From ND to 1D arrays

Use np.ravel (for a 1D view) or np.ndarray.flatten (for a 1D copy) or np.ndarray.flat (for an 1D iterator):

In [12]: a = np.array([[1,2,3], [4,5,6]])

In [13]: b = a.ravel()

In [14]: b
Out[14]: array([1, 2, 3, 4, 5, 6])

Note that ravel() returns a view of a when possible. So modifying b also modifies a. ravel() returns a view when the 1D elements are contiguous in memory, but would return a copy if, for example, a were made from slicing another array using a non-unit step size (e.g. a = x[::2]).

If you want a copy rather than a view, use

In [15]: c = a.flatten()

If you just want an iterator, use np.ndarray.flat:

In [20]: d = a.flat

In [21]: d
Out[21]: <numpy.flatiter object at 0x8ec2068>

In [22]: list(d)
Out[22]: [1, 2, 3, 4, 5, 6]

In [14]: b = np.reshape(a, (np.product(a.shape),))

In [15]: b
Out[15]: array([1, 2, 3, 4, 5, 6])

or, simply:

In [16]: a.flatten()
Out[16]: array([1, 2, 3, 4, 5, 6])

For list of array with different size use following:

import numpy as np

# ND array list with different size
a = [[1],[2,3,4,5],[6,7,8]]

# stack them
b = np.hstack(a)

print(b)

Output:

[1 2 3 4 5 6 7 8]

One of the simplest way is to use flatten(), like this example :

 import numpy as np

 batch_y =train_output.iloc[sample, :]
 batch_y = np.array(batch_y).flatten()

My array it was like this :

    0
0   6
1   6
2   5
3   4
4   3
.
.
.

After using flatten():

array([6, 6, 5, ..., 5, 3, 6])

It's also the solution of errors of this type :

Cannot feed value of shape (100, 1) for Tensor 'input/Y:0', which has shape '(?,)'