I created a simple generator:
def long_time():
for i in range(10):
yield i * 5
a = long_time()
print(next(a))
print(next(a))
Here, I get the outputs 0 and 5, which is OK. But when I try to build the variable a as below:
def long_time():
for i in range(10):
yield i * 5
a = long_time
print(next(a()))
print(next(a()))
Both of the print outputs are 0. What is the difference, should I always call the generator with ()?
When you write
a = long_time
you're just giving a different name to the function long_time.
So writing
print(next(a()))
is the same as writing
print(next(long_time()))
That is, in a single line, you get a generator object by calling long_time(), then pass it to next, getting the generator's first value 0. But since you don't have a reference to that generator, after that line you can't access the object anymore. On your next print... line, you're getting a fresh generator since it involves a new call to long_time().
This is fixed when a isn't a reference to the long_time function, but its return value (the generator) instead:
a = long_time()
[1, 2, 3][0] versus mylist = [1, 2, 3]; mylist[0]: in both cases you get the first element of a list, but in the first example you don't have access to the list afterwards because you created it and extracted its first element in a single line. next_long_time()) is like the former, a = long_time(); next(a) is like the latter.When you call long_time, you get back a fresh generator object. In your second example, a is just another name for the generator function, so each time you call a(), you get a new generator (just as if you had called long_time) that starts at 0.
