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Premature ending of generator in list comprehension

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Premature ending of generator in list comprehension


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8















I'm using generators in list comprehensions, and getting some unexpected behavior with one of the generators ending early. Why does creating the generator outside of the list comprehension cause the behavior to change?



The generator I created is as follows:



def inc_range(a,b):
for i in range(min(a,b), max(a,b) + 1):
yield i


The first way of calling is as follows:



[(i,j) for i in inc_range(1,3) for j in inc_range(4,6)]


This gives me the following result:



[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The second way of calling it is as follows:



a = inc_range(1,3)
b = inc_range(4,6)

[(i,j) for i in a for j in b]


This gives me the following:



[(1, 4), (1, 5), (1, 6)]


Experimenting around, the following two examples gave me the first result:



a = range(1,4)
b = range(4,7)

[(i,j) for i in a for j in b]




a = (i for i in range(1,4))
b = (i for i in range(4,7))

a = list(a)
b = list(b)

[(i,j) for i in a for j in b]


While the following gave me the second result again.



a = (i for i in range(1,4))
b = (i for i in range(4,7))

[(i,j) for i in a for j in b]


What rule am I violating here regarding generators? Why does it make a difference when I assign the generators to variables before using them in a list comprehension, vs. using them directly?



ANSWERS



Check out the following answers which helped me understand what is occurring here:



Alex Yu
mkrieger1










share|improve this question









New contributor




bbminerva is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • unrelated - do not use for x in ...: yield .. - use yield from range(min(a,b), max(a,b) + 1)

    – Patrick Artner
    1 hour ago











  • @PatrickArtner didn't know about yield from, thanks for the tip.

    – bbminerva
    1 hour ago











  • Maybe I'm missing something. Your first way and second way are different only in one thing: j in inc_range(3,6) vs b = inc_range(4,6). Obviously you get different results. Then you make analogous operations with just range and get results equal to "first way" or "second way". What was the point of all this? I'm confused

    – Alex Yu
    1 hour ago











  • @AlexYu that was a typo copying over, they're both inc_range(4,6)

    – bbminerva
    1 hour ago






  • 1





    @smac89 a typo from the question being simplified -- I'll fix it.

    – bbminerva
    1 hour ago


















8















I'm using generators in list comprehensions, and getting some unexpected behavior with one of the generators ending early. Why does creating the generator outside of the list comprehension cause the behavior to change?



The generator I created is as follows:



def inc_range(a,b):
for i in range(min(a,b), max(a,b) + 1):
yield i


The first way of calling is as follows:



[(i,j) for i in inc_range(1,3) for j in inc_range(4,6)]


This gives me the following result:



[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The second way of calling it is as follows:



a = inc_range(1,3)
b = inc_range(4,6)

[(i,j) for i in a for j in b]


This gives me the following:



[(1, 4), (1, 5), (1, 6)]


Experimenting around, the following two examples gave me the first result:



a = range(1,4)
b = range(4,7)

[(i,j) for i in a for j in b]




a = (i for i in range(1,4))
b = (i for i in range(4,7))

a = list(a)
b = list(b)

[(i,j) for i in a for j in b]


While the following gave me the second result again.



a = (i for i in range(1,4))
b = (i for i in range(4,7))

[(i,j) for i in a for j in b]


What rule am I violating here regarding generators? Why does it make a difference when I assign the generators to variables before using them in a list comprehension, vs. using them directly?



ANSWERS



Check out the following answers which helped me understand what is occurring here:



Alex Yu
mkrieger1










share|improve this question









New contributor




bbminerva is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • unrelated - do not use for x in ...: yield .. - use yield from range(min(a,b), max(a,b) + 1)

    – Patrick Artner
    1 hour ago











  • @PatrickArtner didn't know about yield from, thanks for the tip.

