Different behavior between generator expression and first constructing a list then using list-converted...












0














Can these two pieces of codes generate different behavior for downstream applications? In other words, are the returning objects any different ? 



 return (func(i) for i in a_list)


and 



 b_list= 

 for i in a_list:

     b_list.append(func(i))

 return (i for i in b_list)


PS: The second way to build generator is very questionable.






python generator







share|improve this question


















  • 1




    Yes, if the func(i) is influenced by the state of something external.
    – Willem Van Onsem
    Nov 14 '18 at 19:44
















0














Can these two pieces of codes generate different behavior for downstream applications? In other words, are the returning objects any different ? 



 return (func(i) for i in a_list)


and 



 b_list= 

 for i in a_list:

     b_list.append(func(i))

 return (i for i in b_list)


PS: The second way to build generator is very questionable.






python generator







share|improve this question


















  • 1




    Yes, if the func(i) is influenced by the state of something external.
    – Willem Van Onsem
    Nov 14 '18 at 19:44














0












0








0







Can these two pieces of codes generate different behavior for downstream applications? In other words, are the returning objects any different ? 



 return (func(i) for i in a_list)


and 



 b_list= 

 for i in a_list:

     b_list.append(func(i))

 return (i for i in b_list)


PS: The second way to build generator is very questionable.






python generator







share|improve this question













Can these two pieces of codes generate different behavior for downstream applications? In other words, are the returning objects any different ? 



 return (func(i) for i in a_list)


and 



 b_list= 

 for i in a_list:

     b_list.append(func(i))

 return (i for i in b_list)


PS: The second way to build generator is very questionable.






python generator




python generator






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 14 '18 at 19:43









jmf_zaiecp

12117




12117








  • 1




    Yes, if the func(i) is influenced by the state of something external.
    – Willem Van Onsem
    Nov 14 '18 at 19:44














  • 1




    Yes, if the func(i) is influenced by the state of something external.
    – Willem Van Onsem
    Nov 14 '18 at 19:44








1




1




Yes, if the func(i) is influenced by the state of something external.
– Willem Van Onsem
Nov 14 '18 at 19:44




Yes, if the func(i) is influenced by the state of something external.
– Willem Van Onsem
Nov 14 '18 at 19:44












1 Answer
1






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oldest

votes


















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First of all, this code:



b_list= 

for i in a_list:

    b_list.append(func(i))

return (i for i in b_list)


Is entirely the same as this code:



return iter([func(i) for i in a_list])


The difference between these and this:



return (func(i) for i in a_list)


Is that the latter is lazy and the other two are eager - i.e the latter runs func every time its __next__() method is called and the other two run func for every item ia a_list immediately.



So, the answer depends whether func is a pure function - in this case, if it behaves the same regardless of when it is run and independently from external/global variables.



It also depends on whether the items of the list are mutable and if them or the list itself are changed by other code.



If the answers are respectively "yes" and "no", they are same save for their memory footprint.



Finally, as long we're discussing functional concepts, please consider using: 



return map(func, a_list)


Instead of:



return (func(i) for i in a_list)


But only if you're using Python 3's lovely lazy map() and not Python 2's evil eager map().






share|improve this answer























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    First of all, this code:



    b_list= 
    
for i in a_list:
    
    b_list.append(func(i))
    
return (i for i in b_list)


    Is entirely the same as this code:



    return iter([func(i) for i in a_list])


    The difference between these and this:



    return (func(i) for i in a_list)


    Is that the latter is lazy and the other two are eager - i.e the latter runs func every time its __next__() method is called and the other two run func for every item ia a_list immediately.



    So, the answer depends whether func is a pure function - in this case, if it behaves the same regardless of when it is run and independently from external/global variables.



    It also depends on whether the items of the list are mutable and if them or the list itself are changed by other code.



    If the answers are respectively "yes" and "no", they are same save for their memory footprint.



    Finally, as long we're discussing functional concepts, please consider using: 



    return map(func, a_list)


    Instead of:



    return (func(i) for i in a_list)


    But only if you're using Python 3's lovely lazy map() and not Python 2's evil eager map().






    share|improve this answer




























      1














      First of all, this code:



      b_list= 
      
for i in a_list:
      
    b_list.append(func(i))
      
return (i for i in b_list)


      Is entirely the same as this code:



      return iter([func(i) for i in a_list])


      The difference between these and this:



      return (func(i) for i in a_list)


      Is that the latter is lazy and the other two are eager - i.e the latter runs func every time its __next__() method is called and the other two run func for every item ia a_list immediately.



      So, the answer depends whether func is a pure function - in this case, if it behaves the same regardless of when it is run and independently from external/global variables.



      It also depends on whether the items of the list are mutable and if them or the list itself are changed by other code.



      If the answers are respectively "yes" and "no", they are same save for their memory footprint.



      Finally, as long we're discussing functional concepts, please consider using: 



      return map(func, a_list)


      Instead of:



      return (func(i) for i in a_list)


      But only if you're using Python 3's lovely lazy map() and not Python 2's evil eager map().






      share|improve this answer


























        1












        1








        1






        First of all, this code:



        b_list= 
        
for i in a_list:
        
    b_list.append(func(i))
        
return (i for i in b_list)


        Is entirely the same as this code:



        return iter([func(i) for i in a_list])


        The difference between these and this:



        return (func(i) for i in a_list)


        Is that the latter is lazy and the other two are eager - i.e the latter runs func every time its __next__() method is called and the other two run func for every item ia a_list immediately.



        So, the answer depends whether func is a pure function - in this case, if it behaves the same regardless of when it is run and independently from external/global variables.



        It also depends on whether the items of the list are mutable and if them or the list itself are changed by other code.



        If the answers are respectively "yes" and "no", they are same save for their memory footprint.



        Finally, as long we're discussing functional concepts, please consider using: 



        return map(func, a_list)


        Instead of:



        return (func(i) for i in a_list)


        But only if you're using Python 3's lovely lazy map() and not Python 2's evil eager map().






        share|improve this answer














        First of all, this code:



        b_list= 
        
for i in a_list:
        
    b_list.append(func(i))
        
return (i for i in b_list)


        Is entirely the same as this code:



        return iter([func(i) for i in a_list])


        The difference between these and this:



        return (func(i) for i in a_list)


        Is that the latter is lazy and the other two are eager - i.e the latter runs func every time its __next__() method is called and the other two run func for every item ia a_list immediately.



        So, the answer depends whether func is a pure function - in this case, if it behaves the same regardless of when it is run and independently from external/global variables.



        It also depends on whether the items of the list are mutable and if them or the list itself are changed by other code.



        If the answers are respectively "yes" and "no", they are same save for their memory footprint.



        Finally, as long we're discussing functional concepts, please consider using: 



        return map(func, a_list)


        Instead of:



        return (func(i) for i in a_list)


        But only if you're using Python 3's lovely lazy map() and not Python 2's evil eager map().







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 14 '18 at 23:02

























        answered Nov 14 '18 at 22:56









        roeen30

        44629




        44629






























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