>> 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
