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Python: next in for loop

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Python: next in for loop

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6

1

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

share|improve this question

asked 1 hour ago

foosionfoosion

3,325144990

add a comment | 

6

1

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

share|improve this question

asked 1 hour 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

share|improve this question

asked 1 hour ago

foosionfoosion

3,325144990

def gen():
for i in range(10):
 yield i

for g in gen():
 n = next(gen())
 print(g, n)

0 1
2 3

0 0
1 0

share|improve this question

asked 1 hour ago

foosionfoosion

3,325144990

share|improve this question

asked 1 hour ago

foosionfoosion

3,325144990

asked 1 hour ago

foosionfoosion

3,325144990

asked 1 hour ago

foosionfoosion

3,325144990

1 Answer
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5

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)

share|improve this answer

answered 1 hour ago

khelwoodkhelwood

32.2k74465

add a comment | 

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5

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)

share|improve this answer

answered 1 hour ago

khelwoodkhelwood

32.2k74465

add a comment | 

5

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)

share|improve this answer

answered 1 hour 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)

share|improve this answer

answered 1 hour ago

khelwoodkhelwood

32.2k74465

def gen():
 for i in range(10):
 yield i

x = gen()
for g in x:
 n = next(x)
 print(g, n)

share|improve this answer

answered 1 hour ago

khelwoodkhelwood

32.2k74465

answered 1 hour ago

khelwoodkhelwood

32.2k74465

answered 1 hour ago

khelwoodkhelwood

32.2k74465