    – bbminerva
    1 hour ago











  • Maybe I'm missing something. Your first way and second way are different only in one thing: j in inc_range(3,6) vs b = inc_range(4,6). Obviously you get different results. Then you make analogous operations with just range and get results equal to "first way" or "second way". What was the point of all this? I'm confused

    – Alex Yu
    1 hour ago











  • @AlexYu that was a typo copying over, they're both inc_range(4,6)

    – bbminerva
    1 hour ago






  • 1





    @smac89 a typo from the question being simplified -- I'll fix it.

    – bbminerva
    1 hour ago
















8












8








8


1






I'm using generators in list comprehensions, and getting some unexpected behavior with one of the generators ending early. Why does creating the generator outside of the list comprehension cause the behavior to change?



The generator I created is as follows:



def inc_range(a,b):
for i in range(min(a,b), max(a,b) + 1):
yield i


The first way of calling is as follows:



[(i,j) for i in inc_range(1,3) for j in inc_range(4,6)]


This gives me the following result:



[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The second way of calling it is as follows:



a = inc_range(1,3)
b = inc_range(4,6)

[(i,j) for i in a for j in b]


This gives me the following:



[(1, 4), (1, 5), (1, 6)]


Experimenting around, the following two examples gave me the first result:



a = range(1,4)
b = range(4,7)

[(i,j) for i in a for j in b]




a = (i for i in range(1,4))
b = (i for i in range(4,7))

a = list(a)
b = list(b)

[(i,j) for i in a for j in b]


While the following gave me the second result again.



a = (i for i in range(1,4))
b = (i for i in range(4,7))

[(i,j) for i in a for j in b]


What rule am I violating here regarding generators? Why does it make a difference when I assign the generators to variables before using them in a list comprehension, vs. using them directly?



ANSWERS



Check out the following answers which helped me understand what is occurring here:



Alex Yu
mkrieger1










share|improve this question









New contributor




bbminerva is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I'm using generators in list comprehensions, and getting some unexpected behavior with one of the generators ending early. Why does creating the generator outside of the list comprehension cause the behavior to change?



The generator I created is as follows:



def inc_range(a,b):
for i in range(min(a,b), max(a,b) + 1):
yield i


The first way of calling is as follows:



[(i,j) for i in inc_range(1,3) for j in inc_range(4,6)]


This gives me the following result:



[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The second way of calling it is as follows:



a = inc_range(1,3)
b = inc_range(4,6)

[(i,j) for i in a for j in b]


This gives me the following:



[(1, 4), (1, 5), (1, 6)]


Experimenting around, the following two examples gave me the first result:



a = range(1,4)
b = range(4,7)

[(i,j) for i in a for j in b]




a = (i for i in range(1,4))
b = (i for i in range(4,7))

a = list(a)
b = list(b)

[(i,j) for i in a for j in b]


While the following gave me the second result again.



a = (i for i in range(1,4))
b = (i for i in range(4,7))

[(i,j) for i in a for j in b]


What rule am I violating here regarding generators? Why does it make a difference when I assign the generators to variables before using them in a list comprehension, vs. using them directly?



ANSWERS



Check out the following answers which helped me understand what is occurring here:



Alex Yu
mkrieger1







python generator list-comprehension






share|improve this question









New contributor




bbminerva is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




bbminerva is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 33 mins ago







bbminerva













New contributor




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Check out our Code of Conduct.









asked 1 hour ago









bbminervabbminerva

434




434




New contributor




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Check out our Code of Conduct.





New contributor





bbminerva is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






bbminerva is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.













  • unrelated - do not use for x in ...: yield .. - use yield from range(min(a,b), max(a,b) + 1)

    – Patrick Artner
    1 hour ago











  • @PatrickArtner didn't know about yield from, thanks for the tip.

    – bbminerva
    1 hour ago











  • Maybe I'm missing something. Your first way and second way are different only in one thing: j in inc_range(3,6) vs b = inc_range(4,6). Obviously you get different results. Then you make analogous operations with just range and get results equal to "first way" or "second way". What was the point of all this? I'm confused

    – Alex Yu
    1 hour ago











  • @AlexYu that was a typo copying over, they're both inc_range(4,6)

    – bbminerva
    1 hour ago






  • 1





    @smac89 a typo from the question being simplified -- I'll fix it.

