Calculating bounding box of numpy array

Question

I'm struggling with a simple question. I have a numpy array of the form:

[[[ 1152.07507324   430.84799194]
  [ 4107.82910156   413.95199585]
  [ 4127.64941406  2872.32006836]
  [ 1191.71643066  2906.11206055]]]

And I want to calculate the bounding box, meaning, I want to have the leftmost, topmost, rightmost and bottommost point.

This should be a correct solution

[[[ 1152.07507324   413.95199585]
  [ 4127.64941406   413.95199585]
  [ 4127.64941406  2906.11206055]
  [ 1152.07507324  2906.11206055]]]

I developed a nasty function that does the trick, but I'm very unsatisfied with it since its not really pythonic/numpyic

def bounding_box(iterable):
    minimum_x = min(iterable[0], key=lambda x:x[0])[0]
    maximum_x = max(iterable[0], key=lambda x:x[0])[0]
    minimum_y = min(iterable[0], key=lambda x:x[1])[1]
    maximum_y = max(iterable[0], key=lambda x:x[1])[1]

    return numpy.array([[(minimum_x, minimum_y), (maximum_x, minimum_y), (maximum_x, maximum_y), (minimum_x, maximum_y)]], dtype=numpy.float32)

Do you have any idea how to optimize the function above, perhaps using numpy builtins?

Why does your sample array have three levels of square brackets? That makes it looks like a 3D array with shape (1,n,2). Is that intentional? — Warren Weckesser
– Warren Weckesser, Commented Sep 17, 2012 at 2:08

nneonneo · Accepted Answer · 2012-09-16 01:29:55Z

11

Use the numpy.min and numpy.max builtins:

def bounding_box(iterable):
    min_x, min_y = numpy.min(iterable[0], axis=0)
    max_x, max_y = numpy.max(iterable[0], axis=0)
    return numpy.array([(min_x, min_y), (max_x, min_y), (max_x, max_y), (min_x, max_y)])

edited Sep 16, 2012 at 1:29

answered Sep 16, 2012 at 1:22

nneonneo

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5 Comments

DSM Over a year ago

Argv. I was working on a shorter trick to get the Cartesian product of the results, but it's the same basic idea.

nneonneo Over a year ago

itertools.product? Only problem is that it gives the answer in the wrong order.

nneonneo Over a year ago

But if you can get around that and still have short code, by all means, post it. I am curious :)

DSM Over a year ago

I was thinking of something like c = np.array([b.min(0), b.max(0)]); list(np.broadcast(*np.ix_(*c.T))) -- but that second part is too clever, and the order's wrong anyway. I'd much rather do it manually like you and the OP did so that it's instantly obvious what the result is.

Pierre GM Over a year ago

Save yourself a tiny bit of hassle and (1) transform iterable into a 2D array iterable=iterable.squeeze(), (2) use the .min/.max methods instead of the np.min/np.max functions. Otherwise, +1

tal · Accepted Answer · 2015-05-31 23:02:13Z

3

returns the same as previous answer but shorter and cleaner

returns 2*2 ndarray

def bbox(points):
    """
    [xmin xmax]
    [ymin ymax]
    """
    a = zeros((2,2))
    a[:,0] = np.min(points, axis=0)
    a[:,1] = np.max(points, axis=0)
    return a

edited May 31, 2015 at 23:02

answered May 31, 2015 at 15:43

tal

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1 Comment

Ryan Bemrose Over a year ago

Consider adding some text to your answer, to explain what you're doing here and how this is different from the already accepted answer. Answers that are just code blobs aren't very useful.

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