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What is the good python3 equivalent for auto tuple unpacking in lambda?

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What is unpacking of tuple in Python?

Python offers a very powerful tuple assignment tool that maps right hand side arguments into left hand side arguments. THis act of mapping together is known as unpacking of a tuple of values into a norml variable. WHereas in packing, we put values into a regular tuple by means of regular assignment.

How many arguments can Lambda take Python?

Syntax of Lambda in Python You can use as many arguments as you want in a lambda function, but it can have only one expression.

Can Python lambda take two arguments?

Just like a normal function, a Lambda function can have multiple arguments with one expression.

What is tuple packing in Python?

Tuple packing refers to assigning multiple values into a tuple. Tuple unpacking refers to assigning a tuple into multiple variables.


No, there is no other way. You covered it all. The way to go would be to raise this issue on the Python ideas mailing list, but be prepared to argue a lot over there to gain some traction.

Actually, just not to say "there is no way out", a third way could be to implement one more level of lambda calling just to unfold the parameters - but that would be at once more inefficient and harder to read than your two suggestions:

min(points, key=lambda p: (lambda x,y: (x*x + y*y))(*p))

Python 3.8 update

Since the release of Python 3.8, PEP 572 — assignment expressions — have been available as a tool.

So, if one uses a trick to execute multiple expressions inside a lambda - I usually do that by creating a tuple and just returning the last component of it, it is possible to do the following:

>>> a = lambda p:(x:=p[0], y:=p[1], x ** 2 + y ** 2)[-1]
>>> a((3,4))
25

One should keep in mind that this kind of code will seldom be more readable or practical than having a full function. Still, there are possible uses - if there are various one-liners that would operate on this point, it could be worth to have a namedtuple, and use the assignment expression to effectively "cast" the incoming sequence to the namedtuple:

>>> from collections import namedtuple
>>> point = namedtuple("point", "x y")
>>> b = lambda s: (p:=point(*s), p.x ** 2 + p.y ** 2)[-1]

According to http://www.python.org/dev/peps/pep-3113/ tuple unpacking are gone, and 2to3 will translate them like so:

As tuple parameters are used by lambdas because of the single expression limitation, they must also be supported. This is done by having the expected sequence argument bound to a single parameter and then indexing on that parameter:

lambda (x, y): x + y

will be translated into:

lambda x_y: x_y[0] + x_y[1]

Which is quite similar to your implementation.


I don't know any good general alternatives to the Python 2 arguments unpacking behaviour. Here's a couple of suggestion that might be useful in some cases:

  • if you can't think of a name; use the name of the keyword parameter:

    def key(p): # more specific name would be better
        x, y = p
        return x**2 + y**3
    
    result = min(points, key=key)
    
  • you could see if a namedtuple makes your code more readable if the list is used in multiple places:

    from collections import namedtuple
    from itertools import starmap
    
    points = [ (1,2), (2,3)]
    Point = namedtuple('Point', 'x y')
    points = list(starmap(Point, points))
    
    result = min(points, key=lambda p: p.x**2 + p.y**3)
    

While the destructuring arguments was removed in Python3, it was not removed from comprehensions. It is possible to abuse it to obtain similar behavior in Python 3. In essence, we take advantage of the fact that co-routines allow us to turn functions inside out, and yield is not a statement, and hence is allowed within lambdas.

For example:

points = [(1,2), (2,3)]
print(min(points, key=lambda y: next(x*x + y*y for x,y in (lambda a: (yield a))(y))))

In comparison with the accepted answer of using a wrapper, this solution is able to completely destructure the arguments while the wrapper only destructures the first level. That is,

values = [(('A',1),'a'), (('B',0),'b')]
print(min(values, key=lambda y: next(b for (a,b),c in (lambda x: (yield x))(y))))

In comparison to

values = [(('A',1),'a'), (('B',0),'b')]
print(min(points, key=lambda p: (lambda a,b: (lambda x,y: (y))(*a))(*p)))

Alternatively one can also do

values = [(('A',1),'a'), (('B',0),'b')]
print(min(points, key=lambda y: next(b for (a,b),c in [y])))

Or slightly better

print(min(values, key=lambda y: next(b for ((a,b),c) in (y,))))

This is just to suggest that it can be done, and should not be taken as a recommendation.


Consider whether you need to unpack the tuple in the first place:

min(points, key=lambda p: sum(x**2 for x in p))

or whether you need to supply explicit names when unpacking:

min(points, key=lambda p: abs(complex(*p))

I think the better syntax is x * x + y * y let x, y = point, let keyword should be more carefully chosen.

The double lambda is the closest version. lambda point: (lambda x, y: x * x + y * y)(*point)

High order function helper would be useful in case we give it a proper name.

def destruct_tuple(f):
  return lambda args: f(*args)

destruct_tuple(lambda x, y: x * x + y * y)