Calling a lambda with a numpy array

Question

While familiarizing myself with numpy, I noticed an interesting behaviour in numpy arrays:

import numpy as np
arr = np.array([1, 2, 3])
scale = lambda x: x * 3
scale(arr) # Gives array([3, 6, 9])

Contrast this with normal Python lists:

arr = [1, 2, 3]
scale = lambda x: x * 3
scale(arr) # Gives [1, 2, 3, 1, 2, 3, 1, 2, 3]

I'm curious as to how this is possible. Does a numpy array override the multiplication operator or something?

• juanpa.arrivillaga: This has nothing to do with lambda... anyway, numpy arrays override most operators to perform vectorized operations. That's one of the key features of numpy...
• juanpa.arrivillaga: see docs.scipy.org/doc/numpy-dev/user/…

Answer 1

numpy.ndarray overloads the * operator by defining its own __mul__ method. Likewise for +, -, etc. This allows for vector arithmetic.

Answer 2

Its all about Overriding operators in numpy

You can learn numpy.arry here

Let us focus on your lamda function for each;

1. numpy array :

arr = numpy.array([1, 2, 3])
type(arr)
scale = lambda x: x * 3 
scale(arr)

this takes each element from array

2. normal list:

a =[1,2,3]
type(a)
scale = lambda x: x * 3 
scale(a)

this takes full list as x and multiplies the list here itself

Answer 3

These are two different objects which behaves differently when you use * operator on them.

1. In the first case you generate a numpy array. In this case, * operator was overloaded for performing multiplication. i.e. every element will be multiplied by 3.

2. In the second case you generate a list. In this case the * operator is treated as a repetition operator, and the entire list is repeated 3 times.

code example:

type(np.array([1,2,3]))
type([1, 2, 3])

result:

numpy.ndarray
list

• higgs241: I think the types are reversed
• Christian: I now fixed it.