A function which sweep the indexes and doesn't stop at the first encounter

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CA je savais faire Emmanuel. je me demandais si c'était quelque chose de possible avec une compréhension de liste. Ca semble une fonction normale. Si la comprehension de liste balaye tous les éléments de la liste elle devrait être capable d'isoler leur rang. Ce n'était qu'un exemple.

The call to p.index(x) returns the first occurence of x in p (so 1) and it takes linear time in the size of the list.

sage: p=[-100,10,10,10,10,10,10,10,10,10,10]
sage: [p.index(x) for x in p]
[0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

You can also get the index of the next occurence of x, specifying the index at which to start the search

sage: q = [10,20,10,3,4,5,10]
sage: q.index(10)
0
sage: q.index(10, 1)
2
sage: q.index(10, 3)
6

But if you want to know all indices of x, I suggest

sage: [i for i, x in enumerate(q) if x == 10]
[0, 2, 6]

So for your initial example, you can write

sage: [x*1.2^-i for i, x in enumerate(p)]
[-100.000000000000,
 8.33333333333333,
 6.94444444444445,
 5.78703703703704,
 4.82253086419753,
 4.01877572016461,
 3.34897976680384,
 2.79081647233653,
 2.32568039361378,
 1.93806699467815,
 1.61505582889846]

Thanks David you answered my question. But, this elegant solution is slow Wall time: 82.3 µs on my computer. Emmanuel's one is Wall time: 14.5 µs