Reputation: 365
Suppose we have a distance matrix as following:
array([[ 0. , 0.2039889 , 0.25030506, 0.56118992],
[ 0.2039889 , 0. , 0.39916797, 0.6909994 ],
[ 0.25030506, 0.39916797, 0. , 0.63389566],
[ 0.56118992, 0.6909994 , 0.63389566, 0. ]])
How can I get the values of the upper matrix with their indexes like :
1 0.2039889 1 2
2 0.25030506 1 3
3 0.56118992 1 4
4 0.39916797 2 3
5 0.6909994 2 4
6 0.63389566 3 4
Upvotes: 0
Views: 96
Reputation: 61485
Here's one way to solve it using np.triu_indices
# 4 - matrix shape, 1 is diagonal offset
In [67]: idxs = np.triu_indices(4, 1)
In [68]: entries = arr[idxs]
In [69]: one_based_idxs = [ ar+1 for ar in idxs]
In [70]: one_based_idxs
Out[70]: [array([1, 1, 1, 2, 2, 3]), array([2, 3, 4, 3, 4, 4])]
In [71]: entries
Out[71]:
array([ 0.2039889 , 0.25030506, 0.56118992, 0.39916797, 0.6909994 ,
0.63389566])
Upvotes: 0
Reputation:
For the upper triangle, use triu_indices with diagonal offset 1 to exclude the main diagonal (based on the suggestion by closetCoder):
a = # your array
idx = np.triu_indices(a.shape[0], 1) # indices
stacked = np.concatenate((
a[idx].reshape(-1, 1), # reshaped as column of values
np.array(idx).T + 1 # transposed as columns of indices
), axis=1)
Adding 1 because apparently you want 1-based indices.
If the numbers 1,2,3,4,... on the left are also needed, that's another thing to concatenate:
stacked = np.concatenate((
np.arange(idx[0].size).reshape(-1, 1) + 1, # numbers
a[idx].reshape(-1, 1),
np.array(idx).T + 1
), axis=1)
Looks like this:
array([[ 1. , 0.2039889 , 1. , 2. ],
[ 2. , 0.25030506, 1. , 3. ],
[ 3. , 0.56118992, 1. , 4. ],
[ 4. , 0.39916797, 2. , 3. ],
[ 5. , 0.6909994 , 2. , 4. ],
[ 6. , 0.63389566, 3. , 4. ]])
Just for completeness, I'll mention scipy.spatial.distance.squareform which gets the values as above, but not the indexes.
Upvotes: 1