[SciPy-user] comparing two lists/arrays
nicky van foreest
vanforeest@gmail....
Wed Jul 15 12:59:51 CDT 2009
Hi,
Thanks for your advice.
Nicky
2009/7/15 Chris Colbert <sccolbert@gmail.com>:
> this should work for any case:
>
>>>> y = np.random.rand(5)
>>>> x = np.random.rand(5)
>
>>>> y
> array([ 0.21991179, 0.82874802, 0.65327351, 0.02277029, 0.14618527])
>
>>>> x
> array([ 0.46541554, 0.86719123, 0.50618409, 0.13140126, 0.24533278])
>
>>>> (y - x) > 0
> array([False, False, True, False, False], dtype=bool)
>
>
> On Wed, Jul 15, 2009 at 8:53 AM, Robert Cimrman<cimrman3@ntc.zcu.cz> wrote:
>> Scott Sinclair wrote:
>>>> 2009/7/15 Scott Sinclair <scott.sinclair.za@gmail.com>:
>>>>> 2009/7/15 Robert Cimrman <cimrman3@ntc.zcu.cz>:
>>>>> nicky van foreest wrote:
>>>>>> Given two vectors x and y, the (perhaps) common mathematical
>>>>>> definition of x < y is that x_i < y_i for all i. Thus, the
>>>>>> mathematical comparison x <y returns just one boolean, not an array of
>>>>>> booleans for each x_i < y_i. I implemented this behavior as
>>>>>> prod(less(X,Y)) (I use less to be able to deal with lists X and Y
>>>>>> also). Is there perhaps a more straighforward/elegant/readible way to
>>>>>> achieve the same behavior?
>>>>> assuming x, y are numpy arrays: (x < y).all()
>>>> You could do the following to handle the case where they aren't:
>>>>
>>>>>>> import numpy as np
>>>>>>> x = range(10)
>>>>>>> y = range(1, 11)
>>>>>>> np.all(x < y)
>>>> True
>>>
>>> Scratch that
>>>
>>>>>> x < y
>>> True
>>
>> beware!
>>
>> y[2] = -1
>>
>> In [21]: y
>> Out[21]: [1, 2, -1, 4, 5, 6, 7, 8, 9, 10]
>>
>> In [22]: x < y
>> Out[22]: True
>>
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