HansSnah - 5 months ago 37x

Python Question

So, I have been browsing stackoverflow for quite some time now, but I can't seem to find the solution for my problem

Consider this

`import numpy as np`

coo = np.array([[1, 2], [2, 3], [3, 4], [3, 4], [1, 2], [5, 6], [1, 2]])

values = np.array([1, 2, 4, 2, 1, 6, 1])

The coo array contains the (x, y) coordinate positions

x = (1, 2, 3, 3, 1, 5, 1)

y = (2, 3, 4, 4, 2, 6, 2)

and the values array some sort of data for this grid point.

Now I want to get the average of all values for each unique grid point.

For example the coordinate (1, 2) occurs at the positions (0, 4, 6), so for this point I want

`values[[0, 4, 6]]`

How could I get this for all unique grid points?

Answer

You can sort `coo`

with `np.lexsort`

to bring the duplicate ones in succession. Then run `np.diff`

along the rows to get a mask of starts of unique XY's in the sorted version. Using that mask, you can create an ID array that would have the same ID for the duplicates. The ID array can then be used with `np.bincount`

to get the summation of all values with the same ID and also their counts and thus the average values, as the final output. Here's an implementation to go along those lines -

```
# Use lexsort to bring duplicate coo XY's in succession
sortidx = np.lexsort(coo.T)
sorted_coo = coo[sortidx]
# Get mask of start of each unique coo XY
unqID_mask = np.append(True,np.any(np.diff(sorted_coo,axis=0),axis=1))
# Tag/ID each coo XY based on their uniqueness among others
ID = unqID_mask.cumsum()-1
# Get unique coo XY's
unq_coo = sorted_coo[unqID_mask]
# Finally use bincount to get the summation of all coo within same IDs
# and their counts and thus the average values
average_values = np.bincount(ID,values[sortidx])/np.bincount(ID)
```

Sample run -

```
In [65]: coo
Out[65]:
array([[1, 2],
[2, 3],
[3, 4],
[3, 4],
[1, 2],
[5, 6],
[1, 2]])
In [66]: values
Out[66]: array([1, 2, 4, 2, 1, 6, 1])
In [67]: unq_coo
Out[67]:
array([[1, 2],
[2, 3],
[3, 4],
[5, 6]])
In [68]: average_values
Out[68]: array([ 1., 2., 3., 6.])
```

Source (Stackoverflow)

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