I was making a function that could output the results one by one with yield or return all results in a list. I have everything in the right indention, but it always. Does anyone know the reason why?
I am using Python 3.9 if that helps.
def get_ep_info(iter=False):
ep_list_info = list()
ep_nums = [1,2,3,4,5]
for ep_num in ep_nums:
if iter is True:
print("true")
yield ep_num
elif iter is False:
print("false")
ep_list_info.append(ep_num)
if iter is False:
print("false")
return ep_list_info
a = get_ep_info()
print(a)Output:
<generator object get_ep_info at 0x0000022A5D048BA0>
Containing yield makes a function a generator. And since the yield forces the function to be a generator you need to use next or for to get the values and you cannot use return in the function. You could write your function like this:
def get_ep_info(iter=False):
ep_list_info = list()
ep_nums = [1,2,3,4,5]
for ep_num in ep_nums:
if iter is True:
print("true")
yield ep_num
elif iter is False:
print("false")
ep_list_info.append(ep_num)
if iter is False:
print("false")
yield ep_list_info #<- Change from return to yeild
a = next(get_ep_info())
print(a)But it makes far more sense to leave it a generator and use list() to create a list of the returned values.
def get_ep_info():
for ep_num in [1,2,3,4,5]:
yield ep_num
a = list(get_ep_info())
print(a)This is less likely to result in uncontrolled laughter or swearing when people try to use your code
See if this is what you wanted:
def get_ep_info(iterator=False):
ep_list_info = list()
ep_nums = [1,2,3,4,5]
for ep_num in ep_nums:
ep_list_info.append (ep_num)
if iterator is False:
return ep_list_info
else :
return iter(ep_list_info)
a = get_ep_info(True)
print(next (a), next (a))
While this works for the original example:
def get_ep_info(iterator=False):
ep_list_info = list()
ep_nums = [1,2,3,4,5]
for ep_num in ep_nums:
ep_list_info.append (ep_num)
if iterator is False:
return ep_list_info
else :
return iter(ep_list_info)
a = get_ep_info(True)
print(next (a), next (a))It does not create code that acts like a generator. Regardless of the iterator flag all the results are created immediately and the advantages of using a generator are lost. It is better to act like a generator and use something like list() to create a collection from the generator results. That allows getting the benefits of a generator or getting a collection depending on you need.
Thanks to everyone how replyed. This is the code I think I am going to use. I just have never across this before where yielding overrides the return statement. The ep_nums[] is a placeholder for the function that I was working on when I ran into this issue.
def get_ep_info_iter():
# This is placeholder data
ep_nums = [1,2,3,4,5]
for ep_num in ep_nums:
yield ep_num
get_ep_info = lambda : list(get_ep_info_iter())
a = get_ep_info_list()
print(a)
Quote:Containing yield makes a function a generator. And since the yield forces the function to be a generator you need to use next or for to get the values and you cannot use return in the function. You could write your function like this:
def get_ep_info(iter=False):
ep_list_info = list()
ep_nums = [1,2,3,4,5]
for ep_num in ep_nums:
if iter is True:
print("true")
yield ep_num
elif iter is False:
print("false")
ep_list_info.append(ep_num)
if iter is False:
print("false")
yield ep_list_info #<- Change from return to yield
a = next(get_ep_info())
print(a)But it makes far more sense to leave it a generator and use list() to create a list of the returned values.
def get_ep_info():
for ep_num in [1,2,3,4,5]:
yield ep_num
a = list(get_ep_info())
print(a)This is less likely to result in uncontrolled laughter or swearing when people try to use your code
Thanks for the reply, I never knew that list() could take in generators. That's great to know instead of doing a comprehension list
.