Thanks so much Juan! I did not know about this np.digitize command. With this the vectorization of my function then reads as
def ZeroOrderInterpolation(t,table): import numpy as np lookup, values = table[:, 0], table[:, 1:] values = np.concatenate((values[0:1], values), axis=0) indices = np.digitize(t, lookup) return values[indices] Hugs! Seth > On Oct 16, 2017, at 4:29 AM, Juan Nunez-Iglesias <jni.s...@gmail.com> wrote: > > Hi Seth, > > The function you’re looking for is `np.digitize`: > > > In [1]: t = np.array([0.5,1,1.5,2.5,3,10]) > ...: table = np.array([[0,3],[1,0],[2,5],[3,-1]]) > ...: > In [2]: lookup, values = table[:, 0], table[:, 1:] > In [3]: values = np.concatenate((values[0:1], values), axis=0) > In [4]: indices = np.digitize(t, lookup) > In [5]: values[indices] > Out[5]: > array([[ 3], > [ 0], > [ 0], > [ 5], > [-1], > [-1]]) > > > Note the call to concatenate. Depending on how exactly you want your bins to > align, you might need to concatenate at the end or at the start of the > `values` array. > > Hope this helps! > > Juan. > > On 16 Oct 2017, 1:17 PM +1100, Seth Ghandi <seth.ghandi.2...@gmail.com>, > wrote: >> Hi everybody, >> >> I am new to newpy and am trying to define a variant of piecewise or zero >> holder interpolation function, say ZeroOrderInterpolation(t,a), where t is >> an 1D array of size, say p, consisting of real numbers, and a is a 2D array >> of size, say nxm, with first column consisting of increasing real numbers. >> This function should return an array, say y, of size px(m-1) such that >> y[i,:] is equal to >> a[n,1:] if a[n,0] <= t[i], and >> a[k,1:] if k < n and a[k,0] <= t[i] < a[k+1,0]. >> Note that t[0] is assumed to be at least equal to a[0,0]. >> >> I have the following script made of "for loops" and I am trying to vectorize >> it so as to make it faster for large arrays. >> >> def ZeroOrderInterpolation(t,a): >> import numpy as np >> p = t.shape[0] >> n, m = a.shape >> if n == 1: >> return a[0,1:] >> y = np.zeros((p,m-1)) >> for i in range(p): >> if a[n-1,0] <= t[i]: >> y[i] = a[n-1,1:] >> else: >> for j in range(n-1): >> if (a[j,0] <= t[i]) and (t[i] <= a[j+1,0]): >> y[i] = a[j,1:] >> return y >> >> import numpy as np >> t = np.array([0.5,1,1.5,2.5,3,10]) >> table = np.array([[0,3],[1,0],[2,5],[3,-1]]) >> ZeroOrderInterpolation(t,table) >> >> [Out]: array([[ 3.], >> [ 0.], >> [ 0.], >> [ 5.], >> [-1.], >> [-1.]]) >> >> >> Any help for a vectorization "à la numpy" of this fucntion will be >> apprecaited. >> >> Best regards, >> _______________________________________________ >> NumPy-Discussion mailing list >> NumPy-Discussion@python.org >> https://mail.python.org/mailman/listinfo/numpy-discussion > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@python.org > https://mail.python.org/mailman/listinfo/numpy-discussion
_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@python.org https://mail.python.org/mailman/listinfo/numpy-discussion
