Get the 1-norm of a vector in Python

Question

How can I calculate the 1-norm of the difference of two vectors, ||a - b||_1 = sum(|a_i - b_i|) in Python?

a = [1,2,3,4]  
b = [2,3,4,5]

||a - b||_1 = 4  

Answer 1

Python has powerful built-in types, but Python lists are not mathematical vectors or matrices. You could do this with lists, but it will likely be cumbersome for anything more than trivial operations.

If you find yourself needing vector or matrix arithmetic often, the standard in the field is NumPy, which probably already comes packaged for your operating system the way Python also was.

I share the confusion of others about exactly what it is you're trying to do, but perhaps the numpy.linalg.norm function will help:

>>> import numpy
>>> a = numpy.array([1, 2, 3, 4])
>>> b = numpy.array([2, 3, 4, 5])
>>> numpy.linalg.norm((a - b), ord=1)
4

To show how that's working under the covers:

>>> a
array([1, 2, 3, 4])
>>> b
array([2, 3, 4, 5])
>>> (a - b)
array([-1, -1, -1, -1])
>>> numpy.linalg.norm((a - b))
2.0
>>> numpy.linalg.norm((a - b), ord=1)
4

Answer 2

In NumPy, for two vectors a and b, this is just

numpy.linalg.norm(a - b, ord=1)

Answer 3

You appear to be asking for the sum of the differences between the paired components of the two arrays:

>>> A=[1,2,3,4]
>>> B=[2,3,4,5]
>>> sum(abs(a - b) for a, b in zip(A, B))
4

Answer 4

It is not clear what exactly is required here, but here is my guess

a=[1,2,3,4]
b=[2,3,4,5]
def a_b(a,b):
    return sum(map(lambda a:abs(a[0]-a[1]), zip(a,b)))

print a_b(a,b)

Answer 5

Using Numpy you can calculate any norm between two vectors using the linear algebra package.

 import numpy as np
 a = np.array([[2,3,4])
 b = np.array([0,-1,7])
      
 # L1 Norm
 np.linalg.norm(a-b, ord=1)

 # L2 Norm
 np.linalg.norm(a-b, ord=2)

# L3 Norm
 np.linalg.norm(a-b, ord=3)

# Ln Norm
 np.linalg.norm(a-b, ord=n)

Example: