Assignment inside lambda expression in Python

Question

I have a list of objects and I want to remove all objects that are empty except for one, using filter and a lambda expression.

For example if the input is:

[Object(name=""), Object(name="fake_name"), Object(name="")]

...then the output should be:

[Object(name=""), Object(name="fake_name")]

Is there a way to add an assignment to a lambda expression? For example:

flag = True 
input = [Object(name=""), Object(name="fake_name"), Object(name="")] 
output = filter(
    (lambda o: [flag or bool(o.name), flag = flag and bool(o.name)][0]),
    input
)

Solution 1

The assignment expression operator := added in Python 3.8 supports assignment inside of lambda expressions. This operator can only appear within a parenthesized (...), bracketed [...], or braced {...} expression for syntactic reasons. For example, we will be able to write the following:

import sys
say_hello = lambda: (
    message := "Hello world",
    sys.stdout.write(message + "\n")
)[-1]
say_hello()

In Python 2, it was possible to perform local assignments as a side effect of list comprehensions.

import sys
say_hello = lambda: (
    [None for message in ["Hello world"]],
    sys.stdout.write(message + "\n")
)[-1]
say_hello()

However, it's not possible to use either of these in your example because your variable flag is in an outer scope, not the lambda's scope. This doesn't have to do with lambda, it's the general behaviour in Python 2. Python 3 lets you get around this with the nonlocal keyword inside of defs, but nonlocal can't be used inside lambdas.

There's a workaround (see below), but while we're on the topic...

---

In some cases you can use this to do everything inside of a lambda:

(lambda: [
    ['def'
        for sys in [__import__('sys')]
        for math in [__import__('math')]
        for sub in [lambda *vals: None]
        for fun in [lambda *vals: vals[-1]]
        for echo in [lambda *vals: sub(
            sys.stdout.write(u" ".join(map(unicode, vals)) + u"\n"))]
        for Cylinder in [type('Cylinder', (object,), dict(
            __init__ = lambda self, radius, height: sub(
                setattr(self, 'radius', radius),
                setattr(self, 'height', height)),
            volume = property(lambda self: fun(
                ['def' for top_area in [math.pi * self.radius ** 2]],
                self.height * top_area))))]
        for main in [lambda: sub(
            ['loop' for factor in [1, 2, 3] if sub(
                ['def'
                    for my_radius, my_height in [[10 * factor, 20 * factor]]
                    for my_cylinder in [Cylinder(my_radius, my_height)]],
                echo(u"A cylinder with a radius of %.1fcm and a height "
                     u"of %.1fcm has a volume of %.1fcm³."
                     % (my_radius, my_height, my_cylinder.volume)))])]],
    main()])()

A cylinder with a radius of 10.0cm and a height of 20.0cm has a volume of 6283.2cm³.
A cylinder with a radius of 20.0cm and a height of 40.0cm has a volume of 50265.5cm³.
A cylinder with a radius of 30.0cm and a height of 60.0cm has a volume of 169646.0cm³.

Please don't.

---

...back to your original example: though you can't perform assignments to the flag variable in the outer scope, you can use functions to modify the previously-assigned value.

For example, flag could be an object whose .value we set using setattr:

flag = Object(value=True)
input = [Object(name=''), Object(name='fake_name'), Object(name='')] 
output = filter(lambda o: [
    flag.value or bool(o.name),
    setattr(flag, 'value', flag.value and bool(o.name))
][0], input)

[Object(name=''), Object(name='fake_name')]

If we wanted to fit the above theme, we could use a list comprehension instead of setattr:

    [None for flag.value in [bool(o.name)]]

---

But really, in serious code you should always use a regular function definition instead of a lambda if you're going to be doing outer assignment.

flag = Object(value=True)
def not_empty_except_first(o):
    result = flag.value or bool(o.name)
    flag.value = flag.value and bool(o.name)
    return result
input = [Object(name=""), Object(name="fake_name"), Object(name="")] 
output = filter(not_empty_except_first, input)

Solution 2

You cannot really maintain state in a filter/lambda expression (unless abusing the global namespace). You can however achieve something similar using the accumulated result being passed around in a reduce() expression:

>>> f = lambda a, b: (a.append(b) or a) if (b not in a) else a
>>> input = ["foo", u"", "bar", "", "", "x"]
>>> reduce(f, input, [])
['foo', u'', 'bar', 'x']
>>> 

You can, of course, tweak the condition a bit. In this case it filters out duplicates, but you can also use a.count(""), for example, to only restrict empty strings.

