On Wed, Jan 25, 2012 at 15:12, Edward C. Jones <edcjones@comcast.net> wrote:

I have a vector of bits where there are many more zeros than one.  I store the array as a sorted list of the indexes where the bit is one. If the bit array is (0, 1, 0, 0, 0, 1, 1), it is stored as (1, 5, 6). If the bit array, b, has length n, and p is a random permutation of arange(n), then I can permute the bit array using fancy indexing: b[p]. Is there some neat trick I can use to permute an array while leaving it in the list-of-indexes form?  Currently I am doing it with a Python loop but I am looking for a faster way.

Use argsort() to get the "inverse" of the permutation. Then fancy-index the inverse with the list-of-indexes array. [~/scratch] |28> b array([0, 1, 0, 0, 0, 1, 1]) [~/scratch] |29> loi array([1, 5, 6]) [~/scratch] |30> p = np.random.permutation(len(b)) [~/scratch] |31> ps = p.argsort() [~/scratch] |41> p array([2, 3, 5, 4, 6, 1, 0]) [~/scratch] |42> ps array([6, 5, 0, 1, 3, 2, 4]) [~/scratch] |43> ps[loi] array([5, 2, 4]) -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth."   -- Umberto Eco