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
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.]]]])
You can pass more than one arg to shape:
shape : int or sequence of ints Shape of the new array, e.g., (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
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!
