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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 17 hours ago
foosionfoosion
3,330144990
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 17 hours ago
foosionfoosion
3,330144990
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 17 hours ago
foosionfoosion
3,330144990
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 17 hours ago
foosionfoosion
3,330144990
asked 17 hours ago
foosionfoosion
3,330144990
asked 17 hours ago
foosionfoosion
3,330144990
asked 17 hours ago
foosionfoosion
3,330144990
asked 17 hours ago
foosionfoosion
3,330144990
3,330144990
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 17 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 17 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 17 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 17 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 17 hours ago
khelwoodkhelwood
32.2k74465
answered 17 hours ago
khelwoodkhelwood
32.2k74465
answered 17 hours ago
khelwoodkhelwood
32.2k74465
answered 17 hours ago
khelwoodkhelwood
32.2k74465
32.2k74465
add a comment |
add a comment |
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