Is there a way to slice the array below without having to define the row indices i.e. not having to write range(len(X))?
X = np.arange(10*2).reshape((10,2))
L = np.random.randint(0,2,10)
Xs = X[range(len(X)),L]
I thought it was possible to slice with X[:,L] but looks like it's not.
You're probably looking for np.choose:
In [25]: X = np.arange(10*2).reshape((10,2)); X
Out[25]:
array([[ 0, 1],
[ 2, 3],
[ 4, 5],
[ 6, 7],
[ 8, 9],
[10, 11],
[12, 13],
[14, 15],
[16, 17],
[18, 19]])
In [26]: L = np.random.randint(0,2,10); L
Out[26]: array([1, 1, 1, 1, 1, 0, 0, 0, 0, 1])
In [27]: L.choose(X.T)
Out[27]: array([ 1, 3, 5, 7, 9, 10, 12, 14, 16, 19])
In [28]: # or otherwise
In [29]: np.choose(L, X.T)
Out[29]: array([ 1, 3, 5, 7, 9, 10, 12, 14, 16, 19])
Performance note: while this solution is a direct answer to the question, it's quickly becomes not the most optimal with increase of len(X). As of numpy 1.9.0, np.arange approach is faster:
In [17]: %timeit X[range(len(X)), L]
1000 loops, best of 3: 629 µs per loop
In [18]: %timeit X[np.arange(len(X)), L]
10000 loops, best of 3: 78.8 µs per loop
In [19]: %timeit L.choose(X.T)
10000 loops, best of 3: 146 µs per loop
In [20]: X.shape, L.shape
Out[20]: ((10000, 2), (10000,))
np.arange when increasing len(L).You take the diagonal elements of X[:,L] using diag (or diagonal):
np.diag(X[:,L])
Another way to do it is with where:
np.where(L,X[:,1],X[:,0])
diag() may end up returning a view depending on some details, so that N^2 memory usage may stick around for the life of the result unless you explicitly copy it.
where is more efficient for large inputs I think. But it's a bit unsatisfying also, because it doesn't generalize to more than two columns in X. Oh well, it works for the OP's stated case. By the way you can just use L instead of L==1.L. ThanksNote that
In [9]: X[:, L]
Out[9]:
array([[ 1, 1, 0, 0, 1, 0, 1, 0, 1, 0],
[ 3, 3, 2, 2, 3, 2, 3, 2, 3, 2],
[ 5, 5, 4, 4, 5, 4, 5, 4, 5, 4],
[ 7, 7, 6, 6, 7, 6, 7, 6, 7, 6],
[ 9, 9, 8, 8, 9, 8, 9, 8, 9, 8],
[11, 11, 10, 10, 11, 10, 11, 10, 11, 10],
[13, 13, 12, 12, 13, 12, 13, 12, 13, 12],
[15, 15, 14, 14, 15, 14, 15, 14, 15, 14],
[17, 17, 16, 16, 17, 16, 17, 16, 17, 16],
[19, 19, 18, 18, 19, 18, 19, 18, 19, 18]])
And you want the diagonal elements:
So just do:
In [14]: X[:, L].diagonal()
Out[14]: array([ 1, 3, 4, 6, 9, 10, 13, 14, 17, 18])
linspacetakes two arguments minimum: start and end. So your code doesn't run. Is your NumPy different? What version? Also,X[:,L]does work for me, provided that I dolinspace(5, 20, 10*2)or so.np.arange()instead ofrange(). docs.scipy.org/doc/numpy/user/…Lhas shape(10,), but you're using it to index the dimension ofXthat has length 2, not the length 10 one. Is that intentional?