Needless to say, you can do this but you really shouldn't. :)

Lastly, you can do anything in pure Python lambda: http://vanderwijk.info/blog/pure-lambda-calculus-python/

Solution 3

Normal assignment (=) is not possible inside a lambda expression, although it is possible to perform various tricks with setattr and friends.

Solving your problem, however, is actually quite simple:

input = [Object(name=""), Object(name="fake_name"), Object(name="")]
output = filter(
    lambda o, _seen=set():
        not (not o and o in _seen or _seen.add(o)),
    input
    )

which will give you

[Object(Object(name=''), name='fake_name')]

As you can see, it's keeping the first blank instance instead of the last. If you need the last instead, reverse the list going in to filter, and reverse the list coming out of filter:

output = filter(
    lambda o, _seen=set():
        not (not o and o in _seen or _seen.add(o)),
    input[::-1]
    )[::-1]

which will give you

[Object(name='fake_name'), Object(name='')]

One thing to be aware of: in order for this to work with arbitrary objects, those objects must properly implement __eq__ and __hash__ as explained here.

Solution 4

There's no need to use a lambda, when you can remove all the null ones, and put one back if the input size changes:

input = [Object(name=""), Object(name="fake_name"), Object(name="")] 
output = [x for x in input if x.name]
if(len(input) != len(output)):
    output.append(Object(name=""))

Solution 5

UPDATE:

[o for d in [{}] for o in lst if o.name != "" or d.setdefault("", o) == o]

or using filter and lambda:

flag = {}
filter(lambda o: bool(o.name) or flag.setdefault("", o) == o, lst)

Previous Answer

OK, are you stuck on using filter and lambda?

It seems like this would be better served with a dictionary comprehension,

{o.name : o for o in input}.values()

I think the reason that Python doesn't allow assignment in a lambda is similar to why it doesn't allow assignment in a comprehension and that's got something to do with the fact that these things are evaluated on the C side and thus can give us an increase in speed. At least that's my impression after reading one of Guido's essays.

My guess is this would also go against the philosophy of having one right way of doing any one thing in Python.

View more solutions

Comments

[Admin]

No. But you don't need this. Actually I think it would be a pretty obscure way to achive this even if it worked.

[dfb]

Why not just pass a regular old function into filter?

[Cat]

I wanted to use lambda just so it would be a really compact solution. I remember in OCaml I could chain print statements before the return expression, thought this could be replicated in Python

[WestCoastProjects]

It is quite painful to be in the flow of developing a chained pipeilne then realize: "oh I want to create a temp var to make the flow more clear" or "i want to log this intermediate step" : and then you have to jump somewhere else to create a function to do it: and name that function and keep track of it - even though it's used in just one place .

[user202729]

See also Assignment inside lambda expression in Python - Stack Overflow for the special case (++ or -- operator in C)

[halex]

I think you have a small mistake in your code. The second line should be output = [x for x in input if x.name].

[Jeremy]

The last example in this answer doesn't produce the same output as the example, but it looks to me like the example output is incorrect.

[jno]

in short, this boils down to: use .setattr() and alikes (dictionaries should do as well, for instance) to hack side effects into functional code anyway, cool code by @JeremyBanks was shown :)

[JPvdMerwe]

You cannot use locals(), it explicitly says in the documentation that changing it doesn't actually change the local scope (or at least it won't always). globals() on the other hand works as expected.

[JPvdMerwe]

So this isn't completely right. It won't preserve order, nor will it preserve duplicates of non-empty-stringed objects.

[Kaz]

Or just use the original Lisp: (let ((var 42)) (lambda () (setf var 43))).

[jyf1987]

@JPvdMerwe just try it, dont follow the document blindly. and assignment in lambda is breaking rule already

[JPvdMerwe]

It unfortunately only works in the global namespace, in which case you really should be using globals(). pastebin.com/5Bjz1mR4 (tested in both 2.6 and 3.2) proves it.

[MAnyKey]

Order of elements may be important.

[WestCoastProjects]

Thx for the note on the assignment operator !