How to add a new row to an empty numpy array

Using standard Python arrays, I can do the following:

    arr = []
    # arr is now [[1,2,3],[4,5,6]]

However, I cannot do the same thing in numpy. For example:

    arr = np.array([])
    arr = np.append(arr, np.array([1,2,3]))
    arr = np.append(arr, np.array([4,5,6]))
    # arr is now [1,2,3,4,5,6]

I also looked into vstack, but when I use vstack on an empty array, I get:

    ValueError: all the input array dimensions except for the concatenation axis must match exactly

So how do I do append a new row to an empty array in numpy?

The way to "start" the array that you want is:

    arr = np.empty((0,3), int)

Which is an empty array but it has the proper dimensionality.

    >>> arr
    array([], shape=(0, 3), dtype=int64)

Then be sure to append along axis 0:

    arr = np.append(arr, np.array([[1,2,3]]), axis=0)
    arr = np.append(arr, np.array([[4,5,6]]), axis=0)

But, @jonrsharpe is right. In fact, if you're going to be appending in a loop, it would be much faster to append to a list as in your first example, then convert to a numpy array at the end, since you're really not using numpy as intended during the loop:

    In [210]: %%timeit
       .....: l = []
       .....: for i in xrange(1000):
       .....:     l.append([3*i+1,3*i+2,3*i+3])
       .....: l = np.asarray(l)
    1000 loops, best of 3: 1.18 ms per loop

    In [211]: %%timeit
       .....: a = np.empty((0,3), int)
       .....: for i in xrange(1000):
       .....:     a = np.append(a, 3*i+np.array([[1,2,3]]), 0)
    100 loops, best of 3: 18.5 ms per loop

    In [214]: np.allclose(a, l)
    Out[214]: True

The numpythonic way to do it depends on your application, but it would be more like:

    In [220]: timeit n = np.arange(1,3001).reshape(1000,3)
    100000 loops, best of 3: 5.93 µs per loop

    In [221]: np.allclose(a, n)
    Out[221]: True