Python list() function on (non) iterable
I'm facing an issue with the list() function. I have a simple class
class Foo(object):
def __init__(self, value):
self.value = value
And I'm trying to construct a list with the Foo object:
lhs = list(Foo(1))
...
rhs = list(Foo(2))
return lhs + rhs
The desired result should be [Foo(1), Foo(2)]. However I need to use such result in the recursive call, so in the following step this result becomes the argument of the list() function of lhs and I want to append such list by another Foo element:
lhs = list([Foo(1), Foo(2)])
...
rhs = list(Foo(3))
return lhs + rhs
And the final result should be [Foo(1), Foo(2), Foo(3)]. The issue is that the list(Foo(1)) function raises the TypeError because the Foo is not iterable. Is there any way how to fix it?
EDIT
A better use case:
def create_foo(value):
return Foo(value)
def concatenate(lhs, rhs):
return list(lhs) + list(rhs)
concatenate(concatenate(create_foo(1), create_foo(2)), create_foo(3))
python-3.x list
asked Jan 18 at 21:27
marthin23marthin23
102210
add a comment |
I'm facing an issue with the list() function. I have a simple class
class Foo(object):
def __init__(self, value):
self.value = value
And I'm trying to construct a list with the Foo object:
lhs = list(Foo(1))
...
rhs = list(Foo(2))
return lhs + rhs
The desired result should be [Foo(1), Foo(2)]. However I need to use such result in the recursive call, so in the following step this result becomes the argument of the list() function of lhs and I want to append such list by another Foo element:
lhs = list([Foo(1), Foo(2)])
...
rhs = list(Foo(3))
return lhs + rhs
And the final result should be [Foo(1), Foo(2), Foo(3)]. The issue is that the list(Foo(1)) function raises the TypeError because the Foo is not iterable. Is there any way how to fix it?
EDIT
A better use case:
def create_foo(value):
return Foo(value)
def concatenate(lhs, rhs):
return list(lhs) + list(rhs)
concatenate(concatenate(create_foo(1), create_foo(2)), create_foo(3))
python-3.x list
asked Jan 18 at 21:27
marthin23marthin23
102210
3
list((Foo(1),))? Or just[Foo(1)]?
– jonrsharpe
Jan 18 at 21:30
It sounds like you just need an explicit base case for when you have a single item on the left hand side (or no items returning an empty list, depending what you're doing).
– Patrick Haugh
Jan 18 at 21:32
@jonrsharpe I added a better example. The[Foo(1)]would create nested lists which are not the desired result
– marthin23
Jan 18 at 21:51
1
Why are you trying to pass bothFooinstances and sequences ofFooinstances toconcatenate? That seems like a bad idea.
– user2357112
Jan 19 at 6:20
add a comment |
I'm facing an issue with the list() function. I have a simple class
class Foo(object):
def __init__(self, value):
self.value = value
And I'm trying to construct a list with the Foo object:
lhs = list(Foo(1))
...
rhs = list(Foo(2))
return lhs + rhs
The desired result should be [Foo(1), Foo(2)]. However I need to use such result in the recursive call, so in the following step this result becomes the argument of the list() function of lhs and I want to append such list by another Foo element:
lhs = list([Foo(1), Foo(2)])
...
rhs = list(Foo(3))
return lhs + rhs
And the final result should be [Foo(1), Foo(2), Foo(3)]. The issue is that the list(Foo(1)) function raises the TypeError because the Foo is not iterable. Is there any way how to fix it?
EDIT
A better use case:
def create_foo(value):
return Foo(value)
def concatenate(lhs, rhs):
return list(lhs) + list(rhs)
concatenate(concatenate(create_foo(1), create_foo(2)), create_foo(3))
python-3.x list
asked Jan 18 at 21:27
marthin23marthin23
102210
I'm facing an issue with the list() function. I have a simple class
class Foo(object):
def __init__(self, value):
self.value = value
And I'm trying to construct a list with the Foo object:
lhs = list(Foo(1))
...
rhs = list(Foo(2))
return lhs + rhs
The desired result should be [Foo(1), Foo(2)]. However I need to use such result in the recursive call, so in the following step this result becomes the argument of the list() function of lhs and I want to append such list by another Foo element:
lhs = list([Foo(1), Foo(2)])
...
rhs = list(Foo(3))
return lhs + rhs
And the final result should be [Foo(1), Foo(2), Foo(3)]. The issue is that the list(Foo(1)) function raises the TypeError because the Foo is not iterable. Is there any way how to fix it?
