I suspect you are trying to replicate this working list code:

In [56]: x = []                                                                 
In [57]: x.append([1,2])                                                        
In [58]: x                                                                      
Out[58]: [[1, 2]]
In [59]: np.array(x)                                                            
Out[59]: array([[1, 2]])

But with arrays:

In [53]: x = np.empty((2,2),int)                                                
In [54]: x                                                                      
Out[54]: 
array([[73096208, 10273248],
       [       2,       -1]])

Despite the name, the np.empty array is NOT a close of the empty list. It has 4 elements, the shape that you specified.

In [55]: np.append(x, np.array([1,2]), axis=0)                                  
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-55-64dd8e7900e3> in <module>
----> 1 np.append(x, np.array([1,2]), axis=0)

<__array_function__ internals> in append(*args, **kwargs)

/usr/local/lib/python3.6/dist-packages/numpy/lib/function_base.py in append(arr, values, axis)
   4691         values = ravel(values)
   4692         axis = arr.ndim-1
-> 4693     return concatenate((arr, values), axis=axis)
   4694 
   4695 

<__array_function__ internals> in concatenate(*args, **kwargs)

ValueError: all the input arrays must have same number of dimensions, but the array at index 0 has 2 dimension(s) and the array at index 1 has 1 dimension(s)

Note that np.append has passed the task on to np.concatenate. With the axis parameter, that's all this append does. It is NOT a list append clone.

np.concatenate demands consistency in the dimensions of its inputs. One is (2,2), the other (2,). Mismatched dimensions.

np.append is a dangerous function, and not that useful even when used correctly. np.concatenate (and the various stack) functions are useful. But you need to pay attention to shapes. And don't use them iteratively. List append is more efficient for that.

When you got this error, did you look up the np.append, np.empty (and np.concatenate) functions? Read and understand the docs? In the long run SO questions aren't a substitute for reading the documentation.

That is the wrong mental model for using NumPy efficiently. NumPy arrays are stored in contiguous blocks of memory. To append rows or columns to an existing array, the entire array needs to be copied to a new block of memory, creating gaps for the new elements to be stored. This is very inefficient if done repeatedly.

Instead of appending rows, allocate a suitably sized array, and then assign to it row-by-row:

>>> import numpy as np

>>> a = np.zeros(shape=(3, 2))
>>> a
array([[ 0.,  0.],
       [ 0.,  0.],
       [ 0.,  0.]])

>>> a[0] = [1, 2]
>>> a[1] = [3, 4]
>>> a[2] = [5, 6]

>>> a
array([[ 1.,  2.],
       [ 3.,  4.],
       [ 5.,  6.]])

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

1. A Hack

You can create an empty array using the np.empty() function and specify the dimensions to be (0, 0) and then use np.append() to append items later.

>>> a = np.empty((0, 0))
>>> a
array([], shape=(0, 0), dtype=float64)
>>> b = np.append(a, [1, 2])
>>> b
array([1., 2.])

2. However...

The hack above is not advisable, use it with caution. Appending to lists has O(N) complexity, while appending to arrays has O(N^2) (besides different memory use). The proper way should be, then, to append to lists. Note that using list() on numpy arrays to transform them into lists is not correct, as you will get a list of numpy arrays. Instead, use the .tolist() method.

>>> a = np.array([[1, 2], [3, 4]])
>>>
>>> list(a)
[array([1, 2]), array([3, 4])]
>>>
>>> a.tolist()
[[1, 2], [3, 4]]