# [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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>
```