Different behavior between generator expression and first constructing a list then using list-converted...
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
asked Nov 14 '18 at 19:43
jmf_zaiecp
12117
add a comment |
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
asked Nov 14 '18 at 19:43
jmf_zaiecp
12117
1
Yes, if thefunc(i)is influenced by the state of something external.
– Willem Van Onsem
Nov 14 '18 at 19:44
add a comment |
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
asked Nov 14 '18 at 19:43
jmf_zaiecp
12117
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
asked Nov 14 '18 at 19:43
jmf_zaiecp
12117
asked Nov 14 '18 at 19:43
jmf_zaiecp
12117
asked Nov 14 '18 at 19:43
jmf_zaiecp
12117
asked Nov 14 '18 at 19:43
jmf_zaiecp
12117
asked Nov 14 '18 at 19:43
jmf_zaiecp
12117
12117
1
Yes, if thefunc(i)is influenced by the state of something external.
– Willem Van Onsem
Nov 14 '18 at 19:44
add a comment |
1
Yes, if thefunc(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
add a comment |
1 Answer
1
active
oldest
votes
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().
answered Nov 14 '18 at 22:56
roeen30
44629
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
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().
answered Nov 14 '18 at 22:56
roeen30
44629
add a comment |
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().
answered Nov 14 '18 at 22:56
roeen30
44629
add a comment |
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().
answered Nov 14 '18 at 22:56
roeen30
44629
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().
answered Nov 14 '18 at 22:56
roeen30
44629
edited Nov 14 '18 at 23:02
answered Nov 14 '18 at 22:56
roeen30
44629
answered Nov 14 '18 at 22:56
roeen30
44629
answered Nov 14 '18 at 22:56
roeen30
44629
44629
add a comment |
add a comment |
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1
Yes, if the
func(i)is influenced by the state of something external.– Willem Van Onsem
Nov 14 '18 at 19:44