Reverse indices of a sorted list
python, sorting
Solution
The most efficient I can think of (short of possibly looking to `numpy`) that gets rid of the `.index` and can be used for both `idx` and `ridx`:
U=[5,2,3,1,4]
idx = sorted(xrange(len(U)), key=U.__getitem__)
ridx = sorted(xrange(len(U)), key=idx.__getitem__)
# [3, 1, 2, 4, 0] [4, 1, 2, 0, 3]
Problem
I want to return the 'reverse' indices of a sorted list. What I mean by that is: I have an unsorted list `U` and I sort it via `S=sorted(U)`. Now, I can get the sort indices such that `U(idx)=S` - but I want `S(Ridx) = U`. Here a little example: ``` U=[5,2,3,1,4] S=sorted(U) idx = [U.index(S[i]) for i in range(len(U))] >>> idx [3, 1, 2, 4, 0] Ridx = [S.index(U[i]) for i in range(len(U))] >>> Ridx [4, 1, 2, 0, 3] >>>[U[idx[i]] for i in range(len(U))] == S True >>>[S[Ridx[i]] for i in range(len(U))] == U True ``` What I need is an efficient way to get Ridx. Thanks! Edit: All right! I did a little speed test for both of the solutions (@Jon Clements and @Whatang) which answered the question. The script: ``` import datetime as DT import random U=[int(1000*random.random()) for i in xrange(pow(10,8))] S=sorted(U) idx = sorted(xrange(len(U)), key=U.__getitem__) T0 = DT.datetime.now() ridx = sorted(xrange(len(U)), key=idx.__getitem__) print [S[ridx[i]] for i in range(len(U))]==U elapsed = DT.datetime.now()-T0 print str(elapsed) print '===============' T0 = DT.datetime.now() ridx = [ y for (x,y) in sorted(zip(idx, range(len(idx)))) ] print [S[ridx[i]] for i in range(len(U))]==U elapsed = DT.datetime.now()-T0 print str(elapsed) ``` And the results: ``` True 0:02:45.278000 =============== True 0:06:48.889000 ``` Thank you all for the quick and meaningful help!