块矩阵到数组

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!

If you’d like something a bit more readable, you can do this:

A = np.squeeze(np.asarray(M))

Equivalently, you could also do: A = np.asarray(M).reshape(-1), but that’s a bit less easy to read.

result = M.A1

https://docs.scipy.org/doc/numpy-1.14.0/reference/generated/numpy.matrix.A1.html

matrix.A1
1-d base array

A, = np.array(M.T)

depends what you mean by elegance i suppose but thats what i would do

You can try the following variant:

result=np.array(M).flatten()

np.array(M).ravel()

If you care for speed; But if you care for memory:

np.asarray(M).ravel()

Or you could try to avoid some temps with

A = M.view(np.ndarray)
A.shape = -1

First, Mv = numpy.asarray(M.T), which gives you a 4×1 but 2D array.

Then, perform A = Mv[0,:], which gives you what you want. You could put them together, as numpy.asarray(M.T)[0,:].

This will convert the matrix into array

A = np.ravel(M).T

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().

. 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