there's no next() function in a yield generator in python 3
PythonPython 3.xPython Problem Overview
In this question, I have an endless sequence using Python generators. But the same code doesn't work in Python 3 because it seems there is no next() function. What is the equivalent for the next function?
def updown(n):
while True:
for i in range(n):
yield i
for i in range(n - 2, 0, -1):
yield i
uptofive = updown(6)
for i in range(20):
print(uptofive.next())
Python Solutions
Solution 1 - Python
In Python 3, use next(uptofive) instead of uptofive.next().
The built-in next() function also works in Python 2.6 or greater.
Solution 2 - Python
In Python 3, to make syntax more consistent, the next() method was renamed to __next__(). You could use that one. This is explained in PEP 3114.
Following Greg's solution and calling the builtin next() function (which then tries to find an object's __next__() method) is recommended.