[Numpy-discussion] How to Keep An Array Two Dimensional
Hi, I am trying to extract n columns from an 2D array and then operate on the extracted columns. Below is the code: A is an MxN 2D array. u = A[:,:n] #extract the first n columns from A B = np.dot(u, u.T) #take outer product. This code works when n1. However, when n=1, u becomes an 1D array instead of an Mx1 2D array and the code breaks down. I wonder if there is any way to keep u=A[:,:n] an Mxn array no matter what value n takes. I do not want to use matrix because array is more convenient in other places. Thanks, Tom ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] How to Keep An Array Two Dimensional
On Sun, Nov 25, 2012 at 8:24 PM, Tom Bennett tom.benn...@mail.zyzhu.netwrote: Hi, I am trying to extract n columns from an 2D array and then operate on the extracted columns. Below is the code: A is an MxN 2D array. u = A[:,:n] #extract the first n columns from A B = np.dot(u, u.T) #take outer product. This code works when n1. However, when n=1, u becomes an 1D array instead of an Mx1 2D array and the code breaks down. I wonder if there is any way to keep u=A[:,:n] an Mxn array no matter what value n takes. I do not want to use matrix because array is more convenient in other places. Tom, Your example works for me: In [1]: np.__version__ Out[1]: '1.6.2' In [2]: A = arange(15).reshape(3,5) In [3]: A Out[3]: array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14]]) In [4]: u = A[:,:1] In [5]: u Out[5]: array([[ 0], [ 5], [10]]) In [6]: B = np.dot(u, u.T) In [7]: B Out[7]: array([[ 0, 0, 0], [ 0, 25, 50], [ 0, 50, 100]]) Warren Thanks, Tom ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] How to Keep An Array Two Dimensional
Thanks for the quick response. Ah, I see. There is a difference between A[:,:1] and A[:,0]. The former returns an Mx1 2D array whereas the latter returns an M element 1D array. I was using A[:,0] in the code but A[:,:1] in the example. On Sun, Nov 25, 2012 at 8:35 PM, Warren Weckesser warren.weckes...@gmail.com wrote: On Sun, Nov 25, 2012 at 8:24 PM, Tom Bennett tom.benn...@mail.zyzhu.netwrote: Hi, I am trying to extract n columns from an 2D array and then operate on the extracted columns. Below is the code: A is an MxN 2D array. u = A[:,:n] #extract the first n columns from A B = np.dot(u, u.T) #take outer product. This code works when n1. However, when n=1, u becomes an 1D array instead of an Mx1 2D array and the code breaks down. I wonder if there is any way to keep u=A[:,:n] an Mxn array no matter what value n takes. I do not want to use matrix because array is more convenient in other places. Tom, Your example works for me: In [1]: np.__version__ Out[1]: '1.6.2' In [2]: A = arange(15).reshape(3,5) In [3]: A Out[3]: array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14]]) In [4]: u = A[:,:1] In [5]: u Out[5]: array([[ 0], [ 5], [10]]) In [6]: B = np.dot(u, u.T) In [7]: B Out[7]: array([[ 0, 0, 0], [ 0, 25, 50], [ 0, 50, 100]]) Warren Thanks, Tom ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] How to Keep An Array Two Dimensional
On Sun, Nov 25, 2012 at 9:47 PM, Tom Bennett tom.benn...@mail.zyzhu.net wrote: Thanks for the quick response. Ah, I see. There is a difference between A[:,:1] and A[:,0]. The former returns an Mx1 2D array whereas the latter returns an M element 1D array. I was using A[:,0] in the code but A[:,:1] in the example. You'll notice that Python lists and tuples work the same way: foo[0] on a list or tuple gives you the first element whereas foo[:1] gives you a list or tuple containing only the first element. To clarify what's going on in the case of NumPy: when you use the [:, 0] syntax, the interpreter is calling A.__getitem__ with the tuple (slice(None), 0) as the argument. When you use [:, :1], the argument is (slice(None), slice(None, 1)). You can try this out with A[(slice(None), slice(None, 1))] -- it does the same thing (creating index tuples explicitly like this can be very handy in certain cases). The rule that NumPy follows for index tuples is (approximately) that scalar indices always squash the corresponding dimension, whereas slices or iterables (in the case of fancy indexing) preserve the dimension with an appropriate size. Notably, A[:, [0]] will also return an (A.shape[0], 1) array. But the semantics here are different: because using a sequence as an advanced indexing operation, a copy is made, whereas A[:, :1] will return a view. Hope that makes things less mysterious, David ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
