问题
I am working with np.arrays. I am trying to remove the last n elements, where n can be also 1.
n=5
corr=np.full(10,10)
Usually I use this approach with array slicing:
corr=corr[:-n]
But I was thinking of using np.delete to increase the performance:
np.delete(corr,range(-n,0))
But it doesn't work, is there any better solution compare with array slicing? (method able to deal also with case in which n=0, would be a point of advantage)
回答1:
An array is an object with attributes like shape, dtype, and a data buffer. A view like A[:-5] is another array with its own shape, etc, but with a shared data buffer. It's looking at the same data, but only sees a slice.
A[:-5].copy() will appear to be the same, but will have its own data buffer, a copy of selected elements from A.
There's no way of changing the size of the data buffer of A.
np.delete returns a new array with its own data buffer. It uses various methods depending on the shape and delete pattern. It all cases it is a copy, and slower than slicing.
回答2:
Use corr[0:corr.size-n]. this is the faster way since it is only a view.
np.delete is a copy of the remainding elements.
In [9]: %timeit corr[0:corr.size-5]
1000000 loops, best of 3: 1.45 µs per loop
In [10]: %timeit np.delete(corr,range(corr.size-5,corr.size))
10000 loops, best of 3: 145 µs per loop
来源：https://stackoverflow.com/questions/36480086/more-efficient-way-to-remove-last-n-values-fom-np-array