How to Pythonically yield all values from a list?

Suppose I have a list that I wish not to return but to yield values from. What is the most pythonic way to do that?

Here is what I mean. Thanks to some non-lazy computation I have computed the list ['a', 'b', 'c', 'd'], but my code through the project uses lazy computation, so I'd like to yield values from my function instead of returning the whole list.

I currently wrote it as following:

my_list = ['a', 'b', 'c', 'd'] for item in my_list: yield item 

But this doesn't feel pythonic to me.

4 Answers

Since this question doesn't specify; I'll provide an answer that applies in Python >= 3.3

If you need only to return that list, do as Anurag suggests, but if for some reason the function in question really needs to be a generator, you can delegate to another generator; suppose you want to suffix the result list, but only if the list is first exhausted.

def foo(): list_ = ['a', 'b', 'c', 'd'] yield from list_ if something: yield this yield that yield something_else 

In versions of python prior to 3.3, though, you cannot use this syntax; you'll have to use the code as in the question, with a for loop and single yield statement in the body.

Alternatively; you can wrap the generators in a regular function and return the chained result: This also has the advantage of working in python 2 and 3

from itertools import chain def foo(): list_ = ['a', 'b', 'c', 'd'] def _foo_suffix(): if something: yield this yield that yield something_else return chain(list_, _foo_suffix()) 

Use iter to create a list iterator e.g.

return iter(List) 

though if you already have a list, you can just return that, which will be more efficient.

You can build a generator by saying

(x for x in List) 

Best Pythonically way to do this task is

In [1]: my_list = ['a', 'b', 'c', 'd'] In [2]: (x for x in my_list) Out[2]: <generator object <genexpr> at 0x7fd9e4fdc230>