just a solution (not all of them) and the application happen to come up with something like 10k values in the array. don care waiting, but... 2010/7/1 Vincent Davis <vincent@vincentdavis.net>:

On Thu, Jul 1, 2010 at 3:17 AM, Renato Fabbri <renato.fabbri@gmail.com> wrote:

...

hi, i need to find which elements of an array sums up to an specific value

any idea of how to do this?

Not sure if there is a better way but a brut force way would be to

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a array([[ 7.,  5.,  9.,  3.],       [ 7.,  2.,  7.,  8.],       [ 6.,  8.,  3.,  2.]]) alist= a.flatten() alist [7.0, 5.0, 9.0, 3.0, 7.0, 2.0, 7.0, 8.0, 6.0, 8.0, 3.0, 2.0] asolution = [] for r in range(len(alist):     for comb in itertools.combinations(alist, r):         if sum(comb)==TheValue:            asolution.append(comb)

Now just find the comb values in the array.

Like I said kinda brute force. Also depends if you want all solutions or a solution.

Vincent

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best, rf

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to, 2010-07-01 kello 11:46 -0300, Renato Fabbri kirjoitti:

just a solution (not all of them)

and the application happen to come up with something like 10k values in the array. don care waiting, but...

As said, the problem is a well-known one, and it's not really Python or Numpy-specific, so slightly off-topic for this list. Numpy and Scipy don't ship pre-made algorithms for solving these. But anyway, you'll probably find that the brute force algorithm (e.g. the one from Vincent) takes exponential time (and exp(10000) is a big number). So you need to do something more clever. First stop, Wikipedia, http://en.wikipedia.org/wiki/Knapsack_problem http://en.wikipedia.org/wiki/Subset_sum_problem and if you are looking for pre-cooked solutions, second stop stackoverflow, http://stackoverflow.com/search?q=subset+sum+problem Some search words you might want to try on Google: http://www.google.com/search?q=subset%20sum%20dynamic%20programming Generic advice only this time, sorry; I don't have pre-made code for solving this at hand, but hopefully the above links give some pointers for what to do. -- Pauli Virtanen