Reputation: 6707

Numpy matrix to array

I am using numpy. I have a matrix with 1 column and N rows and I want to get an array from with N elements.

For example, if i have M = matrix([[1], [2], [3], [4]]), I want to get A = array([1,2,3,4]).

To achieve it, I use A = np.array(M.T)[0]. Does anyone know a more elegant way to get the same result?

Thanks!

Upvotes: 201

Answers (10)

hpaulj

Reputation: 231665

result = M.A1

https://numpy.org/doc/stable/reference/generated/numpy.matrix.A1.html

matrix.A1
1-d base array

Upvotes: 153

Cephas Svosve

Reputation: 1

Came in a little late, hope this helps someone,

np.array(M.flat)

Upvotes: -1

Siddharth Satpathy

Reputation: 3043

ravel() and flatten() functions from numpy are two techniques that I would try here. I will like to add to the posts made by Joe, Siraj, bubble and Kevad.

Ravel:

A = M.ravel()
print A, A.shape
>>> [1 2 3 4] (4,)

Flatten:

M = np.array([[1], [2], [3], [4]])
A = M.flatten()
print A, A.shape
>>> [1 2 3 4] (4,)

numpy.ravel() is faster, since it is a library level function which does not make any copy of the array. However, any change in array A will carry itself over to the original array M if you are using numpy.ravel().

numpy.flatten() is slower than numpy.ravel(). But if you are using numpy.flatten() to create A, then changes in A will not get carried over to the original array M.

numpy.squeeze() and M.reshape(-1) are slower than numpy.flatten() and numpy.ravel().

%timeit M.ravel()
>>> 1000000 loops, best of 3: 309 ns per loop

%timeit M.flatten()
>>> 1000000 loops, best of 3: 650 ns per loop

%timeit M.reshape(-1)
>>> 1000000 loops, best of 3: 755 ns per loop

%timeit np.squeeze(M)
>>> 1000000 loops, best of 3: 886 ns per loop

Upvotes: 2