Indexing other values in a numpy array during array vectorization (1D) without loops

Question

I have been tasked with applying a "1-2-1" filter to a numpy array and returning an array of the filtered data. Without using for loops while loops or list comprehension.

The "1-2-1" filter maps each point of data to the average of itself twice and its neighbors. For example, if at some point the data contained ...1, 4, 3... then after applying the "1-2-1" filter the 4 would be replaced with (1 + 4 + 4 + 3) / 4 = 12 / 4 = 3.

For example the numpy array [1, 1, 4, 3, 2]
Would after the filter is applied produce a numpy array [1. 1.75 3. 3. 2. ]

Since the end points of the data do not have two neighbors the 1-2-1 filter is only applied to the internal len(data) - 2 points, leaving the end points unchanged.

Essentially I need to access the values before and after a given point during numpy array vectorization. For a array that could be of any length. Which as much as I have googled I cannot work out.

Corralien · Accepted Answer · 2021-04-25 04:56:07Z

1

Pandas solution

s = pd.Series(l)

>>> s.rolling(3, center=True).sum().add(s).div(4).fillna(s).values
array([1.  , 1.75, 3.  , 3.  , 2.  ])

Step by step:

>>> s.rolling(3, center=True).sum().values
array([nan,  6.,  8.,  9., nan])

>>> s.rolling(3, center=True).sum().add(s).values
array([nan,  7., 12., 12., nan])

>>> s.rolling(3, center=True).sum().add(s).div(4).values
array([ nan, 1.75, 3.  , 3.  ,  nan])

>>> s.rolling(3, center=True).sum().add(s).div(4).fillna(s).values
array([1.  , 1.75, 3.  , 3.  , 2.  ])

Numpy solution

a = np.array(l)

>>> np.concatenate([a[:1], np.convolve(a, [1, 2, 1], mode="valid") / 4, a[-1:]])

Step by step:

>>> np.convolve(a, [1, 2, 1], mode="valid")
array([ 7, 12, 12])

>>> np.convolve(a, [1, 2, 1], mode="valid") / 4
array([1.75, 3.  , 3.  ])

>>> np.concatenate([a[:1], np.convolve(a, [1, 2, 1], mode="valid") / 4, a[-1:]])
array([1.  , 1.75, 3.  , 3.  , 2.  ])

Performance

%timeit s.rolling(3, center=True).sum().add(s).div(4).fillna(s).values
706 µs ± 4.62 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit np.concatenate([a[:1], np.convolve(a, [1, 2, 1], mode="valid") / 4, a[-1:]])
10.8 µs ± 22.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

edited Apr 25, 2021 at 4:56

answered Apr 25, 2021 at 4:29

Corralien

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1 Comment

Jaxsonpd Over a year ago

Sadly I think I need to use numpy I’ll give it a shot when I get home. Thanks heaps though

swag2198 · Accepted Answer · 2021-04-25 05:39:43Z

1

You can try something like this:

import numpy as np
from scipy.linalg import circulant

x = np.array([1, 1, 4, 3, 2])

val = np.array([1, 2, 1])
offsets = np.array([0, 1, 2])
col0 = np.zeros(len(x))
col0[offsets] = val

C = circulant(col0).T[:-(len(val) - 1)]
print(C)

C is essentially a circulant matrix which looks like this:

array([[1., 2., 1., 0., 0.],
       [0., 1., 2., 1., 0.],
       [0., 0., 1., 2., 1.]])

Now you can simply compute the filtered output as follows:

y = (C @ x) / 4
print(y)
# array([1.75, 3.  , 3.  ])

answered Apr 25, 2021 at 5:39

swag2198

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2 Comments

Jaxsonpd Over a year ago

I can only use numpy but thanks heaps that seems like a smarter solution then what my quiz is asking for.

swag2198 Over a year ago

I guess if you can figure out a clever way of constructing the circulant matrix using numpy, that will do it!

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