python中的4维零点数组

Question

I want to make an 4 dimensional array of zeros in python. I know how to do this for a square array but I want the lists to have different lengths.

Right now I use this:

numpy.zeros((200,)*4)

Which gives them all length 200 but I would like to have lengths 200,20,100,20 because now I have a lot of zeros in my array that I don't use

Answer 1

You can use np.full :

>>> np.full((200,20,10,20), 0)

numpy.full

Return a new array of given shape and type, filled with fill_value.

Example :

>>> np.full((1,3,2,4), 0)
array([[[[ 0.,  0.,  0.,  0.],
         [ 0.,  0.,  0.,  0.]],

        [[ 0.,  0.,  0.,  0.],
         [ 0.,  0.,  0.,  0.]],

        [[ 0.,  0.,  0.,  0.],
         [ 0.,  0.,  0.,  0.]]]])

Answer 2

You can pass more than one arg to shape:

shape : int or sequence of ints Shape of the new array, eg, (2, 3) or 2.

In [26]: arr = np.zeros((200, 20, 10, 20))

In [27]: arr.shape
Out[27]: (200, 20, 10, 20)

It also seems a lot more efficient when you have large dimensions:

In [43]: timeit arr = np.full((200, 100, 100, 100),0)
1 loops, best of 3: 232 ms per loop

In [44]: timeit arr = np.zeros((200, 100, 100, 100))
100000 loops, best of 3: 12.6 µs per loop
In [45]: timeit arr = np.zeros((500, 100, 100, 100))
100000 loops, best of 3: 13.5 µs per loop    
In [46]: timeit arr = np.full((500, 100, 100, 100),0)
1 loops, best of 3: 1.19 s per loop

Answer 3

As of Python v. 3.50, using the np.full command returns

FutureWarning: in the future, full((1, 3, 2, 4), 0) will return an array of dtype('int32') .

Based on this, I'd plug @Padriac's answer. Both work for now, though!