    – bbminerva
    1 hour ago





















  • unrelated - do not use for x in ...: yield .. - use yield from range(min(a,b), max(a,b) + 1)

    – Patrick Artner
    1 hour ago











  • @PatrickArtner didn't know about yield from, thanks for the tip.

    – bbminerva
    1 hour ago











  • Maybe I'm missing something. Your first way and second way are different only in one thing: j in inc_range(3,6) vs b = inc_range(4,6). Obviously you get different results. Then you make analogous operations with just range and get results equal to "first way" or "second way". What was the point of all this? I'm confused

    – Alex Yu
    1 hour ago











  • @AlexYu that was a typo copying over, they're both inc_range(4,6)

    – bbminerva
    1 hour ago






  • 1





    @smac89 a typo from the question being simplified -- I'll fix it.

    – bbminerva
    1 hour ago



















unrelated - do not use for x in ...: yield .. - use yield from range(min(a,b), max(a,b) + 1)

– Patrick Artner
1 hour ago





unrelated - do not use for x in ...: yield .. - use yield from range(min(a,b), max(a,b) + 1)

– Patrick Artner
1 hour ago













@PatrickArtner didn't know about yield from, thanks for the tip.

– bbminerva
1 hour ago





@PatrickArtner didn't know about yield from, thanks for the tip.

– bbminerva
1 hour ago













Maybe I'm missing something. Your first way and second way are different only in one thing: j in inc_range(3,6) vs b = inc_range(4,6). Obviously you get different results. Then you make analogous operations with just range and get results equal to "first way" or "second way". What was the point of all this? I'm confused

– Alex Yu
1 hour ago





Maybe I'm missing something. Your first way and second way are different only in one thing: j in inc_range(3,6) vs b = inc_range(4,6). Obviously you get different results. Then you make analogous operations with just range and get results equal to "first way" or "second way". What was the point of all this? I'm confused

– Alex Yu
1 hour ago













@AlexYu that was a typo copying over, they're both inc_range(4,6)

– bbminerva
1 hour ago





@AlexYu that was a typo copying over, they're both inc_range(4,6)

– bbminerva
1 hour ago




1




1





@smac89 a typo from the question being simplified -- I'll fix it.

– bbminerva
1 hour ago







@smac89 a typo from the question being simplified -- I'll fix it.

– bbminerva
1 hour ago














3 Answers
3






active

oldest

votes


















3














And the answer is in:



PEP 289 -- Generator Expressions. The details




Only the outermost for-expression is evaluated immediately, the other expressions are deferred until the generator is run:




g = (tgtexp  for var1 in exp1 if exp2 for var2 in exp3 if exp4)


is equivalent to:



def __gen(bound_exp):
for var1 in bound_exp:
if exp2:
for var2 in exp3:
if exp4:
yield tgtexp
g = __gen(iter(exp1))
del __gen


Clarification. Why PEP-289 and not PEP-202?



Because comprehensions on lists works in another way.



This would be example for comprehensions on lists:



a = list(inc_range(1,3))
b = list(inc_range(4,6))

[(i,j) for i in a for j in b]


And there will be no "wonders": everything will be as expected.






share|improve this answer


























  • This would be the answer! Thanks!

    – bbminerva
    51 mins ago








  • 1





    You asked excellent question. I was somewhat confused at first. What you asked - is actually question about generator-expressions not list-comprehensions

    – Alex Yu
    48 mins ago






  • 2





    Good findings. The important point, IMO, is that the inner expression not only is deferred, but it is evaluated again in each iteration of the outer expression.

