On Fri, Feb 5, 2010 at 8:48 AM, Zachary Pincus zachary.pin...@yale.edu 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.
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?
maybe np.in1d(u, u+1) or np.in1d(u,u+1).all() is what you want
help(np.in1d)
Help on function in1d in module numpy.lib.arraysetops:
in1d(ar1, ar2, assume_unique=False)
Test whether each element of a 1D array is also present in a second array.
Returns a boolean array the same length as `ar1` that is True
where an element of `ar1` is in `ar2` and False otherwise.
Josef
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