On 02/09/2011 10:17 AM, Zachary Pincus wrote: >>> As before, the line below does what you said you need, though not >>> maximally efficiently. (Try it in an interpreter...) There may be >>> another way in numpy that doesn't rely on constructing the index >>> array, but this is the first thing that came to mind. >>> >>> last_greater = numpy.arange(arr.shape)[arr>= T][-1] >>> >>> Let's unpack that dense line a bit: >>> >>> mask = arr>= T >>> indices = numpy.arange(arr.shape) >>> above_threshold_indices = indices[mask] >>> last_above_threshold_index = above_threshold_indices[-1] >>> >>> Does this make sense? >> This assumes monotonicity. Is that allowed? > The twice-stated problem was: > >> In a 1-d array, find the first point where all subsequent points >> have values less than a threshold, T. > So that should do the trick... Though Alan's argmax solution is > definitely a better one than indexing an indices array. Same logic and > result though, just more compact. > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion This is similar to Zachary's as it just uses 'np.where'.
>>> x=np.array([5,4,3,6,7,3,2,1]) >>> x array([5, 4, 3, 6, 7, 3, 2, 1]) >>> np.argmax(x>5) # doesn't appear to be correct 3 >>> np.where(x>5)[0][-1]#since np.where gives a tuple and you need the last element 4 This should give the index where all subsequent points are less than some threshold. However, this and Zachary's version fail when if all points are lower than the threshold (eg T=10 in this array). Bruce _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