    – mkrieger1
    45 mins ago











  • @mkrieger1 As I read now Early Binding versus Late Binding there was discussion with pro- and contra- of such behaviour. And using generator-functions changes way of evaluation from first to last (not outermost only)

    – Alex Yu
    32 mins ago





















3














A generator is an iterable object hence when you call it outside of a list it returns the next item alone.



a = inc_range(1,3)
b = inc_range(4,6)
c = inc_range(7,9)

[(i,j,k) for i in a for j in b for k in c]


This will only yield the elements for k in c when run



Hence you need to iterate through all its objects as you have when defining it as an array.



[(i,j,k) for i in inc_range(1,3) for j in inc_range(3,6) for k in inc_range(7,9)]


This forces the generator to yield all values in each of the iterations.






share|improve this answer
























  • I don't think I quite understand the example. All three are iterators, and should iterate until they raise a StopIteration. I thought the for i in a would keep calling a until a StopIteration is thrown, much like what occurs when I turn it into a list first.

    – bbminerva
    1 hour ago













  • @Alex Yu describes this well

    – James Carron
    57 mins ago





















3














To get the desired result, the "inner" generator would have to be run as many times as the "outer" generator yields a value.



But, after the first run, the "inner" generator is exhausted and cannot be run again.



Adding a print illustrates this (simplifying the example):



>>> def inc(a, b):
... for i in range(a, b):
... print(i)
... yield i
...
>>> a = inc(1, 4)
>>> b = inc(4, 7)
>>> [(i, j) for i in a for j in b]
1 # <-- a begins to run
4 # <-- b begins to run
5
6 # <-- b exhausted here
2 # <-- a continued, but not resulting in list item, because lacking value from b
3
[(1, 4), (1, 5), (1, 6)]


The reason why not storing the generators in variables works as expected is because a new "inner" generator is created for each iteration of the "outer" generator. Again, illustrated by some prints:



>>> def inc(a, b):
... print('started', a, b)
... for i in range(a, b):
... yield i
...
>>> [(i, j) for i in inc(1, 4) for j in inc(4, 7)]
started 1 4
started 4 7
started 4 7
started 4 7
[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The reason why using range objects or lists works as expected is because they can be iterated over arbitrarily many times without being exhausted.






share|improve this answer
























  • Thanks for the detailed explanation!

    – bbminerva
    39 mins ago











  • Your point about "again" - is great. I remember now that I had similar confusion as OP many years ago and then forgot till today

    – Alex Yu
    28 mins ago













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














And the answer is in:



PEP 289 -- Generator Expressions. The details




Only the outermost for-expression is evaluated immediately, the other expressions are deferred until the generator is run:




g = (tgtexp  for var1 in exp1 if exp2 for var2 in exp3 if exp4)


is equivalent to:



def __gen(bound_exp):
for var1 in bound_exp:
if exp2:
for var2 in exp3:
if exp4:
yield tgtexp
g = __gen(iter(exp1))
del __gen


Clarification. Why PEP-289 and not PEP-202?



Because comprehensions on lists works in another way.



This would be example for comprehensions on lists:



a = list(inc_range(1,3))
b = list(inc_range(4,6))

[(i,j) for i in a for j in b]


And there will be no "wonders": everything will be as expected.






share|improve this answer


























  • This would be the answer! Thanks!

    – bbminerva
    51 mins ago








  • 1





    You asked excellent question. I was somewhat confused at first. What you asked - is actually question about generator-expressions not list-comprehensions

    – Alex Yu
    48 mins ago






  • 2





    Good findings. The important point, IMO, is that the inner expression not only is deferred, but it is evaluated again in each iteration of the outer expression.

    – mkrieger1
    45 mins ago











  • @mkrieger1 As I read now Early Binding versus Late Binding there was discussion with pro- and contra- of such behaviour. And using generator-functions changes way of evaluation from first to last (not outermost only)

    – Alex Yu
    32 mins ago


















3














And the answer is in:



PEP 289 -- Generator Expressions. The details




Only the outermost for-expression is evaluated immediately, the other expressions are deferred until the generator is run:




g = (tgtexp  for var1 in exp1 if exp2 for var2 in exp3 if exp4)


is equivalent to:



def __gen(bound_exp):
for var1 in bound_exp:
if exp2:
for var2 in exp3:
if exp4:
yield tgtexp
g = __gen(iter(exp1))
del __gen


Clarification. Why PEP-289 and not PEP-202?



