`scipy.spatial.distance.pdist(X)`

gives the pair-wise distances of `X`

, $\operatorname{dist}(X[i], X[i])$ not included. The distances are stored in a dense matrix `D`

.

Question: how to get $\operatorname{dist}(X[i], X[j])$ without expanding this dense matrix by `scipy.spatial.distance.squareform(D)`

?

A good answer from How does condensed distance matrix work? (pdist):

```
def square_to_condensed(i, j, n):
assert i != j, "no diagonal elements in condensed matrix"
if i < j:
i, j = j, i
return n*j - j*(j+1)/2 + i - 1 - j
```

Take $n$ == `X.shape[0]`

== $4$ as an example. In the code, $i$ and $j$ are swapped if $i < j$, so only consider the bottom triangular region, i.e. $i > j$ always holds in the discussion below.

i\j | 0 | 1 | 2 | 3 |
---|---|---|---|---|

0 | - | D[0] | D[1] | D[2] |

1 | D[0] | - | D[3] | D[4] |

2 | D[1] | D[3] | - | D[5] |

3 | D[2] | D[4] | D[5] | - |

Suppose the indexing function is $\operatorname{f}$: $\operatorname{f}(i, j) = k$ when $\operatorname{dist}(X[i], X[j]) = D[k]$

`len(D)`

= ${n \choose 2} = (n-1) + (n-2) + \dots + 1$.

- When $j = 0$, $\operatorname{f}(i, j) = i - j - 1$
- When $j = 1$, $\operatorname{f}(i, j) = (n-1) + (i - j - 1)$
- When $j = 2$, $\operatorname{f}(i, j) = (n-1) + (n-2) + (i - j - 1)$

So the pattern here is:

Further more: