3

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

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

Is there a better way than writing the brackets explicitly?

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  • Also I am not entirely sure whether it is a good way of constructing such a tensor from the bottom up using np.concatenate. Commented Aug 31, 2016 at 1:36
  • What are you going to do with that empty array? Commented Aug 31, 2016 at 1:37
  • Iteratively adding rows of that rank-1 using np.concatenate(a,b,axis=0). Commented Aug 31, 2016 at 1:44
  • 2
    It is a very, very bad idea: if you know the final size of your tensor allocate all of it with np.empty from the get go. If you need to build it "from the bottom up", you are very likely better off using nested Python lists, then calling np.array on the result. Commented Aug 31, 2016 at 10:21

5 Answers 5

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5

You can use empty or zeros.

For example, to create a new array of 2x3, filled with zeros, use: numpy.zeros(shape=(2,3))

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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
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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.]]]]]])
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1

You could directly use the ndarray constructor:

numpy.ndarray(shape=(1,) * 6)

Or the empty variant, since it seems to be more popular:

numpy.empty(shape=(1,) * 6)
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1

This should do it:

x = np.array([])
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