Because comprehensions on lists works in another way.



This would be example for comprehensions on lists:



a = list(inc_range(1,3))
b = list(inc_range(4,6))

[(i,j) for i in a for j in b]


And there will be no "wonders": everything will be as expected.






share|improve this answer


























  • This would be the answer! Thanks!

    – bbminerva
    51 mins ago








  • 1





    You asked excellent question. I was somewhat confused at first. What you asked - is actually question about generator-expressions not list-comprehensions

    – Alex Yu
    48 mins ago






  • 2





    Good findings. The important point, IMO, is that the inner expression not only is deferred, but it is evaluated again in each iteration of the outer expression.

    – mkrieger1
    45 mins ago











  • @mkrieger1 As I read now Early Binding versus Late Binding there was discussion with pro- and contra- of such behaviour. And using generator-functions changes way of evaluation from first to last (not outermost only)

    – Alex Yu
    32 mins ago
















3












3








3







And the answer is in:



PEP 289 -- Generator Expressions. The details




Only the outermost for-expression is evaluated immediately, the other expressions are deferred until the generator is run:




g = (tgtexp  for var1 in exp1 if exp2 for var2 in exp3 if exp4)


is equivalent to:



def __gen(bound_exp):
for var1 in bound_exp:
if exp2:
for var2 in exp3:
if exp4:
yield tgtexp
g = __gen(iter(exp1))
del __gen


Clarification. Why PEP-289 and not PEP-202?



Because comprehensions on lists works in another way.



This would be example for comprehensions on lists:



a = list(inc_range(1,3))
b = list(inc_range(4,6))

[(i,j) for i in a for j in b]


And there will be no "wonders": everything will be as expected.






share|improve this answer















And the answer is in:



PEP 289 -- Generator Expressions. The details




Only the outermost for-expression is evaluated immediately, the other expressions are deferred until the generator is run:




g = (tgtexp  for var1 in exp1 if exp2 for var2 in exp3 if exp4)


is equivalent to:



def __gen(bound_exp):
for var1 in bound_exp:
if exp2:
for var2 in exp3:
if exp4:
yield tgtexp
g = __gen(iter(exp1))
del __gen


Clarification. Why PEP-289 and not PEP-202?



Because comprehensions on lists works in another way.



This would be example for comprehensions on lists:



a = list(inc_range(1,3))
b = list(inc_range(4,6))

[(i,j) for i in a for j in b]


And there will be no "wonders": everything will be as expected.







share|improve this answer














share|improve this answer



share|improve this answer








edited 48 mins ago

























answered 57 mins ago









Alex YuAlex Yu

2,0011522




2,0011522













  • This would be the answer! Thanks!

    – bbminerva
    51 mins ago








  • 1





    You asked excellent question. I was somewhat confused at first. What you asked - is actually question about generator-expressions not list-comprehensions

    – Alex Yu
    48 mins ago






  • 2





    Good findings. The important point, IMO, is that the inner expression not only is deferred, but it is evaluated again in each iteration of the outer expression.

    – mkrieger1
    45 mins ago











  • @mkrieger1 As I read now Early Binding versus Late Binding there was discussion with pro- and contra- of such behaviour. And using generator-functions changes way of evaluation from first to last (not outermost only)

    – Alex Yu
    32 mins ago





















  • This would be the answer! Thanks!

    – bbminerva
    51 mins ago








  • 1





    You asked excellent question. I was somewhat confused at first. What you asked - is actually question about generator-expressions not list-comprehensions

    – Alex Yu
    48 mins ago






  • 2





    Good findings. The important point, IMO, is that the inner expression not only is deferred, but it is evaluated again in each iteration of the outer expression.

