I have a 2D Numpy array that consists of (X,Y,Z,A) values, where (X,Y,Z) are Cartesian coordinates in 3D space, and A is some value at that location. As an example..
__X__|__Y__|__Z__|__A_
13 | 7 | 21 | 1.5
9 | 2 | 7 | 0.5
15 | 3 | 9 | 1.1
13 | 7 | 21 | 0.9
13 | 7 | 21 | 1.7
15 | 3 | 9 | 1.1
Is there an efficient way to find all the unique combinations of (X,Y), and add their values? For example, the total for (13,7) would be (1.5+0.9+1.7), or 4.1.
scipy.sparse matrix takes this kind of information, but for just 2d
sparse.coo_matrix((data, (row, col)))
where row and col are indices like your X,Y and Z. It sums duplicates.
The first step to doing that is a lexical sort of the indices. That puts points with matching coordinates next to each other.
The actually grouping and summing is done, I believe, in compiled code. Part of the difficulty in doing that fast in numpy terms is that there will be a variable number of elements in each group. Some will be unique, others might have 3 or more.
Python itertools has a groupby tool. Pandas also has grouping functions. I can also imagine using a default_dict to group and sum values.
The ufunc reduceat might also work, though it's easier to use in 1d than 2 or 3.
If you are ignoring the Z, the sparse coo_matrix approach may be easiest.
In [2]: X=np.array([13,9,15,13,13,15])
In [3]: Y=np.array([7,2,3,7,7,3])
In [4]: A=np.array([1.5,0.5,1.1,0.9,1.7,1.1])
In [5]: M=sparse.coo_matrix((A,(X,Y)))
In [15]: M.sum_duplicates()
In [16]: M.data
Out[16]: array([ 0.5, 2.2, 4.1])
In [17]: M.row
Out[17]: array([ 9, 15, 13])
In [18]: M.col
Out[18]: array([2, 3, 7])
In [19]: M
Out[19]:
<16x8 sparse matrix of type '<class 'numpy.float64'>'
with 3 stored elements in COOrdinate format>
Here's what I had in mind with lexsort
In [32]: Z=np.array([21,7,9,21,21,9])
In [33]: xyz=np.stack([X,Y,Z],1)
In [34]: idx=np.lexsort([X,Y,Z])
In [35]: idx
Out[35]: array([1, 2, 5, 0, 3, 4], dtype=int32)
In [36]: xyz[idx,:]
Out[36]:
array([[ 9, 2, 7],
[15, 3, 9],
[15, 3, 9],
[13, 7, 21],
[13, 7, 21],
[13, 7, 21]])
In [37]: A[idx]
Out[37]: array([ 0.5, 1.1, 1.1, 1.5, 0.9, 1.7])
When sorted like this it becomes more evident that the Z coordinate is 'redundant', at least for this purpose.
Using reduceat to sum groups:
In [40]: np.add.reduceat(A[idx],[0,1,3])
Out[40]: array([ 0.5, 2.2, 4.1])
(for now I just eyeballed the [0,1,3] list)
Approach #1
Get each row as a view, thus converting each into a scalar each and then use np.unique to tag each row as a minimum scalar starting from (0......n), withnas no. of unique scalars based on the uniqueness among others and finally usenp.bincount` to perform the summing of the last column based on the unique scalars obtained earlier.
Here's the implementation -
def get_row_view(a):
void_dt = np.dtype((np.void, a.dtype.itemsize * np.prod(a.shape[1:])))
a = np.ascontiguousarray(a)
return a.reshape(a.shape[0], -1).view(void_dt).ravel()
def groupby_cols_view(x):
a = x[:,:2].astype(int)
a1D = get_row_view(a)
_, indx, IDs = np.unique(a1D, return_index=1, return_inverse=1)
return np.c_[x[indx,:2],np.bincount(IDs, x[:,-1])]
Approach #2
Same as approach #1, but instead of working with the view, we will generate equivalent linear index equivalent for each row and thus reducing each row to a scalar. Rest of the workflow is same as with the first approach.
The implementation -
def groupby_cols_linearindex(x):
a = x[:,:2].astype(int)
a1D = a[:,0] + a[:,1]*(a[:,0].max() - a[:,1].min() + 1)
_, indx, IDs = np.unique(a1D, return_index=1, return_inverse=1)
return np.c_[x[indx,:2],np.bincount(IDs, x[:,-1])]
Sample runs
In [80]: data
Out[80]:
array([[ 2. , 5. , 1. , 0.40756048],
[ 3. , 4. , 6. , 0.78945661],
[ 1. , 3. , 0. , 0.03943097],
[ 2. , 5. , 7. , 0.43663582],
[ 4. , 5. , 0. , 0.14919507],
[ 1. , 3. , 3. , 0.03680583],
[ 1. , 4. , 8. , 0.36504428],
[ 3. , 4. , 2. , 0.8598825 ]])
In [81]: groupby_cols_view(data)
Out[81]:
array([[ 1. , 3. , 0.0762368 ],
[ 1. , 4. , 0.36504428],
[ 2. , 5. , 0.8441963 ],
[ 3. , 4. , 1.64933911],
[ 4. , 5. , 0.14919507]])
In [82]: groupby_cols_linearindex(data)
Out[82]:
array([[ 1. , 3. , 0.0762368 ],
[ 1. , 4. , 0.36504428],
[ 3. , 4. , 1.64933911],
[ 2. , 5. , 0.8441963 ],
[ 4. , 5. , 0.14919507]])
