Elegant way to perform tuple arithmetic

Question:

What is the most elegant and concise way (without creating my own class with operator overloading) to perform tuple arithmetic in Python 2.7?

Lets say I have two tuples:

a = (10, 10)
b = (4, 4)

My intended result is

c = a - b = (6, 6)

I currently use:

c = (a[0] - b[0], a[1] - b[1])

I also tried:

c = tuple([(i - j) for i in a for j in b])

but the result was (6, 6, 6, 6). I believe the above works as a nested for loops resulting in 4 iterations and 4 values in the result.

Asked By: user1737647

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Answers:

Use zip and a generator expression:

c = tuple(x-y for x, y in zip(a, b))

Demo:

>>> a = (10, 10)
>>> b = (4, 4)
>>> c = tuple(x-y for x, y in zip(a, b))
>>> c
(6, 6)

Use itertools.izip for a memory efficient solution.

help on zip:

>>> print zip.__doc__
zip(seq1 [, seq2 [...]]) -> [(seq1[0], seq2[0] ...), (...)]

Return a list of tuples, where each tuple contains the i-th element
from each of the argument sequences.  The returned list is truncated
in length to the length of the shortest argument sequence.
Answered By: Ashwini Chaudhary

One option would be,

>>> from operator import sub
>>> c = tuple(map(sub, a, b))
>>> c
(6, 6)

And itertools.imap can serve as a replacement for map.

Of course you can also use other functions from operator to add, mul, div, etc.

But I would seriously consider moving into another data structure since I don’t think this type of problem is fit for tuples

Answered By: Jared

If you’re looking for fast, you can use numpy:

>>> import numpy
>>> numpy.subtract((10, 10), (4, 4))
array([6, 6])

and if you want to keep it in a tuple:

>>> tuple(numpy.subtract((10, 10), (4, 4)))
(6, 6)
Answered By: vroomfondel

This can also be done just as nicely without an import at all, although lambda is often undesirable:

tuple(map(lambda x, y: x - y, a, b))

If you are looking to get the distance between two points on say a 2d coordinate plane you should use the absolute value of the subtraction of the pairs.

tuple(map(lambda x ,y: abs(x - y), a, b))
Answered By: miigotu

my element-wise tuple arithmetic helper

supported operations: +, -, /, *, d

operation = ‘d’ calculates distance between two points on a 2D coordinate plane

def tuplengine(tuple1, tuple2, operation):
    """ 
    quick and dirty, element-wise, tuple arithmetic helper,
    created on Sun May 28 07:06:16 2017
    ...
    tuple1, tuple2: [named]tuples, both same length
    operation: '+', '-', '/', '*', 'd'
    operation 'd' returns distance between two points on a 2D coordinate plane (absolute value of the subtraction of pairs)
    """
    assert len(tuple1) == len(tuple2), "tuple sizes doesn't match, tuple1: {}, tuple2: {}".format(len(tuple1), len(tuple2))
    assert isinstance(tuple1, tuple) or tuple in type(tuple1).__bases__, "tuple1: not a [named]tuple"
    assert isinstance(tuple2, tuple) or tuple in type(tuple2).__bases__, "tuple2: not a [named]tuple"
    assert operation in list("+-/*d"), "operation has to be one of ['+','-','/','*','d']"
    return eval("tuple( a{}b for a, b in zip( tuple1, tuple2 ))".format(operation)) 
    if not operation == "d" 
      else eval("tuple( abs(a-b) for a, b in zip( tuple1, tuple2 ))")
Answered By: internety

JFYI, execution time in my laptop in 100000 times iteration

np.subtract(a, b) : 0.18578505516052246

tuple(x - y for x, y in zip(a, b)) :
0.09348797798156738

tuple(map(lambda x, y: x - y, a, b)) : 0.07900381088256836

from operator import sub tuple(map(sub, a, b)) : 0.044342041015625

operator looks more elegant for me.

Answered By: Kohei Kawasaki

As an addition to Kohei Kawasaki’s answer, for speed, the original solution was actually the fastest (For a length two tuple at least).

>>> timeit.timeit('tuple(map(add, a, b))',number=1000000,setup='from operator import add; a=(10,11); b=(1,2)')
0.6502681339999867
>>> timeit.timeit('(a[0] - b[0], a[1] - b[1])',number=1000000,setup='from operator import add; a=(10,11); b=(1,2)')
0.19015854899998885
>>> 
Answered By: user5650823

Since in your question there only examples of 2-number-tuples, for such coordinate-like tuples, you may be good with simple built-in “complex” class:

>>> a=complex(7,5)
>>> b=complex(1,2)
>>> a-b
>>> c=a-b
>>> c
(6+3j)
>>> c.real
6.0
>>> c.imag
3.0
Answered By: Rustam A.
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