[SciPy-user] comparing two lists/arrays
Chris Colbert
sccolbert@gmail....
Wed Jul 15 09:13:27 CDT 2009
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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