dict.pop or dict.get and evaluation

Question

Consider the following code: ``` >>> def default_answer(): ... print "Default was required!" ... return 100 ... >>> g = { 'name': 'Jordan', 'age': 35 } >>> result = g.get('age', default_answer()) Default was required! >>> result = g.pop('age', default_answer()) Default was required! ``` Notice that whether `g` contains the expected key or not, default_answer is called. This makes sense programmatically but if `default_answer` was computationally expensive this would be a hassle (to run a lot of code to create a value that was going to be thrown away). The only way I can think of writing this without always calling `default_answer` is: ``` result = g.pop('age', False) or default_answer() ``` Is this the best way to do it? (Note that I'm aware that replacing `default_answer` with an object with lazy evaluation would also solve part of this problem, but that's outside the scope of this question).

Accepted Answer

No, this is not the best way to do this, since the result of `g.pop()` might be "falsy". Use `try`/`except` instead:

```
try:
    result = g.pop('age')
except KeyError:
    result = default_answer()
```

This idiom is called EAFP and is usually preferred over LBYL for various reasons:

It avoids looking up the key twice (which is not an issue with an actual `dict`, but might be suboptimal with other mappings).

It avoids a race condition in threaded code.

Many people find it easier to read.

dict.pop or dict.get and evaluation

Solution

Problem