Trying to vectorize iterative calculation with numpy

Question

I am trying to make some piece of code more efficient by using the vectorized form in numpy. Let me show you an example so you know what I mean.

Given the following code:

a = np.zeros([4,4])
a[0] = [1., 2., 3., 4.]
for i in range(len(a)-1):
    a[i+1] = 2*a[i]
print a

It outputs

[[  1.   2.   3.   4.]
 [  2.   4.   6.   8.]
 [  4.   8.  12.  16.]
 [  8.  16.  24.  32.]]

When I now try to vectorize the code like this:

a = np.zeros([4,4])
a[0] = [1., 2., 3., 4.]
a[1:] = 2*a[0:-1]
print a

I just get the first iteration correct:

[[ 1.  2.  3.  4.]
 [ 2.  4.  6.  8.]
 [ 0.  0.  0.  0.]
 [ 0.  0.  0.  0.]]

Is it possible to write the code above efficiently in a vectorized form (where the next iteration always accesses the previous iteration) or do I have to keep the for loop?

Community · Accepted Answer · 2017-05-23 12:24:06Z

A linear recurrence such as this can be computed using scipy.signal.lfilter:

In [19]: from scipy.signal import lfilter

In [20]: num = np.array([1.0])

In [21]: alpha = 2.0

In [22]: den = np.array([1.0, -alpha])

In [23]: a = np.zeros((4,4))

In [24]: a[0,:] = [1,2,3,4]

In [25]: lfilter(num, den, a, axis=0)
Out[25]: 
array([[  1.,   2.,   3.,   4.],
       [  2.,   4.,   6.,   8.],
       [  4.,   8.,  12.,  16.],
       [  8.,  16.,  24.,  32.]])

See the following for more details: python recursive vectorization with timeseries, Recursive definitions in Pandas

Note that using lfilter really only makes sense if you are solving a nonhomogeneous problem such as x[i+1] = alpha*x[i] + u[i], where u is a given input array. For the simple recurrence a[i+1] = alpha*a[i], you can use the exact solution a[i] = a[0]*alpha**i. The solution for multiple initial values can be vectorized using broadcasting. For example,

In [271]: alpha = 2.0

In [272]: a0 = np.array([1, 2, 3, 4])

In [273]: n = 5

In [274]: a0 * (alpha**np.arange(n).reshape(-1, 1))
Out[274]: 
array([[  1.,   2.,   3.,   4.],
       [  2.,   4.,   6.,   8.],
       [  4.,   8.,  12.,  16.],
       [  8.,  16.,  24.,  32.],
       [ 16.,  32.,  48.,  64.]])

Frank M · Accepted Answer · 2014-12-19 15:24:54Z

5

Numpy's vector calculations act on the vector, not as a sequence of steps, so you have to vectorize the entire expression. For example:

np.multiply(np.arange(1,5), 2**np.arange(0,4)[np.newaxis].T)

To address the "final" question, yes you have to keep the for loop if you want to do a sequential calculation. You might make it more efficient with map or [... for ...] but optimizing that way takes a lot of trial and error. The beauty of thinking in vectorial terms and using Numpy to implement is that you get a result efficiently without all the trial and error.

The cumsum and cumprod functions can do something similar to what you're asking for. Instead of 2**np.arange(...), you can get the same thing from

np.multiply(np.arange(1,5), np.cumprod([1,2,2,2,])[np.newaxis].T)

edited Dec 19, 2014 at 15:24

answered Dec 19, 2014 at 15:18

Frank M

1,59015 silver badges16 bronze badges

Comments

basic_bgnr · Accepted Answer · 2014-12-20 05:07:44Z

2

looking at your result matrix

result = [[ 1, 2,   3, 4,]
          [ 2, 4,   6, 8,]
          [ 4, 8,  12, 16,]
          [ 8, 16, 24, 32,]]

it can be deconstructed into product(elem-wise) of two matrix as

a = [[1, 2, 3, 4],
    [1, 2, 3, 4],
    [1, 2, 3, 4],
    [1, 2, 3, 4]]

b = [[1, 1, 1, 1],
     [2, 2, 2, 2],
     [4, 4, 4, 4]
     [8, 8, 8, 8]]

result = a * b

you can calculate this type of operation using the meshgrid function

aa, bb = np.meshgrid(np.array([1.0, 2.0, 3.0, 4.0]),
                     np.array([1.0, 2.0, 4.0, 8.0]))
result = aa * bb

answered Dec 20, 2014 at 5:07

basic_bgnr

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