EDIT
A better use case:
def create_foo(value):
return Foo(value)
def concatenate(lhs, rhs):
return list(lhs) + list(rhs)
concatenate(concatenate(create_foo(1), create_foo(2)), create_foo(3))
python-3.x list
python-3.x list
asked Jan 18 at 21:27
marthin23marthin23
102210
asked Jan 18 at 21:27
marthin23marthin23
102210
edited Jan 18 at 21:42
marthin23
asked Jan 18 at 21:27
marthin23marthin23
102210
asked Jan 18 at 21:27
marthin23marthin23
102210
asked Jan 18 at 21:27
marthin23marthin23
102210
102210
3
list((Foo(1),))? Or just[Foo(1)]?
– jonrsharpe
Jan 18 at 21:30
It sounds like you just need an explicit base case for when you have a single item on the left hand side (or no items returning an empty list, depending what you're doing).
– Patrick Haugh
Jan 18 at 21:32
@jonrsharpe I added a better example. The[Foo(1)]would create nested lists which are not the desired result
– marthin23
Jan 18 at 21:51
1
Why are you trying to pass bothFooinstances and sequences ofFooinstances toconcatenate? That seems like a bad idea.
– user2357112
Jan 19 at 6:20
add a comment |
3
list((Foo(1),))? Or just[Foo(1)]?
– jonrsharpe
Jan 18 at 21:30
It sounds like you just need an explicit base case for when you have a single item on the left hand side (or no items returning an empty list, depending what you're doing).
– Patrick Haugh
Jan 18 at 21:32
@jonrsharpe I added a better example. The[Foo(1)]would create nested lists which are not the desired result
– marthin23
Jan 18 at 21:51
1
Why are you trying to pass bothFooinstances and sequences ofFooinstances toconcatenate? That seems like a bad idea.
– user2357112
Jan 19 at 6:20
3
3
list((Foo(1),))? Or just [Foo(1)]?– jonrsharpe
Jan 18 at 21:30
list((Foo(1),))? Or just [Foo(1)]?– jonrsharpe
Jan 18 at 21:30
It sounds like you just need an explicit base case for when you have a single item on the left hand side (or no items returning an empty list, depending what you're doing).
– Patrick Haugh
Jan 18 at 21:32
It sounds like you just need an explicit base case for when you have a single item on the left hand side (or no items returning an empty list, depending what you're doing).
– Patrick Haugh
Jan 18 at 21:32
@jonrsharpe I added a better example. The
[Foo(1)] would create nested lists which are not the desired result– marthin23
Jan 18 at 21:51
@jonrsharpe I added a better example. The
[Foo(1)] would create nested lists which are not the desired result– marthin23
Jan 18 at 21:51
1
1
Why are you trying to pass both
Foo instances and sequences of Foo instances to concatenate? That seems like a bad idea.– user2357112
Jan 19 at 6:20
Why are you trying to pass both
Foo instances and sequences of Foo instances to concatenate? That seems like a bad idea.– user2357112
Jan 19 at 6:20
add a comment |
1 Answer
1
active
oldest
votes
You could just check for lists and use .extend to concatenate, or .append to add non-list items:
class Foo:
def __init__(self, value):
self.value = value
def __repr__(self):
return f'Foo({self.value})'
def concat(*args):
result =
for arg in args:
if isinstance(arg,list):
result.extend(arg)
else:
result.append(arg)
return result
a = concat(Foo(1),Foo(2),Foo(3))
print(a)
b = concat(Foo(6),Foo(7))
print(b)
c = concat(a,Foo(4),Foo(5),b)
print(c)
Output:
[Foo(1), Foo(2), Foo(3)]
[Foo(6), Foo(7)]
[Foo(1), Foo(2), Foo(3), Foo(4), Foo(5), Foo(6), Foo(7)]
answered Jan 19 at 6:53
Mark TolonenMark Tolonen
92.7k12112176
Brilliant solution! Thanks a lot
– marthin23
Jan 20 at 19:57
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 could just check for lists and use .extend to concatenate, or .append to add non-list items:
