Dictionary of points and values to array

Question

I have a dictionary of coordinate pairs and integers that go with those pairs. The first problem is that some of the points are negative.

{(0, 0): 1, (-1, -1): 2, (1, -1): 3, (1, 1): 4}

to 

+---+---+---+
| 0 | 0 | 4 |
+---+---+---+
| 0 | 1 | 0 |
+---+---+---+
| 2 | 0 | 3 |
+---+---+---+

I believe I have to adjust all the pairs so they are greater than zero so I can use them as row, column pairs for a matrix. Then the insertion is the next. I believe that python only has nested lists not matrices so I would want a numpy array.

Your example has odd bounds (3 x 3) so the center is unambiguous. How do you define the center of a matrix with even bounds? Suppose you have a matrix 4 x 6; what is the center for relative coordinates to that? — dawg
– dawg, Commented Apr 26, 2014 at 16:10
It doesn't necessarily have to be centred. It's just like moving points from the Cartesian plane so that it is all in quadrant 1. — Ben Longo
– Ben Longo, Commented Apr 26, 2014 at 19:33

Ashwini Chaudhary · Accepted Answer · 2014-04-26 16:28:27Z

1

Using pure Python:

def solve(d):

    x_min, y_min = map(min, zip(*d))
    x_max, y_max = map(max, zip(*d)) 

    arr = [[0]*(x_max-x_min+1) for _ in xrange(y_max-y_min+1)]

    for i, y in enumerate(xrange(y_min, y_max+1)):
        for j, x in enumerate(xrange(x_min, x_max+1)):
            arr[i][j] = d.get((x, y), 0)
    return arr[::-1]

Output:

solve({(0, 0): 1, (-1, -1): 2, (1, -1): 3, (1, 1): 4})
Out[80]: 
[[0, 0, 4],
[0, 1, 0],
[2, 0, 3]]

solve({(0, 0): 1, (-1, -1): 2, (1, -1): 3, (1, 1): 4, (2, 2):30, (-3, -4):100})
Out[82]: 
[[0, 0, 0, 0, 0, 30],
 [0, 0, 0, 0, 4, 0],
 [0, 0, 0, 1, 0, 0],
 [0, 0, 2, 0, 3, 0],
 [0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0],
 [100, 0, 0, 0, 0, 0]]

edited Apr 26, 2014 at 16:28

answered Apr 26, 2014 at 16:16

Ashwini Chaudhary

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1 Comment

Ashwini Chaudhary Over a year ago

@BenLongo A built-in function in Python 2. Use range() just in case you're on Python 3.

mblw · Accepted Answer · 2014-04-26 16:35:05Z

import numpy as np


def make_array(data):
    # In your example row is the second index and col is the first.
    # Also positive row indexes go in up direction.
    c, r = np.array(zip(*data.keys()))

    rows = r.max()-r.min()+1
    cols = c.max()-c.min()+1

    result = np.zeros((rows, cols), dtype=int)

    for k, v in data.iteritems():
        # Minus before first index required for 
        # the last row contain 2, 0, 3 in the example.
        # Also numpy successfully handle negative indexes
        # and inversion not required
        result[-k[1]+1, k[0]+1] = v

    return result

Your test case:

data = {(0, 0): 1, (-1, -1): 2, (1, -1): 3, (1, 1): 4}
print make_array(data)

Result:

[[0 0 4]
 [0 1 0]
 [2 0 3]]

Example with different rows and columns count:

data = {(0, 0): 1, (-1, -1): 2, (1, -1): 3, (1, 1): 4, (2, 1): 5}
print make_array(data)

Result:

   ----------- "-First" column
  |      ----- Second column
  |     |
[[0 0 4 5]     <-- First row
 [0 1 0 0]     <-- Zero row
 [2 0 3 0]]    <-- "-First" row

Daniel · Accepted Answer · 2014-04-26 16:11:05Z

0

import numpy
s = {(0, 0): 1, (-1, -1): 2, (1, -1): 3, (1, 1): 4}
x = numpy.array([k+(v,) for k,v in s.iteritems()])
x[:,0]-=x[:,0].min()
x[:,1]-=x[:,1].min()
w = numpy.zeros((x[:,0].max()+1,x[:,1].max()+1))
w[x[:,:2].T.tolist()]=x[:,2]

the resut:

>>> w
array([[ 2.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 3.,  0.,  4.]])

answered Apr 26, 2014 at 16:11

Daniel

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