使用numpy.tile重复2D数组

Question

I have 3 arrays:

e = np.array(range(3,100))
dRdE = np.load('arr_25.npy')

The npy file contains an array with random values but is of the same length as e. I then take the outer product of dRdE with another array.

s = np.array(range(1,100))
dRdE = np.outer(s, dRdE)

And so dRdE is now 2D.

I want e to repeat the number of times of each element in dRdE . I could use dRdE before when it was 1D and of the same length as e using numpy repeat. The code I had before was:

earray = np.repeat(e, dRdE)

But since dRdE is no longer 1D I thought I might be able to use np.tile but I'm not sure exactly how to.

The context might help, so I have an energy varying between 3keV to 100keV and I have an array containing the number of events ( dRdE ) for each energy between that range. But now I want to introduce a third variable that is just some factor of dRdE between 1 to 100. So for example if:

dRdE= [1,2,1,2..],[2,4,2,4..],[3,6,3,6]

I want:

earray=([3,4,4,5,6,6,...],[3,3,4,4,4,4,5,5,6,6,6,6,...],[3,3,3,4...])

Any help would be appreciated!

Answer 1

I asked about variable length sublists because that is a clear indicator that some sort of iteration is required. There aren't many numpy operations that produce a list or tuple of differing length arrays.

Looking into tile a bit more (it's not something I use everyday), I don't think it helps at all. Its reps are one value per dimension , not per element .

But lets explore repeat .

In [73]: vals=[[1,2,3,4],[11,12,13,14],[21,22,23,24]]
In [74]: reps=[[1,2,1,2],[2,3,2,4],[3,6,3,6]]
In [75]: valA=np.array(vals)
In [76]: repA=np.array(reps)

An iterative solution:

In [77]: [np.repeat(v,r) for v,r in zip(valA,repA)]
Out[77]: 
[array([1, 2, 2, 3, 4, 4]),
 array([11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 14]),
 array([21, 21, 21, 22, 22, 22, 22, 22, 22, 23, 23, 23, 24, 24, 24, 24, 24,
       24])]

We could also apply repeat once

In [78]: np.repeat(valA.flat,repA.flat)
Out[78]: 
array([ 1,  2,  2,  3,  4,  4, 11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 14,
       21, 21, 21, 22, 22, 22, 22, 22, 22, 23, 23, 23, 24, 24, 24, 24, 24,
       24])

Now if only we could split that array into 3 arrays, each 6,11,18 elements long.

In [79]: [len(np.repeat(v,r)) for v,r in zip(valA,repA)]
Out[79]: [6, 11, 18]

In [80]: np.split(np.repeat(valA.flat,repA.flat),[6,17])
Out[80]: 
[array([1, 2, 2, 3, 4, 4]),
 array([11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 14]),
 array([21, 21, 21, 22, 22, 22, 22, 22, 22, 23, 23, 23, 24, 24, 24, 24, 24,
       24])]

I'm sure there's a faster way way of calculating [6,17] list.

But when I do a timeit I find that the list comprehension is faster (10x) than the split. It's even faster that the single repeat. Timings will vary with larger arrays, but getting around the list iteration does not look promising.