reduce one dimension within an ndarray

Question

I have points coordinates stored in a 3-dimensional array:
(UPD. the array is actually numpy -derived ndarray , sorry for the confusion in the initial version)

a = [ [[11,12]], [[21,22]], [[31,32]], [[41,42]] ]

you see that each coordinate pair is stored as nested 2-d array like [[11,12]] , while I would like it to be [11,12] , ie my array should have this content:

b = [ [11,12], [21,22], [31,32], [41,42] ]

So, how to get from a to b form? For now my solution is to create a list and then convert it to an array with numpy :

b = numpy.array([p[0] for p in a])

This works but I assume there must be a simpler and cleaner way...

UPD. originally I tried to do a simple comprehension: b = [p[0] for p in a] - but then b turned out to be a list, not an array - I assume that's because the original a array is ndarray from numpy

Answer 1

If you do want to use numpy:

b = np.array(a)[:, 0, :]

This will be faster than a comprehension.

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Well... I certainly thought it would be

a = np.random.random((100_000, 1, 2)).tolist()

%timeit np.array([x[0] for x in a])
41.1 ms ± 304 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit np.array(a)[:, 0, :]
57.6 ms ± 1.27 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit x = np.array(a); x.shape = len(a), 2
58.2 ms ± 381 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

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edit

Oh if its a numpy array then definitely use this method. Or use .squeeze() if you're sure it's not empty.

Answer 2

这是使用列表理解的另一种解决方案：

b = [x[0] for x in a]

Answer 3

If you are going to use numpy later, then it's best to avoid the list comprehension. Also it's always good practice to automate things as much as possible, so instead of manually selecting the singleton dimension just let numpy take care of: b=numpy.array(a).squeeze() Unless there are other singleton dimensions that you need to keep.

Answer 4

In order to flatten a "nested 2-d array like" as you call them, you just need to get the first element. arr[0]

Apply this concept in several ways:

• list comprehension (most performing) : flatter_a_compr = [e[0] for e in a]

• iterating (second best performing):

   b =[] for e in a: b.append(e[0]) 

• lambda (un-Pythonic): flatter_a = list(map(lambda e : e[0], a))

• numpy (worst performing) : flatter_a_numpy = np.array(a)[:, 0, :]