class Foo:
def __init__(self, value):
self.value = value
def __repr__(self):
return f'Foo({self.value})'
def concat(*args):
result =
for arg in args:
if isinstance(arg,list):
result.extend(arg)
else:
result.append(arg)
return result
a = concat(Foo(1),Foo(2),Foo(3))
print(a)
b = concat(Foo(6),Foo(7))
print(b)
c = concat(a,Foo(4),Foo(5),b)
print(c)
Output:
[Foo(1), Foo(2), Foo(3)]
[Foo(6), Foo(7)]
[Foo(1), Foo(2), Foo(3), Foo(4), Foo(5), Foo(6), Foo(7)]
answered Jan 19 at 6:53
Mark TolonenMark Tolonen
92.7k12112176
Brilliant solution! Thanks a lot
– marthin23
Jan 20 at 19:57
add a comment |
You could just check for lists and use .extend to concatenate, or .append to add non-list items:
class Foo:
def __init__(self, value):
self.value = value
def __repr__(self):
return f'Foo({self.value})'
def concat(*args):
result =
for arg in args:
if isinstance(arg,list):
result.extend(arg)
else:
result.append(arg)
return result
a = concat(Foo(1),Foo(2),Foo(3))
print(a)
b = concat(Foo(6),Foo(7))
print(b)
c = concat(a,Foo(4),Foo(5),b)
print(c)
Output:
[Foo(1), Foo(2), Foo(3)]
[Foo(6), Foo(7)]
[Foo(1), Foo(2), Foo(3), Foo(4), Foo(5), Foo(6), Foo(7)]
answered Jan 19 at 6:53
Mark TolonenMark Tolonen
92.7k12112176
Brilliant solution! Thanks a lot
– marthin23
Jan 20 at 19:57
add a comment |
You could just check for lists and use .extend to concatenate, or .append to add non-list items:
class Foo:
def __init__(self, value):
self.value = value
def __repr__(self):
return f'Foo({self.value})'
def concat(*args):
result =
for arg in args:
if isinstance(arg,list):
result.extend(arg)
else:
result.append(arg)
return result
a = concat(Foo(1),Foo(2),Foo(3))
print(a)
b = concat(Foo(6),Foo(7))
print(b)
c = concat(a,Foo(4),Foo(5),b)
print(c)
Output:
[Foo(1), Foo(2), Foo(3)]
[Foo(6), Foo(7)]
[Foo(1), Foo(2), Foo(3), Foo(4), Foo(5), Foo(6), Foo(7)]
answered Jan 19 at 6:53
Mark TolonenMark Tolonen
92.7k12112176
You could just check for lists and use .extend to concatenate, or .append to add non-list items:
class Foo:
def __init__(self, value):
self.value = value
def __repr__(self):
return f'Foo({self.value})'
def concat(*args):
result =
for arg in args:
if isinstance(arg,list):
result.extend(arg)
else:
result.append(arg)
return result
a = concat(Foo(1),Foo(2),Foo(3))
print(a)
b = concat(Foo(6),Foo(7))
print(b)
c = concat(a,Foo(4),Foo(5),b)
print(c)
Output:
[Foo(1), Foo(2), Foo(3)]
[Foo(6), Foo(7)]
[Foo(1), Foo(2), Foo(3), Foo(4), Foo(5), Foo(6), Foo(7)]
answered Jan 19 at 6:53
Mark TolonenMark Tolonen
92.7k12112176
answered Jan 19 at 6:53
Mark TolonenMark Tolonen
92.7k12112176
answered Jan 19 at 6:53
Mark TolonenMark Tolonen
92.7k12112176
answered Jan 19 at 6:53
Mark TolonenMark Tolonen
92.7k12112176
92.7k12112176
Brilliant solution! Thanks a lot
– marthin23
Jan 20 at 19:57
add a comment |
Brilliant solution! Thanks a lot
– marthin23
Jan 20 at 19:57
Brilliant solution! Thanks a lot
– marthin23
Jan 20 at 19:57
Brilliant solution! Thanks a lot
– marthin23
Jan 20 at 19:57
add a comment |
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3
list((Foo(1),))? Or just[Foo(1)]?– jonrsharpe
Jan 18 at 21:30
It sounds like you just need an explicit base case for when you have a single item on the left hand side (or no items returning an empty list, depending what you're doing).
– Patrick Haugh
Jan 18 at 21:32
@jonrsharpe I added a better example. The
[Foo(1)]would create nested lists which are not the desired result– marthin23
Jan 18 at 21:51
1
Why are you trying to pass both
Fooinstances and sequences ofFooinstances toconcatenate? That seems like a bad idea.– user2357112
Jan 19 at 6:20