> I'm having some trouble here. I have a list of numpy arrays. I > want to > know if an array 'u' is in the list.
Try: any(numpy.all(u == l) for l in array_list) standard caveats about float comparisons apply; perhaps any(numpy.allclose(u, l) for l in array_list) is more appropriate in certain circumstances. Can of course replace the first 'any' with 'all' or 'sum' to get different kinds of information, but using 'any' is equivalent to the 'in' query that you wanted. Why the 'in' operator below fails is that behind the scenes, 'u not in [u+1]' causes Python to iterate through the list testing each element for equality with u. Except that as the error states, arrays don't support testing for equality because such tests are ambiguous. (cf. many threads about this.) Zach On Feb 5, 2010, at 6:47 AM, Neal Becker wrote: > I'm having some trouble here. I have a list of numpy arrays. I > want to > know if an array 'u' is in the list. > > As an example, > u = np.arange(10) > > : u not in [u+1] > --------------------------------------------------------------------------- > ValueError Traceback (most recent > call last) > > /home/nbecker/raysat/test/<ipython console> in <module>() > > ValueError: The truth value of an array with more than one element is > ambiguous. Use a.any() or a.all() > > What would be the way to do this? > > _______________________________________________ > 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
