[Numpy-discussion] Addressing arrays

[Brett Olsen]

On Mon, Jan 30, 2012 at 11:31 AM, Ted To <rainexpected at theo.to> wrote:
> On 01/30/2012 12:13 PM, Brett Olsen wrote:
>> On Mon, Jan 30, 2012 at 10:57 AM, Ted To <rainexpected at theo.to> wrote:
>>> Sure thing.  To keep it simple suppose I have just a two dimensional
>>> array (time,output):
>>> [(1,2),(2,3),(3,4)]
>>> I would like to look at all values of output for which, for example time==2.
>>>
>>> My actual application has a six dimensional array and I'd like to look
>>> at the contents using one or more of the first three dimensions.
>>>
>>> Many thanks,
>>> Ted
>>
>> Couldn't you just do something like this with boolean indexing:
>>
>> In [1]: import numpy as np
>>
>> In [2]: a = np.array([(1,2),(2,3),(3,4)])
>>
>> In [3]: a
>> Out[3]:
>> array([[1, 2],
>>        [2, 3],
>>        [3, 4]])
>>
>> In [4]: mask = a[:,0] == 2
>>
>> In [5]: mask
>> Out[5]: array([False,  True, False], dtype=bool)
>>
>> In [6]: a[mask,1]
>> Out[6]: array([3])
>>
>> ~Brett
>
> Thanks!  That works great if I only want to search over one index but I
> can't quite figure out what to do with more than a single index.  So
> suppose I have a labeled, multidimensional array with labels 'month',
> 'year' and 'quantity'.  a[['month','year']] gives me an array of indices
> but "a[['month','year']]==(1,1960)" produces "False".  I'm sure I simply
> don't know the proper syntax and I apologize for that -- I'm kind of new
> to numpy.
>
> Ted

You'd want to update your mask appropriately to get everything you
want to select, one criteria at a time e.g.:
mask = a[:,0] == 1
mask &= a[:,1] == 1960

Alternatively:
mask = (a[:,0] == 1) & (a[:,1] == 1960)
but be careful with the parens, & and | are normally high-priority
bitwise operators and if you leave the parens out, it will try to
bitwise-and 1 and a[:,1] and throw an error.

If you've got a ton of parameters, you can combine these more
aesthetically with:
mask = (a[:,[0,1]] == [1, 1960]).all(axis=1)

~Brett