Understanding lambda in Python and using it to pass multiple arguments

Question

After reading everything I can find on lambda expressions in Python, I still don't understand how to make it do what I want.

Everyone uses the example:

lambda x, y : x + y

Why do you need to state both x and y before the :? Also how do you make it return multiple arguments?

For example:

self.buttonAdd_1 = Button(self, text='+', command=lambda : self.calculate(self.buttonOut_1.grid_info(), 1))

This works just fine. But the following code does not:

self.entry_1.bind("<Return>", lambda : self.calculate(self.buttonOut_1.grid_info(), 1))

It yields the error:

TypeError: () takes no arguments (1 given)

"Also how do you make it return multiple arguments?" This does not make sense, because "argument" is not what you call something that gets returned. — Karl Knechtel
– Karl Knechtel, Commented Mar 10, 2023 at 7:00

phipsgabler · Accepted Answer · 2012-04-27 06:30:48Z

184

Why do you need to state both 'x' and 'y' before the ':'?

Because a lambda is (conceptually) the same as a function, just written inline. Your example is equivalent to

def f(x, y) : return x + y

just without binding it to a name like f.

Also how do you make it return multiple arguments?

The same way like with a function. Preferably, you return a tuple:

lambda x, y: (x+y, x-y)

Or a list, or a class, or whatever.

The thing with self.entry_1.bind should be answered by Demosthenex.

edited Apr 27, 2012 at 6:30

answered Apr 27, 2012 at 5:43

phipsgabler

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4 Comments

phipsgabler Over a year ago

To really get the least bit out of them, you might try some functional programming, which is an awesome experience when you begin to understand it, and probably will make you a better programmer.</propaganda>

Revolucion for Monica Over a year ago

You are my god and your propaganda is my daily bread. Yet, I tried to it apply on a tuple : (train["pred_idx_cos"],train["target"]).apply(lambda x,y: get_result(x, y)) and it seems it doesn't work. Please, save my soul from the mist of procedural programing

Mass Dot Net Over a year ago

@RevolucionforMonica A bit of a late reply, but here's a Python 3 REPL code example of how to do it: repl.it/@foobar123/ScarceWhimsicalMainframe

Lu Kas Over a year ago

Probably good for someone asking this type of question, to explain that even a normal function, when returning multiple variables with e.g. return a, b, c, actually just returns a tuple (a, b, c). And the typical statement using a function a, b, c = fun(x,y) just unpacks the tuple that was returned.

Demosthenex · Accepted Answer · 2012-04-27 05:38:03Z

14

I believe bind always tries to send an event parameter. Try:

self.entry_1.bind("<Return>", lambda event: self.calculate(self.buttonOut_1.grid_info(), 1))

You accept the parameter and never use it.

answered Apr 27, 2012 at 5:38

Demosthenex

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1 Comment

Demosthenex Over a year ago

You can also try "event=None" to give it a default value, then the function can be used for bind and the button.

kindall · Accepted Answer · 2018-02-21 07:54:37Z

7

Why do you need to state both x and y before the :?

Because it's a function definition and it needs to know what parameters the function accepts, and in what order. It can't just look at the expression and use the variables names in that, because some of those names you might want to use existing local or global variable values for, and even if it did that, it wouldn't know what order it should expect to get them.

Your error message means that Tk is calling your lambda with one argument, while your lambda is written to accept no arguments. If you don't need the argument, just accept one and don't use it. (Demosthenex has the code, I would have posted it but was beaten to it.)

edited Feb 21, 2018 at 7:54

user8554766

answered Apr 27, 2012 at 5:43

kindall

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Comments

AlbertoAndreotti · Accepted Answer · 2014-04-17 22:36:14Z

4

Why do you need to state both 'x' and 'y' before the ':'?

You could actually in some situations(when you have only one argument) do not put the x and y before ":".

>>> flist = []
>>> for i in range(3):
...     flist.append(lambda : i)

but the i in the lambda will be bound by name, so,

>>> flist[0]()
2
>>> flist[2]()
2
>>>

different from what you may want.

answered Apr 17, 2014 at 22:36

AlbertoAndreotti

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