# How can I initialize an empty Numpy array with a given number of dimensions?

## Question:

I basically want to initialize an empty 6-tensor, like this:

```
a = np.array([[[[[[]]]]]])
```

Is there a better way than writing the brackets explicitly?

## Answers:

Iteratively adding rows of that rank-1 using np.concatenate(a,b,axis=0)

Don’t. Creating an array iteratively is slow, since it has to create a new array at each step. Plus `a`

and `b`

have to match in all dimensions except the concatenation one.

```
np.concatenate((np.array([[[]]]),np.array([1,2,3])), axis=0)
```

will give you dimensions error.

The only thing you can concatenate to such an array is an array with size 0 dimenions

```
In [348]: np.concatenate((np.array([[]]),np.array([[]])),axis=0)
Out[348]: array([], shape=(2, 0), dtype=float64)
In [349]: np.concatenate((np.array([[]]),np.array([[1,2]])),axis=0)
------
ValueError: all the input array dimensions except for the concatenation axis must match exactly
In [354]: np.array([[]])
Out[354]: array([], shape=(1, 0), dtype=float64)
In [355]: np.concatenate((np.zeros((1,0)),np.zeros((3,0))),axis=0)
Out[355]: array([], shape=(4, 0), dtype=float64)
```

To work iteratively, start with a empty list, and `append`

to it; then make the array at the end.

`a = np.zeros((1,1,1,1,1,0))`

could be concatenated on the last axis with another `np.ones((1,1,1,1,1,n))`

array.

```
In [363]: np.concatenate((a,np.array([[[[[[1,2,3]]]]]])),axis=-1)
Out[363]: array([[[[[[ 1., 2., 3.]]]]]])
```

You can do something like `np.empty(shape = [1] * (dimensions - 1) + [0])`

.

Example:

```
>>> a = np.array([[[[[[]]]]]])
>>> b = np.empty(shape = [1] * 5 + [0])
>>> a.shape == b.shape
True
```

This should do it:

```
x = np.array([])
```