Amelio Vazquez-Reina Amelio Vazquez-Reina - 1 month ago 8x
Python Question

From ND to 1D arrays

Say I have an array


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

array([[1, 2, 3],
[4, 5, 6]])

I would like to convert it to a 1D array (i.e. a column vector):

b = np.reshape(a, (1,np.product(a.shape)))

but this returns

array([[1, 2, 3, 4, 5, 6]])

which is not the same as:

array([1, 2, 3, 4, 5, 6])

I can take the first element of this array to manually convert it to a 1D array:

b = np.reshape(a, (1,np.product(a.shape)))[0]

but this requires me to know how many dimensions the original array has (and concatenate [0]'s when working with higher dimensions)

Is there a dimensions-independent way of getting a column/row vector from an arbitrary ndarray?


Use np.ravel (for a 1D view) or np.flatten (for a 1D copy) or np.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.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]