    – mkrieger1
    45 mins ago











  • @mkrieger1 As I read now Early Binding versus Late Binding there was discussion with pro- and contra- of such behaviour. And using generator-functions changes way of evaluation from first to last (not outermost only)

    – Alex Yu
    32 mins ago



















This would be the answer! Thanks!

– bbminerva
51 mins ago







This would be the answer! Thanks!

– bbminerva
51 mins ago






1




1





You asked excellent question. I was somewhat confused at first. What you asked - is actually question about generator-expressions not list-comprehensions

– Alex Yu
48 mins ago





You asked excellent question. I was somewhat confused at first. What you asked - is actually question about generator-expressions not list-comprehensions

– Alex Yu
48 mins ago




2




2





Good findings. The important point, IMO, is that the inner expression not only is deferred, but it is evaluated again in each iteration of the outer expression.

– mkrieger1
45 mins ago





Good findings. The important point, IMO, is that the inner expression not only is deferred, but it is evaluated again in each iteration of the outer expression.

– mkrieger1
45 mins ago













@mkrieger1 As I read now Early Binding versus Late Binding there was discussion with pro- and contra- of such behaviour. And using generator-functions changes way of evaluation from first to last (not outermost only)

– Alex Yu
32 mins ago







@mkrieger1 As I read now Early Binding versus Late Binding there was discussion with pro- and contra- of such behaviour. And using generator-functions changes way of evaluation from first to last (not outermost only)

– Alex Yu
32 mins ago















3














A generator is an iterable object hence when you call it outside of a list it returns the next item alone.



a = inc_range(1,3)
b = inc_range(4,6)
c = inc_range(7,9)

[(i,j,k) for i in a for j in b for k in c]


This will only yield the elements for k in c when run



Hence you need to iterate through all its objects as you have when defining it as an array.



[(i,j,k) for i in inc_range(1,3) for j in inc_range(3,6) for k in inc_range(7,9)]


This forces the generator to yield all values in each of the iterations.






share|improve this answer
























  • I don't think I quite understand the example. All three are iterators, and should iterate until they raise a StopIteration. I thought the for i in a would keep calling a until a StopIteration is thrown, much like what occurs when I turn it into a list first.

    – bbminerva
    1 hour ago













  • @Alex Yu describes this well

    – James Carron
    57 mins ago


















3














A generator is an iterable object hence when you call it outside of a list it returns the next item alone.



a = inc_range(1,3)
b = inc_range(4,6)
c = inc_range(7,9)

[(i,j,k) for i in a for j in b for k in c]


This will only yield the elements for k in c when run



Hence you need to iterate through all its objects as you have when defining it as an array.



[(i,j,k) for i in inc_range(1,3) for j in inc_range(3,6) for k in inc_range(7,9)]


This forces the generator to yield all values in each of the iterations.






share|improve this answer
























  • I don't think I quite understand the example. All three are iterators, and should iterate until they raise a StopIteration. I thought the for i in a would keep calling a until a StopIteration is thrown, much like what occurs when I turn it into a list first.

    – bbminerva
    1 hour ago













  • @Alex Yu describes this well

    – James Carron
    57 mins ago
















3












3








3







A generator is an iterable object hence when you call it outside of a list it returns the next item alone.



a = inc_range(1,3)
b = inc_range(4,6)
c = inc_range(7,9)

[(i,j,k) for i in a for j in b for k in c]


This will only yield the elements for k in c when run



Hence you need to iterate through all its objects as you have when defining it as an array.



[(i,j,k) for i in inc_range(1,3) for j in inc_range(3,6) for k in inc_range(7,9)]


This forces the generator to yield all values in each of the iterations.






share|improve this answer













A generator is an iterable object hence when you call it outside of a list it returns the next item alone.



a = inc_range(1,3)
b = inc_range(4,6)
c = inc_range(7,9)

[(i,j,k) for i in a for j in b for k in c]


This will only yield the elements for k in c when run



Hence you need to iterate through all its objects as you have when defining it as an array.



