python: for loop compact representation

Question

Python, Numpy

Is there a more compact way to operate on array elements, without having to use the standard for loop.?

For example, consider the function below:

filterData(A):
    B = numpy.zeros(len(A));
    B[0] = (A[0] + A[1])/2.0;
    for i in range(1, len(A)): 
        B[i] = (A[i]-A[i-1])/2.0;
    return B;

check out [np.diff](http://docs.scipy.org/doc/numpy/reference/generated/numpy.diff.html), which works on both numpy arrays and python native arrays — Cam.Davidson.Pilon, Dec 30 '12 at 03:37
I think you should look at this question here: http://stackoverflow.com/questions/1156087/python-search-in-lists-of-lists — Diego Garcia, Dec 30 '12 at 03:39

score 5 · Answer 1 · answered Dec 30 '12 at 03:38

5

Numpy has a diff operator that works on both numpy arrays and Python native arrays. You can rewrite your code as:

def filterData(A):
    B = numpy.zeros(len(A));
    B[1:] = np.diff( A )/2.0
    B[0] = (A[0] + A[1])/2.0;
    return B

answered Dec 30 '12 at 03:38

Cam.Davidson.Pilon

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score 1 · Answer 2 · answered Feb 10 '13 at 14:57

There's also numpy.ediff1d, which allows you to explicitly prepend or append to the diff using the to_end and to_begin parameters, e.g.:

>>> import numpy as np
>>> a = np.arange(10.)
>>> diff = np.ediff1d(a,to_begin = a[:2].sum()) / 2.
>>> diff
array([ 0.5,  0.5,  0.5,  0.5,  0.5,  0.5,  0.5,  0.5,  0.5,  0.5])
>>> diff.size
10

python: for loop compact representation

2 Answers2