Fill 2D numpy array from three 1D numpy arrays

Question

Is there an efficient way of creating a 2D array of the values from unsorted coordinate points (i.e. not all lons and/or lats are ascending or descending) without using loops?

Example Data

lats = np.array([45.5,45.5,45.5,65.3,65.3,65.3,43.2,43.2,43.2,65.3])
lons = np.array([102.5,5.5,116.2,102.5,5.5,116.2,102.5,5.5,116.2,100])
vals = np.array([3,4,5,6,7,7,9,1,0,4])

Example Output
Each column represents a unique longitude (102.5, 5.5, 116.2, & 100) and each column represents a unique latitude (45.5,65.3, & 43.2).

([ 3, 4, 5, NaN],
 [ 6, 7, 7, 4],
 [ 9, 1, 0, NaN])

Though, it isn't so straight forward because I don't necessarily know how many duplicates of each lon or lat there are which determines the shape of the array.

Update:
I had the data arranged incorrectly for my question. I have arranged it now, so they are all unique pairs and there is an additional data point to demonstrate how the data should be arranged when NaNs are present.

Answer 1

The example you have posted makes very little sense, and it doesn't allow any reasonable way to specify missing data. I am guessing here, but the only reasonable thing you may be dealing with seems to be something like this :

>>> lats = np.array([43.2, 43.2, 43.2, 45.5, 45.5, 45.5, 65.3, 65.3, 65.3])
>>> lons = np.array([5.5, 102.5, 116.2, 5.5, 102.5, 116.2, 5.5, 102.5, 116.2])
>>> vals = np.array([3, 4, 5, 6, 7, 7, 9, 1, 0])

Where the value in vals[j] comes from latitude lats[j] and longitude lons[j], but the data may come scrambled, as in :

>>> indices = np.arange(9)
>>> np.random.shuffle(indices)
>>> lats = lats[indices]
>>> lons = lons[indices]
>>> vals = vals[indices]
>>> lats
array([ 45.5,  43.2,  65.3,  45.5,  43.2,  65.3,  45.5,  65.3,  43.2])
>>> lons
array([   5.5,  116.2,  102.5,  116.2,    5.5,  116.2,  102.5,    5.5,  102.5])
>>> vals
array([6, 5, 1, 7, 3, 0, 7, 9, 4])

You can get this arranged into an array as follows:

>>> lat_vals, lat_idx = np.unique(lats, return_inverse=True)
>>> lon_vals, lon_idx = np.unique(lons, return_inverse=True)
>>> vals_array = np.empty(lat_vals.shape + lon_vals.shape)
>>> vals_array.fill(np.nan) # or whatever yor desired missing data flag is
>>> vals_array[lat_idx, lon_idx] = vals
>>> vals_array
array([[ 3.,  4.,  5.],
       [ 6.,  7.,  7.],
       [ 9.,  1.,  0.]])

Answer 2

If you're creating a 2D array, then all arrays will have to have the same number of points. If this is true, you can simply do

out = np.vstack((lats, lons, vals))

---

edit

I think this might be what you're after, it matches your question at least :)

xsize = len(np.unique(lats))
ysize = len(np.unique(lons))

and then if your data is very well behaved

out = [vals[i] for i, (x, y) in enumerate(zip(lats, lons))]
out = np.asarray(out).reshape((xsize, ysize))

Answer 3

import numpy as np

lats = np.array([45.5,45.5,45.5,65.3,65.3,65.3,43.2,43.2,43.2,65.3])
lons = np.array([102.5,5.5,116.2,102.5,5.5,116.2,102.5,5.5,116.2,100])
vals = np.array([3,4,5,6,7,7,9,1,0,4])


def unique_order(seq): 
    # http://www.peterbe.com/plog/uniqifiers-benchmark (Dave Kirby)
    # Order preserving
    seen = set()
    return [x for x in seq if x not in seen and not seen.add(x)]

unique_lats, idx_lats = np.unique(lats, return_inverse=True)
unique_lons, idx_lons = np.unique(lons, return_inverse=True)
perm_lats = np.argsort(unique_order(lats))
perm_lons = np.argsort(unique_order(lons))

result = np.empty((len(unique_lats), len(unique_lons)))
result.fill(np.nan)
result[perm_lats[idx_lats], perm_lons[idx_lons]] = vals
print(result)

yields

[[  3.   4.   5.  nan]
 [  6.   7.   7.   4.]
 [  9.   1.   0.  nan]]