[(i,j,k) for i in inc_range(1,3) for j in inc_range(3,6) for k in inc_range(7,9)]


This forces the generator to yield all values in each of the iterations.







share|improve this answer












share|improve this answer



share|improve this answer










answered 1 hour ago









James CarronJames Carron

618




618













  • I don't think I quite understand the example. All three are iterators, and should iterate until they raise a StopIteration. I thought the for i in a would keep calling a until a StopIteration is thrown, much like what occurs when I turn it into a list first.

    – bbminerva
    1 hour ago













  • @Alex Yu describes this well

    – James Carron
    57 mins ago





















  • I don't think I quite understand the example. All three are iterators, and should iterate until they raise a StopIteration. I thought the for i in a would keep calling a until a StopIteration is thrown, much like what occurs when I turn it into a list first.

    – bbminerva
    1 hour ago













  • @Alex Yu describes this well

    – James Carron
    57 mins ago



















I don't think I quite understand the example. All three are iterators, and should iterate until they raise a StopIteration. I thought the for i in a would keep calling a until a StopIteration is thrown, much like what occurs when I turn it into a list first.

– bbminerva
1 hour ago







I don't think I quite understand the example. All three are iterators, and should iterate until they raise a StopIteration. I thought the for i in a would keep calling a until a StopIteration is thrown, much like what occurs when I turn it into a list first.

– bbminerva
1 hour ago















@Alex Yu describes this well

– James Carron
57 mins ago







@Alex Yu describes this well

– James Carron
57 mins ago













3














To get the desired result, the "inner" generator would have to be run as many times as the "outer" generator yields a value.



But, after the first run, the "inner" generator is exhausted and cannot be run again.



Adding a print illustrates this (simplifying the example):



>>> def inc(a, b):
... for i in range(a, b):
... print(i)
... yield i
...
>>> a = inc(1, 4)
>>> b = inc(4, 7)
>>> [(i, j) for i in a for j in b]
1 # <-- a begins to run
4 # <-- b begins to run
5
6 # <-- b exhausted here
2 # <-- a continued, but not resulting in list item, because lacking value from b
3
[(1, 4), (1, 5), (1, 6)]


The reason why not storing the generators in variables works as expected is because a new "inner" generator is created for each iteration of the "outer" generator. Again, illustrated by some prints:



>>> def inc(a, b):
... print('started', a, b)
... for i in range(a, b):
... yield i
...
>>> [(i, j) for i in inc(1, 4) for j in inc(4, 7)]
started 1 4
started 4 7
started 4 7
started 4 7
[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The reason why using range objects or lists works as expected is because they can be iterated over arbitrarily many times without being exhausted.






share|improve this answer
























  • Thanks for the detailed explanation!

    – bbminerva
    39 mins ago











  • Your point about "again" - is great. I remember now that I had similar confusion as OP many years ago and then forgot till today

    – Alex Yu
    28 mins ago


















3














To get the desired result, the "inner" generator would have to be run as many times as the "outer" generator yields a value.



But, after the first run, the "inner" generator is exhausted and cannot be run again.



Adding a print illustrates this (simplifying the example):



>>> def inc(a, b):
... for i in range(a, b):
... print(i)
... yield i
...
>>> a = inc(1, 4)
>>> b = inc(4, 7)
>>> [(i, j) for i in a for j in b]
1 # <-- a begins to run
4 # <-- b begins to run
5
6 # <-- b exhausted here
2 # <-- a continued, but not resulting in list item, because lacking value from b
3
[(1, 4), (1, 5), (1, 6)]


The reason why not storing the generators in variables works as expected is because a new "inner" generator is created for each iteration of the "outer" generator. Again, illustrated by some prints:



>>> def inc(a, b):
... print('started', a, b)
... for i in range(a, b):
... yield i
...
>>> [(i, j) for i in inc(1, 4) for j in inc(4, 7)]
started 1 4
started 4 7
started 4 7
started 4 7
[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The reason why using range objects or lists works as expected is because they can be iterated over arbitrarily many times without being exhausted.






share|improve this answer
























  • Thanks for the detailed explanation!

