[python] Subtracting 2 lists in Python

Right now I have vector3 values represented as lists. is there a way to subtract 2 of these like vector3 values, like

[2,2,2] - [1,1,1] = [1,1,1]

Should I use tuples?

If none of them defines these operands on these types, can I define it instead?

If not, should I create a new vector3 class?

If your lists are a and b, you can do:

map(int.__sub__, a, b)

But you probably shouldn't. No one will know what it means.

A slightly different Vector class.

class Vector( object ):
    def __init__(self, *data):
        self.data = data
    def __repr__(self):
        return repr(self.data) 
    def __add__(self, other):
        return tuple( (a+b for a,b in zip(self.data, other.data) ) )  
    def __sub__(self, other):
        return tuple( (a-b for a,b in zip(self.data, other.data) ) )

Vector(1, 2, 3) - Vector(1, 1, 1)

The combination of map and lambda functions in Python is a good solution for this kind of problem:

a = [2,2,2]
b = [1,1,1]
map(lambda x,y: x-y, a,b)

zip function is another good choice, as demonstrated by @UncleZeiv

For the one who used to code on Pycharm, it also revives others as well.

 import operator
 Arr1=[1,2,3,45]
 Arr2=[3,4,56,78]
 print(list(map(operator.sub,Arr1,Arr2)))

I'd have to recommend NumPy as well

Not only is it faster for doing vector math, but it also has a ton of convenience functions.

If you want something even faster for 1d vectors, try vop

It's similar to MatLab, but free and stuff. Here's an example of what you'd do

from numpy import matrix
a = matrix((2,2,2))
b = matrix((1,1,1))
ret = a - b
print ret
>> [[1 1 1]]

Boom.

This answer shows how to write "normal/easily understandable" pythonic code.

I suggest not using zip as not really everyone knows about it.


The solutions use list comprehensions and common built-in functions.

Alternative 1 (Recommended):

a = [2, 2, 2]
b = [1, 1, 1]
result = [a[i] - b[i] for i in range(len(a))]

Recommended as it only uses the most basic functions in Python

Alternative 2:

a = [2, 2, 2]
b = [1, 1, 1]
result = [x - b[i] for i, x in enumerate(a)]

Alternative 3 (as mentioned by BioCoder):

a = [2, 2, 2]
b = [1, 1, 1]
result = list(map(lambda x, y: x - y, a, b))

Here's an alternative to list comprehensions. Map iterates through the list(s) (the latter arguments), doing so simulataneously, and passes their elements as arguments to the function (the first arg). It returns the resulting list.

import operator
map(operator.sub, a, b)

This code because has less syntax (which is more aesthetic for me), and apparently it's 40% faster for lists of length 5 (see bobince's comment). Still, either solution will work.

import numpy as np
a = [2,2,2]
b = [1,1,1]
np.subtract(a,b)

If you have two lists called 'a' and 'b', you can do: [m - n for m,n in zip(a,b)]

If you plan on performing more than simple one liners, it would be better to implement your own class and override the appropriate operators as they apply to your case.

Taken from Mathematics in Python:

class Vector:

  def __init__(self, data):
    self.data = data

  def __repr__(self):
    return repr(self.data)  

  def __add__(self, other):
    data = []
    for j in range(len(self.data)):
      data.append(self.data[j] + other.data[j])
    return Vector(data)  

x = Vector([1, 2, 3])    
print x + x

arr1=[1,2,3]
arr2=[2,1,3]
ls=[arr2-arr1 for arr1,arr2 in zip(arr1,arr2)]
print(ls)
>>[1,-1,0]

Try this:

list(array([1,2,3])-1)