On Wed, Feb 9, 2011 at 1:18 PM, Zachary Pincus <zachary.pin...@yale.edu> wrote: >> >>>> In a 1-d array, find the first point where all subsequent points >>>> have values >>>> less than a threshold. >> >> This doesn't imply monotonicity. >> Suppose with have a sin curve, and I want to find the last trough. Or >> a business cycle and I want to find the last recession. >> >> Unless my english deteriorated recently. >> > > Not sure that I follow? I read the statement as: > find the smallest non-negative index i such that array[j] < t for all > i <= j < len(array) > > for which the various solutions proposed work, except for the corner > case where array.min() >= t. > > I'm not sure where monotonicity comes into play, unless one interprets > the "all subsequent points" clause as "some number of subsequent > points" or something... > > For those sort of problems, I usually wind up smoothing the array > [optional], and then using scipy.ndimage.maximum to calculate peak > values within a given window, and then find the points that equal the > peak values -- this works pretty robustly for peak/trough detection in > arbitrary dimension.
(Sorry, I should reply when I'm not able to actually pay attention) your code works fine with a hump, I didn't read the example carefully enough >>> x = np.arange(10,0,-1) >>> x2 = 25-(x-5)**2 >>> x2 array([ 0, 9, 16, 21, 24, 25, 24, 21, 16, 9]) >>> T=20 >>> np.argmax(x2<T) 0 >>> np.arange(x2.shape[0])[x2 >= T][-1]+1 8 >>> len(x)-np.argmax(x2[::]>T) + 1 8 Josef > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