    – bbminerva
    39 mins ago











  • Your point about "again" - is great. I remember now that I had similar confusion as OP many years ago and then forgot till today

    – Alex Yu
    28 mins ago
















3












3








3







To get the desired result, the "inner" generator would have to be run as many times as the "outer" generator yields a value.



But, after the first run, the "inner" generator is exhausted and cannot be run again.



Adding a print illustrates this (simplifying the example):



>>> def inc(a, b):
... for i in range(a, b):
... print(i)
... yield i
...
>>> a = inc(1, 4)
>>> b = inc(4, 7)
>>> [(i, j) for i in a for j in b]
1 # <-- a begins to run
4 # <-- b begins to run
5
6 # <-- b exhausted here
2 # <-- a continued, but not resulting in list item, because lacking value from b
3
[(1, 4), (1, 5), (1, 6)]


The reason why not storing the generators in variables works as expected is because a new "inner" generator is created for each iteration of the "outer" generator. Again, illustrated by some prints:



>>> def inc(a, b):
... print('started', a, b)
... for i in range(a, b):
... yield i
...
>>> [(i, j) for i in inc(1, 4) for j in inc(4, 7)]
started 1 4
started 4 7
started 4 7
started 4 7
[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The reason why using range objects or lists works as expected is because they can be iterated over arbitrarily many times without being exhausted.






share|improve this answer













To get the desired result, the "inner" generator would have to be run as many times as the "outer" generator yields a value.



But, after the first run, the "inner" generator is exhausted and cannot be run again.



Adding a print illustrates this (simplifying the example):



>>> def inc(a, b):
... for i in range(a, b):
... print(i)
... yield i
...
>>> a = inc(1, 4)
>>> b = inc(4, 7)
>>> [(i, j) for i in a for j in b]
1 # <-- a begins to run
4 # <-- b begins to run
5
6 # <-- b exhausted here
2 # <-- a continued, but not resulting in list item, because lacking value from b
3
[(1, 4), (1, 5), (1, 6)]


The reason why not storing the generators in variables works as expected is because a new "inner" generator is created for each iteration of the "outer" generator. Again, illustrated by some prints:



>>> def inc(a, b):
... print('started', a, b)
... for i in range(a, b):
... yield i
...
>>> [(i, j) for i in inc(1, 4) for j in inc(4, 7)]
started 1 4
started 4 7
started 4 7
started 4 7
[(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)]


The reason why using range objects or lists works as expected is because they can be iterated over arbitrarily many times without being exhausted.







share|improve this answer












share|improve this answer



share|improve this answer










answered 54 mins ago









mkrieger1mkrieger1

4,62821932




4,62821932













  • Thanks for the detailed explanation!

    – bbminerva
    39 mins ago











  • Your point about "again" - is great. I remember now that I had similar confusion as OP many years ago and then forgot till today

    – Alex Yu
    28 mins ago





















  • Thanks for the detailed explanation!

    – bbminerva
    39 mins ago











  • Your point about "again" - is great. I remember now that I had similar confusion as OP many years ago and then forgot till today

    – Alex Yu
    28 mins ago



















Thanks for the detailed explanation!

– bbminerva
39 mins ago





Thanks for the detailed explanation!

– bbminerva
39 mins ago













Your point about "again" - is great. I remember now that I had similar confusion as OP many years ago and then forgot till today

– Alex Yu
28 mins ago







Your point about "again" - is great. I remember now that I had similar confusion as OP many years ago and then forgot till today

– Alex Yu
28 mins ago












bbminerva is a new contributor. Be nice, and check out our Code of Conduct.










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bbminerva is a new contributor. Be nice, and check out our Code of Conduct.
















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