On Wed, Apr 14, 2010 at 3:12 PM, Anne Archibald <peridot.face...@gmail.com> wrote: > On 14 April 2010 16:56, Keith Goodman <kwgood...@gmail.com> wrote: >> On Wed, Apr 14, 2010 at 12:39 PM, Nikolaus Rath <nikol...@rath.org> wrote: >>> Keith Goodman <kwgood...@gmail.com> writes: >>>> On Wed, Apr 14, 2010 at 8:49 AM, Keith Goodman <kwgood...@gmail.com> wrote: >>>>> On Wed, Apr 14, 2010 at 8:16 AM, Nikolaus Rath <nikol...@rath.org> wrote: >>>>>> Hello, >>>>>> >>>>>> How do I best find out the indices of the largest x elements in an >>>>>> array? >>>>>> >>>>>> Example: >>>>>> >>>>>> a = [ [1,8,2], [2,1,3] ] >>>>>> magic_function(a, 2) == [ (0,1), (1,2) ] >>>>>> >>>>>> Since the largest 2 elements are at positions (0,1) and (1,2). >>>>> >>>>> Here's a quick way to rank the data if there are no ties and no NaNs: >>>> >>>> ...or if you need the indices in order: >>>> >>>>>> shape = (3,2) >>>>>> x = np.random.rand(*shape) >>>>>> x >>>> array([[ 0.52420123, 0.43231286], >>>> [ 0.97995333, 0.87416228], >>>> [ 0.71604075, 0.66018382]]) >>>>>> r = x.reshape(-1).argsort().argsort() >>> >>> I don't understand why this works. Why do you call argsort() twice? >>> Doesn't that give you the indices of the sorted indices? >> >> It is confusing. Let's look at an example: >> >>>> x = np.random.rand(4) >>>> x >> array([ 0.37412289, 0.68248559, 0.12935131, 0.42510212]) >> >> If we call argsort once we get the index that will sort x: >> >>>> idx = x.argsort() >>>> idx >> array([2, 0, 3, 1]) >>>> x[idx] >> array([ 0.12935131, 0.37412289, 0.42510212, 0.68248559]) >> >> Notice that the first element of idx is 2. That's because element x[2] >> is the min of x. But that's not what we want. We want the first >> element to be the rank of the first element of x. So we need to >> shuffle idx around so that the order aligns with x. How do we do that? >> We sort it! > > Unless I am mistaken, what you are doing here is inverting the > permutation returned by the first argsort. The second argsort is an n > log n method, though, and permutations can be inverted in linear time: > > ix = np.argsort(X) > ixinv = np.zeros_like(ix) > ixinv[ix] = np.arange(len(ix)) > > This works because if ix is a permutation and ixinv is its inverse, > A = B[ix] > is the same as > A[ixinv] = B > This also means that you can often do without the actual inverse by > moving the indexing operation to the other side of the equal sign. > (Not in the OP's case, though.)
That is very nice. And very fast for large arrays: >> x = np.random.rand(4) >> timeit idx = x.argsort().argsort() 1000000 loops, best of 3: 1.45 us per loop >> timeit idx = x.argsort(); idxinv = np.zeros_like(idx); idxinv[idx] = >> np.arange(len(idx)) 100000 loops, best of 3: 9.52 us per loop >> x = np.random.rand(1000) >> timeit idx = x.argsort().argsort() 10000 loops, best of 3: 112 us per loop >> timeit idx = x.argsort(); idxinv = np.zeros_like(idx); idxinv[idx] = >> np.arange(len(idx)) 10000 loops, best of 3: 82.9 us per loop >> x = np.random.rand(100000) >> timeit idx = x.argsort().argsort() 10 loops, best of 3: 20.4 ms per loop >> timeit idx = x.argsort(); idxinv = np.zeros_like(idx); idxinv[idx] = >> np.arange(len(idx)) 100 loops, best of 3: 13.2 ms per loop > I'll also add that if the OP needs the top m for 1<m<<n, sorting the > whole input array is not the most efficient algorithm; there are > priority-queue-based schemes that are asymptotically more efficient, > but none exists in numpy. Since numpy's sorting is quite fast, I > personally would just use the sorting. Partial sorting would find a lot of uses in the numpy community (like in median). _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
