[SciPy-User] "inverting" an array

[Eric Hermes]

On 2/4/2014 1:52 PM, Warren Weckesser wrote:
>
> On Tue, Feb 4, 2014 at 2:33 PM, nicky van foreest 
> <vanforeest at gmail.com <mailto:vanforeest at gmail.com>> wrote:
>
>     Hi,
>
>     I am wondering whether a shortcut exists in numpy/scipy for the
>     following problem.  The values in an array represent the number of
>     customers that arrive in a certain time slot, e.g.,
>
>     a = [0,4,7,3,1,5, 0,0,0,]
>
>     means that in time slot 1 4 customers arrive, in time slot 2 seven
>     arrive, and so on. Now I want to "invert" this array to compute
>     the arrival time of the i-th customer. Thus, customer 2 arrives in
>     time slot 1, customer 6 in time slot 2, and so on. For this
>     problem I wrote the following function:
>
>     a = [0,4,7,3,1,5, 0,0,0,]
>     A = np.cumsum(a)
>
>     def invert(A):
>         Ainv = np.empty(A[-1])
>         aprev=0
>         for i, a in enumerate(A):
>             Ainv[aprev:a] = i
>             aprev = a
>         return Ainv
>
>
>     Ainv= invert(A)
>
>     print a
>     print A
>     print Ainv
>
>     The output is
>
>     [0, 4, 7, 3, 1, 5, 0, 0, 0]
>     [ 0  4 11 14 15 20 20 20 20]
>     [ 1.  1.  1.  1.  2.  2.  2.  2.  2.  2.  2.  3.  3.  3.  4.  5.
>      5.  5.
>       5.  5.]
>
>     Does anybody know whether this code can be made faster, or whether
>     a numpy/scipy function exists that establishes this in one go?
>
>     thanks
>
>     Nicky
>
>
> You can use `np.repeat`:
> In [10]: a
> Out[10]: [0, 4, 7, 3, 1, 5, 0, 0, 0]
>
> In [11]: np.repeat(np.arange(len(a)), a)
> Out[11]: array([1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 5, 5, 5, 
> 5, 5])
>
>
> Warren
>
I came up with a version that only uses python intrinsics:

def invert(a):
     ainv = []
     for i, n in enumerate(a):
         ainv += [i]*n
     return ainv

Eric
>
>
>
>
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-- 
Eric Hermes
J.R. Schmidt Group
Chemistry Department
University of Wisconsin - Madison

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