Hi, I've a small problem for which I cannot find a solution and I'm quite sure there is an obvious one: I've an array Z (any dtype) with some data. I've a (sorted) array I (of integer, same size as Z) that tells me the index of Z[i] (if necessary, the index can be stored in Z). Now, I have an arbitrary sequence S of indices (in the sense of I), how do I build the corresponding data ? Here is a small example: Z = [(0,0), (1,1), (2,2), (3,3), (4,4)) I = [0, 20, 23, 24, 37] S = [ 20,20,0,24] -> Result should be [(1,1), (1,1), (0,0),(3,3)] S = [15,15] -> Wrong (15 not in I) but ideally, I would like this to be converted to [(0,0), (0,0)] Any idea ? Nicolas
Hi Nicolas On Thu, Aug 7, 2014 at 1:16 PM, Nicolas P. Rougier <Nicolas.Rougier@inria.fr> wrote:
Here is a small example:
Z = [(0,0), (1,1), (2,2), (3,3), (4,4)) I = [0, 20, 23, 24, 37]
S = [ 20,20,0,24] -> Result should be [(1,1), (1,1), (0,0),(3,3)]
S = [15,15] -> Wrong (15 not in I) but ideally, I would like this to be converted to [(0,0), (0,0)]
First try: Z = np.array([(0,0), (1,1), (2,2), (3,3), (4,4)]) I = np.array([0, 20, 23, 24, 37]) S = np.array([ 20,20,0,24,15]) out = np.zeros((len(S), len(Z[0]))) mask = (S[:, np.newaxis] == I) item, coord = np.where(mask) out[item, :] = Z[coord] Perhaps there's a neater way of doing it! Stéfan
Nice ! Thanks Stéfan. I will add it to the numpy 100 problems. Nicolas On 07 Aug 2014, at 13:31, Stéfan van der Walt <stefan@sun.ac.za> wrote:
Hi Nicolas
On Thu, Aug 7, 2014 at 1:16 PM, Nicolas P. Rougier <Nicolas.Rougier@inria.fr> wrote:
Here is a small example:
Z = [(0,0), (1,1), (2,2), (3,3), (4,4)) I = [0, 20, 23, 24, 37]
S = [ 20,20,0,24] -> Result should be [(1,1), (1,1), (0,0),(3,3)]
S = [15,15] -> Wrong (15 not in I) but ideally, I would like this to be converted to [(0,0), (0,0)]
First try:
Z = np.array([(0,0), (1,1), (2,2), (3,3), (4,4)]) I = np.array([0, 20, 23, 24, 37]) S = np.array([ 20,20,0,24,15])
out = np.zeros((len(S), len(Z[0]))) mask = (S[:, np.newaxis] == I) item, coord = np.where(mask) out[item, :] = Z[coord]
Perhaps there's a neater way of doing it!
Stéfan _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Am 07.08.2014 um 13:16 schrieb Nicolas P. Rougier <Nicolas.Rougier@inria.fr>:
Hi,
I've a small problem for which I cannot find a solution and I'm quite sure there is an obvious one:
I've an array Z (any dtype) with some data. I've a (sorted) array I (of integer, same size as Z) that tells me the index of Z[i] (if necessary, the index can be stored in Z).
Now, I have an arbitrary sequence S of indices (in the sense of I), how do I build the corresponding data ?
Here is a small example:
Z = [(0,0), (1,1), (2,2), (3,3), (4,4)) I = [0, 20, 23, 24, 37]
S = [ 20,20,0,24] -> Result should be [(1,1), (1,1), (0,0),(3,3)]
S = [15,15] -> Wrong (15 not in I) but ideally, I would like this to be converted to [(0,0), (0,0)]
Any idea ?
