Create matrix from list of values within Dictionary

Question

I wish to turn the following dictionary into a matrix where the first and second values of the dictionary are the column and row values. Where the matrix is true I want there to be a '1' and when it is false I want a '0'.

{0: [2, 5.0], 1: [6, 7.0], 2: [6, 8.0], 3: [5, 6.0], 4: [1, 5.0], 5: [3, 4.0], 6: [4, 5.0]}

The desired output would look something like this

        1   2   3   4   5   6   7   8

1       0   0   0   0   1   0   0   0
2       0   0   0   0   1   0   0   0
3       0   0   0   1   0   0   0   0
4       0   0   1   0   1   0   0   0
5       1   1   0   1   0   1   0   0
6       0   0   0   0   1   0   1   1
7       0   0   0   0   0   1   0   0
8       0   0   0   0   0   1   0   0

Thanks heaps, any pointers would be awesome!

Danielle.

Sorry, I don't understand the format of your dictionary; could you explain a bit more how you're getting from there to the desired output? What does "first and second values of the dictionary" refer to? — Danica
– Danica, Commented Aug 18, 2012 at 3:57
Where the matrix is true? That doesn't make any sense. Can you please explain what constitutes true and false for each element? — Aesthete
– Aesthete, Commented Aug 18, 2012 at 4:05
@Aesthete, I believe the idea is that an integrer value 1 is True and 0 is False. Though, your question is legitimate considering the vagueness of the question (and lack of full explanation of the input -> output relationship. — sberry
– sberry, Commented Aug 18, 2012 at 4:33
It's confusing.. Is 0: [2, 5.0], supposed to be 0: [2, 5, 0], where the last value specifyies True or False? Why is one matrix index an int and the other a float? — Aesthete
– Aesthete, Commented Aug 18, 2012 at 4:39

sberry · Accepted Answer · 2012-08-18 04:46:32Z

Here is one that matches your desired output, though I think using @Antimony's answer as a basis (numpy) is likely the way to go (at least more-so than this answer)

N = 8

d = {0: [2, 5.0], 1: [6, 7.0], 2: [6, 8.0], 3: [5, 6.0], 4: [1, 5.0], 5: [3, 4.0], 6: [4, 5.0]}

m = [0] * (N ** 2 + 1)

for x, y in d.values():
    m[x + int(y - 1) * N] = 1
    m[int(y) + (x - 1) * N] = 1

print " ".ljust(9),
print "   ".join(map(str, range(1, N + 1)))
print
for i in range(1, N ** 2 + 1):
    if i % N == 1:
        print "%d  ".ljust(10) % (i / N + 1),
    val = m[i]
    print "%d  " % val,
    if not i % N:
        print

OUTPUT

          1   2   3   4   5   6   7   8

1         0   0   0   0   1   0   0   0  
2         0   0   0   0   1   0   0   0  
3         0   0   0   1   0   0   0   0  
4         0   0   1   0   1   0   0   0  
5         1   1   0   1   0   1   0   0  
6         0   0   0   0   1   0   1   1  
7         0   0   0   0   0   1   0   0  
8         0   0   0   0   0   1   0   0

Based on @DSM's comment to @Antimony's answer, here is one using numpy:

import numpy
N = 8
m = numpy.zeros((N,N))

d = {0: [2, 5.0], 1: [6, 7.0], 2: [6, 8.0], 3: [5, 6.0], 4: [1, 5.0], 5: [3, 4.0], 6: [4, 5.0]}

for i,j in d.itervalues(): 
    m[i-1,j-1] = 1
    m[j-1,i-1] = 1

print " ".ljust(9),
print "   ".join(map(str, range(1, N + 1)))
print
for i, line in enumerate(m.tolist()):
    print "%s %s" % (("%s".ljust(10) % (i+1),"   ".join(map(str, map(int, line)))))

OUTPUT

          1   2   3   4   5   6   7   8

1         0   0   0   0   1   0   0   0
2         0   0   0   0   1   0   0   0
3         0   0   0   1   0   0   0   0
4         0   0   1   0   1   0   0   0
5         1   1   0   1   0   1   0   0
6         0   0   0   0   1   0   1   1
7         0   0   0   0   0   1   0   0
8         0   0   0   0   0   1   0   0

Antimony · Accepted Answer · 2012-08-18 04:19:18Z

0

Your given desired output makes no sense, but here's something that does what I'm guessing you actually want. You can trim off the 0th row and column by doing m = m[1:,1:]

>>> d = {0: [2, 5.0], 1: [6, 7.0], 2: [6, 8.0], 3: [5, 6.0], 4: [1, 5.0], 5: [3, 4.0], 6: [4, 5.0]}
>>> dims = [1 + int(x) for x in map(max, zip(*d.values()))]
>>> m = numpy.zeros(dims, dtype=int)
>>> for v in map(tuple, d.values()):
    m[v] = 1

>>> m
array([[0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 1]])

answered Aug 18, 2012 at 4:19

Antimony

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1 Comment

DSM Over a year ago

I think the desired output does make sense, even though the description is very unclear. I think it's equivalent to m = numpy.zeros((8,8)); for i,j in d.itervalues(): m[i-1,j-1] = 1; m[j-1,i-1] = 1.

Aesthete · Accepted Answer · 2012-08-18 04:51:34Z

0

Using numbers as dictionary keys doesn't make any sense. You should just use a list.

vals = [True, True, False, True] # Assuming this is a list of n*n elements
matrix = [] # This will be a two-dimensional array, or matrix.

This will produce a square matrix for arbitrary sized lists:

from math import sqrt
size = sqrt(len(vals))
for x in range(0, size):
    matrix.append(list(vals[x*size:x*size+size]))

# matrix == [[True, True], [False, True]]
# matrix[0][0] == True
# int(matrix[0][0]) == 1

edited Aug 18, 2012 at 4:51

answered Aug 18, 2012 at 4:34

Aesthete

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