I have a generator function which I want to call from another function and return the generator obtained. I can see two approaches here -
Note that the below functions are simple dummy functions to illustrate the purpose. Please don't come up with better ways to implement those functions itself.
Method 1def fun_a(n):
for i in range(n):
yield i+10
def fun_b(n):
if n < 0:
yield None
return
yield fun_a(n)
and use it as list(list(fun_b(10))[0])
to get [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
def fun_b(n):
if n < 0:
yield None
return
for i in fun_a(n):
yield i
Then list(fun_b(10))
can give me [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
.
While method 1 seems ok'ish, I don't want to return a list of list because in other cases I am returning a list and I don't want to mess up my code. Method 2 is just inefficient.
What would be a really good approach to deal with this situation?
I you are using Python 3.3 or higher, you can use the yield from
syntax introduced in PEP 380:
PEP 380 adds the yield from expression, allowing a generator to delegate part of its operations to another generator. This allows a section of code containing yield to be factored out and placed in another generator. Additionally, the subgenerator is allowed to return with a value, and the value is made available to the delegating generator.
>>> def fun_a(n):
for i in range(n):
yield i+10
>>> def fun_b(n):
if n < 0:
yield None
return
yield from fun_a(n)
>>> list(fun_b(10))
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>>
If not, then you'll just have to use the for item in iterable:
syntax.
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