If I is sorted, I would propose to use a bisection algorithm, faster than linear search: Z = array([(0,0), (1,1), (2,2), (3,3), (4,4)]) I = array([0, 20, 23, 24, 37]) S = array([ 20,20,0,24,15,27]) a = zeros(S.shape,dtype=int) b = a + S.shape[0]-1 for i in range(int(log2(S.shape[0]))+2): c = (a+b)>>1 sel = I[c]<=S a[sel] = c[sel] b[~sel] = c[~sel] Z[c] If I[c] != S, then there is no corresponding index entry in I to match S. Gregor
Am 07.08.2014 um 13:59 schrieb Gregor Thalhammer <gregor.thalhammer@gmail.com>:
Am 07.08.2014 um 13:16 schrieb Nicolas P. Rougier <Nicolas.Rougier@inria.fr>:
Hi,
I've a small problem for which I cannot find a solution and I'm quite sure there is an obvious one:
I've an array Z (any dtype) with some data. I've a (sorted) array I (of integer, same size as Z) that tells me the index of Z[i] (if necessary, the index can be stored in Z).
Now, I have an arbitrary sequence S of indices (in the sense of I), how do I build the corresponding data ?
Here is a small example:
Z = [(0,0), (1,1), (2,2), (3,3), (4,4)) I = [0, 20, 23, 24, 37]
S = [ 20,20,0,24] -> Result should be [(1,1), (1,1), (0,0),(3,3)]
S = [15,15] -> Wrong (15 not in I) but ideally, I would like this to be converted to [(0,0), (0,0)]
Any idea ?
If I is sorted, I would propose to use a bisection algorithm, faster than linear search:
Z = array([(0,0), (1,1), (2,2), (3,3), (4,4)]) I = array([0, 20, 23, 24, 37]) S = array([ 20,20,0,24,15,27])
a = zeros(S.shape,dtype=int) b = a + S.shape[0]-1 for i in range(int(log2(S.shape[0]))+2): c = (a+b)>>1 sel = I[c]<=S a[sel] = c[sel] b[~sel] = c[~sel]
or even simpler: c = searchsorted(I, S) Z[c] Gregor
Z[c]
If I[c] != S, then there is no corresponding index entry in I to match S.
Gregor
Oh thanks, I would never have imagined a one-line solution... Here are the benchmarks: In [2]: %timeit stefan(S) 100000 loops, best of 3: 10.8 µs per loop In [3]: %timeit gregor(S) 10000 loops, best of 3: 48.1 µs per loop In [4]: %timeit gregor2(S) 100000 loops, best of 3: 3.23 µs per loop Nicolas On 07 Aug 2014, at 14:04, Gregor Thalhammer <gregor.thalhammer@gmail.com> wrote:
Am 07.08.2014 um 13:59 schrieb Gregor Thalhammer <gregor.thalhammer@gmail.com>:
Am 07.08.2014 um 13:16 schrieb Nicolas P. Rougier <Nicolas.Rougier@inria.fr>:
Hi,
I've a small problem for which I cannot find a solution and I'm quite sure there is an obvious one:
I've an array Z (any dtype) with some data. I've a (sorted) array I (of integer, same size as Z) that tells me the index of Z[i] (if necessary, the index can be stored in Z).
Now, I have an arbitrary sequence S of indices (in the sense of I), how do I build the corresponding data ?
Here is a small example:
Z = [(0,0), (1,1), (2,2), (3,3), (4,4)) I = [0, 20, 23, 24, 37]
S = [ 20,20,0,24] -> Result should be [(1,1), (1,1), (0,0),(3,3)]
S = [15,15] -> Wrong (15 not in I) but ideally, I would like this to be converted to [(0,0), (0,0)]
Any idea ?
If I is sorted, I would propose to use a bisection algorithm, faster than linear search:
Z = array([(0,0), (1,1), (2,2), (3,3), (4,4)]) I = array([0, 20, 23, 24, 37]) S = array([ 20,20,0,24,15,27])
a = zeros(S.shape,dtype=int) b = a + S.shape[0]-1 for i in range(int(log2(S.shape[0]))+2): c = (a+b)>>1 sel = I[c]<=S a[sel] = c[sel] b[~sel] = c[~sel]
or even simpler: c = searchsorted(I, S) Z[c]
Gregor
Z[c]
If I[c] != S, then there is no corresponding index entry in I to match S.
Gregor
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participants (3)
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Gregor Thalhammer -
Nicolas P. Rougier -
Stéfan van der Walt