[SciPy-User] Choosing sub array based on values in a column
Zachary Pincus
zachary.pincus@yale....
Tue Mar 23 06:06:03 CDT 2010
Y[['var3','var4']][Y['var1']==0 and Y['var2']==0]
should be
Y[['var3','var4']][(Y['var1']==0) & (Y['var2']==0)]
Long story, explained elsewhere, but python's 'and' is expected to
coerce into a single boolean True or False, which is ambiguous for an
array ('any' vs. 'all' interpretation), so it just raises an error
instead. To do elementwise logical operations like you want, use the
bitwise operators & | ^ on boolean arrays, but note the == operator
has lower precedence so parenthesization is a must. Or use
logical_and(), logical_or(), etc.
Zach
On Mar 22, 2010, at 11:00 PM, Vincent Davis wrote:
> I feel kinda stupid as I think this must be easier than I am making
> it. Below my question you will see an answer I got to a question
> that I thought I would be able to complete the last steps my self I
> was wrong :)
>
> So if I have an array
>
>
> Y = np.rec.array([(1.0, 0.0, 3.0, 3.5), (0.0, 0.0, 6.0, 6.5), (1.0,
> 1.0, 9.0, 9.5)], dtype=[('var1', '<f8'), ('var2', '<f8'), ('var3',
> '<f8'), ('var4', '<f8')])
>
>
> do this works like I would expect
>
> Y[['var3','var4']][Y['var1']==1]
>
> >>>array([(3.0, 3.5), (9.0, 9.5)],
>
> dtype=[('var3', '<f8'), ('var4', '<f8')])
>
> But I would like to do this,
>
>
> >>> Y[['var3','var4']][Y['var1']==0 and Y['var2']==0]
>
> Traceback (most recent call last):
>
> File "<string>", line 1, in <fragment>
>
> ValueError: The truth value of an array with more than one element
> is ambiguous. Use a.any() or a.all()
>
> I tried some any() and all() combination but nothing worked. What s
> the right way to go about this?
>
>
>
> Thanks
>
> Vincent
>
>
>
>
>
> This answer was received on the mail list from Skipper Seabold If
> you have a rec array
>
> Y = np.rec.array([(1.0, 2.0, 3.0), (4.0, 5.0, 6.0), (7.0, 8.0,
> 9.0)],dtype=[('var1', ''var2','var3'])
>
> You can access the rows like,Y[['var1','var2','var3']]Note the list
> within [].
>
> If you want a "normal" array, I like this way that Pierre recently
> pointed out. 3 is the number of columns, and it fills in the
> numberof rows.
>
> Y[['var1','var2','var3']].view((float,3))
>
> note the tuple for the view, if they're all floats. Taking a
> viewmight not work if var# have different types, like ints and floats.
>
> If you want the mean of the rows (mean over the columns axis = 1)
> Y[['var1','var2','var3']].view((float,3)).mean(1)Some
> shortcuts.Y[list(Y.dtype.names)].view((float,len(Y.dtype))).mean(1)
>
> Also, for now, the columns will given back to you in the order
> they'rein in the array no matter which way you ask for them. A patch
> hasbeen submitted for returning the order you ask that I hope gets
> picked up...
>
>
> Vincent Davis
> 720-301-3003
> vincent@vincentdavis.net
>
> my blog | LinkedIn
>
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