On Fri, Mar 19, 2010 at 11:17 AM, Keith Goodman <kwgood...@gmail.com> wrote:
> On Fri, Mar 19, 2010 at 7:53 AM, gerardob <gberbeg...@gmail.com> wrote:
>>
>> Hello, i would like to produce lists of lists 1's and 0's.
>>
>> For example, to produce the list composed of:
>>
>> L = [[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]
>>
>> I just need to do the following:
>>
>> n=4
>> numpy.eye(n,dtype=int).tolist()
>>
>> I would like to know a simple way to generate a list containing all the
>> lists having two 1's at each element.
>>
>> Example, n = 4
>> L2 = [[1,1,0,0],[1,0,1,0],[1,0,0,1],[0,1,1,0],[0,1,0,1],[0,0,1,1]]
>>
>> Any ideas?
>> Thanks.
>
> Here's the brute force way:
>
>>> for i in range(4):
> ....: for j in range(i+1, 4):
> ....: x = np.zeros(4)
> ....: x[i] = 1
> ....: x[j] = 1
> ....: print x
> ....:
> ....:
> [ 1. 1. 0. 0.]
> [ 1. 0. 1. 0.]
> [ 1. 0. 0. 1.]
> [ 0. 1. 1. 0.]
> [ 0. 1. 0. 1.]
> [ 0. 0. 1. 1.]
here are two numpy version, but they don't take any shortcuts
>>> a= np.array(list(np.ndindex(*2*np.ones(4))))
>>> a
array([[0, 0, 0, 0],
[0, 0, 0, 1],
[0, 0, 1, 0],
[0, 0, 1, 1],
[0, 1, 0, 0],
[0, 1, 0, 1],
[0, 1, 1, 0],
[0, 1, 1, 1],
[1, 0, 0, 0],
[1, 0, 0, 1],
[1, 0, 1, 0],
[1, 0, 1, 1],
[1, 1, 0, 0],
[1, 1, 0, 1],
[1, 1, 1, 0],
[1, 1, 1, 1]])
>>> a[a.sum(1)==2]
array([[0, 0, 1, 1],
[0, 1, 0, 1],
[0, 1, 1, 0],
[1, 0, 0, 1],
[1, 0, 1, 0],
[1, 1, 0, 0]])
>>> [list(r) for r in np.ndindex(*2*np.ones(4)) if sum(r)==2]
[[0, 0, 1, 1], [0, 1, 0, 1], [0, 1, 1, 0], [1, 0, 0, 1], [1, 0, 1, 0],
[1, 1, 0, 0]]
Josef
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