Filling matrix with array of coordinates in python

Question

I currently have 2 arrays corresponding to coordinates (X,Y) and a third array corresponding to the value at this point in 2d space. It is coded as such instead of being a matrix, since it is a pretty sparse matrix (not every point possesses a value). Now I want to reconstruct the matrix in order to plot the values with matplotlib.imshow().

So far, the easiest way I have is to do a for loop, like this:

X = [1, 1, 3, 5];
Y = [2, 2, 3, 7];
Z = [0.3, -0.5, 1, 1];
matrix = np.zeros([10,10])

for i in range(len(Z)):
    matrix[X[i],Y[i]] = Z[i]

I mean, it is not terrible, but I fear the big arrays. Is there a function that would take the 1st and 2nd inputs as the 1st and 2nd coordinates respectively with the 3rd input as the value at these coordinates? Or would there be something similar?

Assuming I read this correctly, if you need a sparse way of representing points, why not a set? If you need to have values associated with them, why not a dict? — user2085282
– user2085282, Commented Oct 26, 2015 at 3:53
Thanks for the reply. It is indeed stored in a dictionary, but I re-framed it in terms of arrays as it is simpler to explain. — PhABC
– PhABC, Commented Oct 26, 2015 at 4:21

Anand S Kumar · Accepted Answer · 2015-10-26 04:17:15Z

6

For what you are doing, you can simply use the lists directly (without) loop. Example -

matrix[X,Y] = Z

Demo -

In [3]: X = [1, 1, 3, 5];

In [4]: Y = [2, 2, 3, 7];

In [5]: Z = [0.3, -0.5, 1, 1];

In [6]: matrix = np.zeros([10,10])

In [7]: matrix[X,Y] = Z

In [8]: matrix
Out[8]:
array([[ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. , -0.5,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  1. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  1. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ],
       [ 0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ,  0. ]])

In [9]: matrix1 = np.zeros([10,10])

In [10]: for i in range(len(Z)):
   ....:     matrix1[X[i],Y[i]] = Z[i]

In [13]: (matrix1 == matrix).all()  #Just to show its equal to OP's `for` loop method.
Out[13]: True

Timing tests -

In [24]: X = np.arange(1000)

In [25]: Y = np.arange(1000)

In [26]: Z = np.random.rand(1000)

In [27]: %%timeit
   ....: matrix = np.zeros([1000,1000])
   ....: matrix[X,Y] = Z
   ....:
1000 loops, best of 3: 834 µs per loop

In [28]: %%timeit
   ....: matrix1 = np.zeros([1000,1000])
   ....: for i in range(len(Z)):
   ....:     matrix1[X[i],Y[i]] = Z[i]
   ....:
The slowest run took 6.47 times longer than the fastest. This could mean that an intermediate result is being cached
1000 loops, best of 3: 1.43 ms per loop

The vectorized method would be faster when dealing with large arrays (and Z is large).

If Z is small, then it would be faster to use the for loop method.

edited Oct 26, 2015 at 4:17

answered Oct 26, 2015 at 4:01

Anand S Kumar

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2 Comments

PhABC Over a year ago

I am not sure why I thought python wouldn't handle that; I didn't even try it. Vectorization is exactly what I was looking for. Thank you

Anand S Kumar Over a year ago

@PhC-PhD Also, its numpy that handles the vectorization correctly, this would not work with normal python lists. matrix needs to be a numpy array or numpy matrix for this to work.

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