I am quite new to python and have read lots of SO questions on this topic however none of them answers my needs.
I end up with an ndarray:
[[1, 2, 3]
[4, 5, 6]]
Now I want to pad each element (e.g. [1, 2, 3]) with a tailored padding just for that element. Of course I could do it in a for loop and append each result to a new ndarray but isn't there a faster and cleaner way I could apply this over the whole ndarray at once?
I imagined it could work like:
myArray = [[1, 2, 3]
[4, 5, 6]]
paddings = [(1, 2),
(2, 1)]
myArray = np.pad(myArray, paddings, 'constant')
But of course this just outputs:
[[0 0 0 0 0 0 0 0 0]
[0 0 1 2 3 0 0 0 0]
[0 0 3 4 5 0 0 0 0]
[0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0]]
Which is not what i need. The target result would be:
[[0 1 2 3 0 0]
[0 0 4 5 6 0]]
How can I achieve this using numpy?
Here is a loop based solution but with creating a zeros array as per the dimensions of input array and paddings. Explanation in comments:
In [192]: myArray
Out[192]:
array([[1, 2, 3],
[4, 5, 6]])
In [193]: paddings
Out[193]:
array([[1, 2],
[2, 1]])
# calculate desired shape; needed for initializing `padded_arr`
In [194]: target_shape = (myArray.shape[0], myArray.shape[1] + paddings.shape[1] + 1)
In [195]: padded_arr = np.zeros(target_shape, dtype=np.int32)
In [196]: padded_arr
Out[196]:
array([[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0]], dtype=int32)
After this, we can use a for loop to slot fill the sequences from myArray, based on the values from paddings:
In [199]: for idx in range(paddings.shape[0]):
...: padded_arr[idx, paddings[idx, 0]:-paddings[idx, 1]] = myArray[idx]
...:
In [200]: padded_arr
Out[200]:
array([[0, 1, 2, 3, 0, 0],
[0, 0, 4, 5, 6, 0]], dtype=int32)
The reason we've to resort to a loop based solution is because numpy.lib.pad() doesn't yet support this sort of padding, even with all available additional modes and keyword arguments that it already provides.