Creating an X by Y dimension array of (a,b) points in Python 2.7

Question

I've been oddly bashing my head against this problem for several hours, and would appreciate any help!

I would like to create a (for example) 100x100 array in which each index is a (x,y) coordinate. The overall goal here is the following:

I have x,y coordinates and would like to arrange them in a 2D space so that I can use the np.diagonal function to return the (x,y) coordinates along a line. I'll then use those (x,y) points to compare particular values.

The first step here is actually creating the array and I just can't seem to do it.

Could you provide some example of the code you are trying to run? i.e. arr[(x,y)], arr[x][y], arr[x,y], ... — Farmer Joe
– Farmer Joe, Commented Nov 3, 2014 at 18:32
You need to use a 3D array of shape (100, 100, 2), e.g. a = np.empty((100, 100, 2)). Calling np.diagonal(a) will return an array of shape (2, 100). — Jaime
– Jaime, Commented Nov 3, 2014 at 20:04
Hmmm, I think I can use Jamie's suggestion and follow it up by using zip(np.diagonal(a)[0], np.diagonal(a)[1]) to get a single list of (x,y) coordinates. All I need to do is to make sure the X and Y values are input correctly into the 3D array. What I don't understand is why I need 100 boxes of 100x2 in order to do this. — Ben
– Ben, Commented Nov 4, 2014 at 19:36

Mike · Accepted Answer · 2014-11-03 18:54:49Z

1

I'm not sure about the numpy part of your request, but you can create the array like so:

coords = [[(y,x) for x in range(100)] for y in range(100)]
>>> coords[50][2]
(50,2)

edited Nov 3, 2014 at 18:54

answered Nov 3, 2014 at 18:42

Mike

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ssm · Accepted Answer · 2014-11-04 02:38:03Z

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If you just want the values along the diagonal, why dont you just create a 1D list?

import numpy as np
xs = np.linspace(1,10,100) # assuming x goes form 1 to 10
ys = np.linspace(2,3, 100) # assuming y goes from 2 to 3
xy = zip(xs, ys)

You no longer need the 2d array and then call the diagonal.

answered Nov 4, 2014 at 2:38

ssm

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1 Comment

Ben Over a year ago

So the issue here is that calling the diagonal returns a single value, not every value on the diagonal which is what I need

Chichi Lalescu · Accepted Answer · 2014-11-06 11:59:08Z

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Working on Jaime's suggestion:

>>> x, y = numpy.mgrid[0:100, 0:100]
>>> z = numpy.array([x, y]).transpose([1,2,0])
>>> z[50, 2]
array([50,  2])

EDIT: Given an array of points p, of the shape (2, P), this is how you would find out which of these points are underneath diagonal n:

>>> d = numpy.diagonal(z, n)
>>> cond0 = p[0, None] < d[0, :, None]
>>> cond1 = p[1, None] < d[1, :, None]
>>> good_indices_full = numpy.where(numpy.logical_and(cond0, cond1))
>>> good_indices = good_indices_full[1]

(I prefer to work with "good_indices", i.e. write stuff like p[:, good_indices], rather than the full tuple of arrays that numpy.where gives back).

edited Nov 6, 2014 at 11:59

answered Nov 4, 2014 at 2:14

Chichi Lalescu

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2 Comments

Ben Over a year ago

So, this returns a single value ([50,2]) which is a problem since I will later be evaluating whether a particular point is above or below the appropriate point on the diagonal. To do this, I need every value that falls on that diagonal and not just the first value.

Chichi Lalescu Over a year ago

for me, numpy.diagonal(z) returns an array of shape (2, 100), that you should be able to use in your comparison.

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