# How to add a new row to an empty numpy array

Using standard Python arrays, I can do the following:

```
arr = []
arr.append([1,2,3])
arr.append([4,5,6])
# 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
```

From: stackoverflow.com/q/22392497