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Python: next in for loop
Calling an external command in PythonWhat are metaclasses in Python?Finding the index of an item given a list containing it in PythonDifference between append vs. extend list methods in PythonHow can I safely create a nested directory in Python?Does Python have a ternary conditional operator?How to get the current time in PythonAccessing the index in 'for' loops?Iterating over dictionaries using 'for' loopsDoes Python have a string 'contains' substring method?
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I want to use next to skip one or more items returned from a generator. Here is a simplified example designed to skip one item per loop (in actual use, I'd test n and depending on the result, may repeat the next() and the generator is from a package I don't control):
def gen():
for i in range(10):
yield i
for g in gen():
n = next(gen())
print(g, n)
I expected the result to be
0 1
2 3
etc.
Instead I got
0 0
1 0
etc.
What am I doing wrong?
python generator next
asked 5 hours ago
foosionfoosion
3,325144990
add a comment |
I want to use next to skip one or more items returned from a generator. Here is a simplified example designed to skip one item per loop (in actual use, I'd test n and depending on the result, may repeat the next() and the generator is from a package I don't control):
def gen():
for i in range(10):
yield i
for g in gen():
n = next(gen())
print(g, n)
I expected the result to be
0 1
2 3
etc.
Instead I got
0 0
1 0
etc.
What am I doing wrong?
python generator next
asked 5 hours ago
foosionfoosion
3,325144990
add a comment |
I want to use next to skip one or more items returned from a generator. Here is a simplified example designed to skip one item per loop (in actual use, I'd test n and depending on the result, may repeat the next() and the generator is from a package I don't control):
def gen():
for i in range(10):
yield i
for g in gen():
n = next(gen())
print(g, n)
I expected the result to be
0 1
2 3
etc.
Instead I got
0 0
1 0
etc.
What am I doing wrong?
python generator next
asked 5 hours ago
foosionfoosion
3,325144990
I want to use next to skip one or more items returned from a generator. Here is a simplified example designed to skip one item per loop (in actual use, I'd test n and depending on the result, may repeat the next() and the generator is from a package I don't control):
def gen():
for i in range(10):
yield i
for g in gen():
n = next(gen())
print(g, n)
I expected the result to be
0 1
2 3
etc.
Instead I got
0 0
1 0
etc.
What am I doing wrong?
python generator next
python generator next
asked 5 hours ago
foosionfoosion
3,325144990
asked 5 hours ago
foosionfoosion
3,325144990
asked 5 hours ago
foosionfoosion
3,325144990
asked 5 hours ago
foosionfoosion
3,325144990
asked 5 hours ago
foosionfoosion
3,325144990
3,325144990
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You're making a new generator each time you call gen(). Each new generator starts from 0.
Instead, you can call it once and capture the return value.
def gen():
for i in range(10):
yield i
x = gen()
for g in x:
n = next(x)
print(g, n)
answered 5 hours ago
khelwoodkhelwood
32.2k74465
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You're making a new generator each time you call gen(). Each new generator starts from 0.
Instead, you can call it once and capture the return value.
def gen():
for i in range(10):
yield i
x = gen()
for g in x:
n = next(x)
print(g, n)
answered 5 hours ago
khelwoodkhelwood
32.2k74465
add a comment |
You're making a new generator each time you call gen(). Each new generator starts from 0.
Instead, you can call it once and capture the return value.
def gen():
for i in range(10):
yield i
x = gen()
for g in x:
n = next(x)
print(g, n)
answered 5 hours ago
khelwoodkhelwood
32.2k74465
add a comment |
You're making a new generator each time you call gen(). Each new generator starts from 0.
Instead, you can call it once and capture the return value.
def gen():
for i in range(10):
yield i
x = gen()
for g in x:
n = next(x)
print(g, n)
answered 5 hours ago
khelwoodkhelwood
32.2k74465
You're making a new generator each time you call gen(). Each new generator starts from 0.
Instead, you can call it once and capture the return value.
def gen():
for i in range(10):
yield i
x = gen()
for g in x:
n = next(x)
print(g, n)
answered 5 hours ago
khelwoodkhelwood
32.2k74465
answered 5 hours ago
khelwoodkhelwood
32.2k74465
answered 5 hours ago
khelwoodkhelwood
32.2k74465
answered 5 hours ago
khelwoodkhelwood
32.2k74465
32.2k74465
add a comment |
add a comment |
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