How can I use lambda as function parameter?

Question

The author of the top answer of this question uses a lamda function as parameter for this function:

>>> d1 = {'one':1, 'both':3, 'falsey_one':False, 'falsey_both':None}
>>> d2 = {'two':2, 'both':30, 'falsey_two':None, 'falsey_both':False}


def dict_ops(d1, d2, setop):
...     """Apply set operation `setop` to dictionaries d1 and d2
... 
...     Note: In cases where values are present in both d1 and d2, the value from
...     d1 will be used.
...     """
...     return {k:d1.get(k,k in d1 or d2[k]) for k in setop(set(d1), set(d2))}

Like this:

>>> print "d1 - d2:", dict_ops(d1, d2, lambda x,y: x-y)

Which returns:

d1 - d2: {'falsey_one': False, 'one': 1}

I tried to do the same thing but not as a function, because I want to understand how the setop part works.

{k:d1.get(k,k in d1 or d2[k]) for k in lambda d1,d2: set(d1) - set(d2)}

However this code returns a syntax error.

But this works:

l = lambda d1,d2: set(d1) - set(d2)
{k:d1.get(k,k in d1 or d2[k]) for k in l(d1,d2)}

Why does the second solution work, but the first one dont?

If I call the dict_ops function with these parameters (d1, d2, lambda x,y: x-y) how does setop(set(d1) - set(d2) look like?

You're not calling the lambda. (lambda d1,d2: set(d1) - set(d2))(d1, d2) would work. — jonrsharpe
– jonrsharpe, Commented Jun 10, 2020 at 14:52
So I need to provide the lambda function with paramters? Was my mistake that I just 'wrote' a lambda function without providing parameters? — gython
– gython, Commented Jun 10, 2020 at 14:56
That's part of it. It would be syntactically valid if you just surrounded it with parentheses, but you'd get a TypeError ('function' object is not iterable) instead. You can define a lambda function without calling it, that's just not what you actually want to do here. — jonrsharpe
– jonrsharpe, Commented Jun 10, 2020 at 14:57

Tom Karzes · Accepted Answer · 2020-06-10 14:57:59Z

1

If you do an exact substitution, it will work. You have:

{... for k in setop(set(d1), set(d2))}

The value passed in for setop is:

lambda x,y: x-y

A direct substitution of setop is:

{... for k in (lambda x,y: x-y)(set(d1), set(d2))|

which will work.

For the second version, you have:

{... for k in l(d1,d2)}

where l is:

lambda d1,d2: set(d1) - set(d2)

A direct substitution here produces:

{... for k in (lambda d1,d2: set(d1) - set(d2))(d1,d2)}

which will also work.

answered Jun 10, 2020 at 14:57

Tom Karzes

24.3k3 gold badges28 silver badges46 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

How can I use lambda as function parameter?

1 Answer 1

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related