# Transposing a 'sparse', randomly-ordered array

This note is about a little programming difficulty I ran into recently. I'm putting it here so I can point people to it while trying to figure out if this is either (a) already known, (b) easy to solve with advanced indexing, (c) in fact not a real problem, or (d) actually hard.

**Update:** So, it appears that this is just a sparse matrix operation, and one way to solve it would be to transition everything to sparse matrices. The `scipy.sparse`

package would be quite useful for this, and should support the format described below with `coo_matrix`

. For my usecase, I briefly used a `numba`

implementation, given below, and then found a way to avoid having to use the transpose entirely. 😅 I'm leaving this note up in case anyone else runs into a similar problem!

## Task

Let's say we have a `numpy`

array `u`

with dimensions `[n, m, …]`

, with `n>m`

. From now on we'll ignore the additional indices `…`

in `u`

, as we're interested in operations on the first two indices.

The entries in `u`

correspond to the non-zero entries of an implicit, large `[n, n, …]`

`numpy`

array `U`

. We know the following about `U`

: For each index pair `[i, k]`

, if `U[i, k]`

is nonzero, `U[k, i]`

is also nonzero, and for each `i`

there are at most `m`

non-zero entries `k`

in that row/column. Therefore, `u`

contains all the relevant information about the non-zero elements of `U`

. Each row in `m`

may be padded with zero entries at the end to make `u`

not ragged.

There is one more important twist: the entries in each row of `u`

are ordered in some arbitrary, essentially random way. To track this, we have an index array `idx`

with dimension `[n, m]`

. Each row in `i`

contains the column indices (`k`

above) in `U`

corresponding to the entries in `u`

at that position. The index `-1`

is reserved for padded entries. So for instance, if a row `i`

of `u`

is `[1, 2, 3]`

, corresponding to `[i, k3]`

, `[i, k1]`

, `[i, k2]`

in `U`

, then `idx[i] = [k3, k1, k2]`

.

Our task is to compute the "transpose" `t`

of `u`

, in the sense that `t`

has the same relationship to `transpose(U)`

as `u`

has to `U`

: if the `[i, j]`

entry of `u`

corresponds to `[i, k]`

in `U`

, then `[i, j]`

of `t`

should correspond to `[k, i]`

.

The entry `[i, j]`

of `u`

belongs in row `idx[i, j] = jj`

of the transpose. The difficulty is figuring out the *column* index: We need to know which index in `u[jj, :]`

belongs to `i`

. We'll call this "reverse" index `ii`

in the code.

For my particular usecase, we'll need to compute this transpose many times for different `u`

, but can expect `idx`

to stay the same: the values we are transposing might change, but not the underlying structure.

### Example

This is rather confusing, so here is an example.

```
u = np.array([[3, 0],
[2, 1],
[4, 0]], dtype=int)
idx = np.array([[1, -1],
[2, 0],
[1, -1]], dtype=int)
```

with the result of our `transpose`

:

```
t = [[1, 0],
[4, 3],
[2, 0]]
```

## Naive solution

Here is a naive pure-python implementation.

```
def naive_transpose(u, idx):
t = np.zeros_like(u)
for i in range(u.shape[0]):
for j in range(u.shape[1]):
jj = idx[i, j]
if jj == -1:
break
for kk in range(u.shape[1]):
if idx[jj, ii] == i:
break
t[i, j] = u[jj, ii]
return t
```

Clearly, this is inefficient: We run through each row of the `idx`

array many times to search for a match. But it works!

`numba`

+ indexing solution

This solution has two parts: First, we make an `[n, m, 2]`

transpose index array that contains `idx`

in the `[:, :, 0]`

entries, and the "reverse" indices in `[:, :, 1]`

. This part is implemented using `numba`

, which provides a just-in-time (`jit`

) compiler for a subset of `python`

, speeding up the `for`

loops considerably.

We then use some ✨ advanced indexing ✨ to collect the entries of `t`

out of `u`

based on the indices in our transpose index array.

This approach has the advantage of separating out the complicated index-finding part from the actual array operations: We'll only need to compute the transpose index array once, and then can re-use it for different `u`

s with the same structure.

```
import numba
import numpy as np
def get_transpose_idx(idx):
transpose = np.zeros((*idx.shape[:2], 2), dtype=int)
return _get_transpose_idx(idx.astype(int), transpose)
@numba.jit(nopython=True)
def _get_transpose_idx(idx, transpose):
n, m = idx.shape
for i in range(n):
for j in range(m):
# what entry are we inverting?
jj = idx[i, j]
if jj == -1:
ii = -1
else:
ii = find(idx[jj], i)
transpose[i, j] = [jj, ii]
return transpose
@numba.jit(nopython=True)
def find(array, target):
n = len(array)
for i in range(n):
if array[i] == target:
return i
return -1
```

(For better performance one needs to either find a way to make use of accumulated knowledge about "reverse" indices when going through `idx`

, or optimise the lookup, for example by switching to a `dict`

, where average lookup complexity is better.)

The transpose itself can be achieved via

```
t = u[t_idx[:, :, 0], t_idx[:, :, 1], ...]
```

adding `:`

in place of `...`

for additional dimensions of `u`

.

This approach should be reasonably fast, but please run your own benchmarks if you end up using it!