 4D array. However, your images are all 3D.
So, first you need to add a leading singleton dimension to all images, and then to concatenate them along this axis:
imgs = [i_[None, ...] for i_ in orig_images] # add singleton dim to all images
x = np.concatenate(imgs, axis=0) # stack along the first axis
Edit:
Based on Mad Phyiscist's comment, it seems like using np.stack is more appropriate here: np.stack takes care of adding the leading singleton dimension for you:
x = np.stack(orig_images, axis=0)

Delete 2d subarray from 3d array in numpy

In numpy I have a 3d array and I would ike to remove some of the 2d subarrays. Think about it like this:
r = range(27)
arr = np.reshape(r, (3,3,3))
del = [[0,1,2],[0,0,2]]
flatSeam = np.ravel_multi_index(del, arr.shape)
arr = np.delete(arr, flatSeam)
So at the end I would like to have an array of the shape (3,2,3) without the elements 00, 10, 22 from the original array. My problem is that I acn not use ravel_multi_index for this, because my indices are 2d and the array shape is 3d, so the wrong indices are calculated (the code above also does not execute because the indices array and the shape have to be the same size).
Do you have any ideas how I can achieve this?

Here's an approach using advanced-indexing -
# arr: Input array, rm_idx : 2-row list/array of indices to be removed
m,n,p = arr.shape
mask = np.asarray(rm_idx[1])[:,None] != np.arange(n)
out = arr[np.arange(m)[:,None],np.where(mask)[1].reshape(m,-1)]
Alternatively, with boolean-indexing -
out = arr.reshape(-1,p)[mask.ravel()].reshape(m,-1,p)
A bit less memory-intensive approach as we try to avoid creating 2D mask -
vmask = ~np.in1d(np.arange(m*n),rm_idx[1] + n*np.arange(m))
out = arr.reshape(-1,p)[vmask].reshape(m,-1,p)

why unable to concatenate two arrays in python?

I have two arrays
>>> array1.shape
(97, 195)
>>> array2.shape
(195,)
>>> array1 = numpy.concatenate((array1, array2), axis=0)
when I perform concatenate operation it shows an error
ValueError: all the input arrays must have same number of dimensions
is that the second array shape (195,) creating problem?

Just make both have the same dimensions and the same size except along the axis to be concatenated:
np.concatenate((array1, array2[np.newaxis,...]), axis=0)

In order for this to work, you need array2 to actually be 2d.
array1 = numpy.concatenate((array1, array2.reshape((1,195)))
should work

Another easy way to achieve the array concatenation that you’re looking for is to use Numpy’s vstack function as follows:
array1 = np.vstack([array1, array2])