How to force a lambda function to return an object rather than the None value of the last method on the object?

Question

Sometimes I cannot use a lambda function on an object because the last method applied to the object returns None. What is the trick to be able to return the object itself instead of the method returned value?

Example, I want to get the updated copy of the dictionary instead of None:

dict_update = lambda d, upd: d.copy().update(upd)

d = dict(lw=3, ls='-')
upd = dict(c='C0', lw=4)
d1 = dict_update(d, upd)
for v in [d, upd, d1]: print(v)

{'lw': 3, 'ls': '-'}
{'c': 'C0', 'lw': 4}
None

This question is similar to: How do I merge two dictionaries in a single expression in Python?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Ignatius Reilly
– Ignatius Reilly, Commented Sep 27, 2024 at 18:53
The question is valid, even if the example use of the lambda expression is an anti-pattern. — chepner
– chepner, Commented Sep 27, 2024 at 18:55
@IgnatiusReilly They said that's an Example. Of the more general thing they want. — no comment
– no comment, Commented Sep 27, 2024 at 18:55

trincot · Accepted Answer · 2024-09-27 18:49:38Z

2

You could use dict unpacking and packing:

dict_update = lambda d, upd: {**d, **upd}

answered Sep 27, 2024 at 18:49

trincot

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2 Comments

no comment Over a year ago

Or dict_update = dict.__or__.

no comment Over a year ago

@mins How does this answer anything other than your example?

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