Yes, thanks, np.in1d is what I needed. I didn't know how to find that.
It still seems counterintuitive to me that
indexes = np.where( records.integer_field in values )
does not work whereas
indexes = np.where( records.integer_field > 5 )
does.
In one case numpy is overriding the > operator; it's not checking if an array
is
greater than 5, but whether each element in the array is greater than 5.
>From a naive user's point of view, not knowing much about the difference
>between
> and in from a python point of view, it seems like in would get overridden the
same way.
________________________________
From: Christopher Barker <[email protected]>
To: Discussion of Numerical Python <[email protected]>
Sent: Fri, May 27, 2011 5:48:37 PM
Subject: Re: [Numpy-discussion] finding elements that match any in a set
On 5/27/11 9:48 AM, Michael Katz wrote:
> I have a numpy array, records, with named fields including a field named
> "integer_field". I have an array (or list) of values of interest, and I
> want to get the indexes where integer_field has any of those values.
>
> Because I can do
>
> indexes = np.where( records.integer_field > 5 )
>
> I thought I could do
>
> indexes = np.where( records.integer_field in values )
>
> But that doesn't work. (As a side question I'm interested in why that
> doesn't work, when values is a python list.)
that doesn't work because the python list "in" operator doesn't
understand arrays -- so it is looking ot see if the entire array is in
the list. actually, it doesn't even get that far:
In [16]: a in l
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
/Users/chris.barker/<ipython console> in <module>()
ValueError: The truth value of an array with more than one element is
ambiguous. Use a.any() or a.all()
The ValueError results because it was decided that numpy array should
not have a boolean value to avoid confusion -- i.e. is na array true
whenever it is non-empty (like a list), or when all it's elements are
true, or????
When I read this question, I thought -- hmmm, numpy needs something like
"in", as the usual way: np.any(), would require a loop in this case.
Then I read Skipper's message:
On 5/27/11 9:55 AM, Skipper Seabold wrote:
> Check out this recent thread. I think the proposed class does what you
> want. It's more efficient than in1d, if values is small compared to
> the length of records.
So that class may be worthwhile, but I think np.in1d is exactly what you
are looking for:
indexes = np.in1d( records.integer_field, values )
Funny I'd never noticed that before.
-Chris
--
Christopher Barker, Ph.D.
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