get object on which a method was called in Python
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0
Is there a way in Python to get a reference to an object on which a method was called?
And in case it is, is it possible even in a nested way?
my_class.py:
from modules import math_ops
class A():
def __init__(self):
self.math_ops = math_ops.B()
self.number = 1
modules/math_ops.py:
class B():
def add_1():
where_to_add = # Get instance of A() object
where_to_add.number += 1
To execute this:
>>> a = A()
>>> a.math_ops.add_1()
And get this:
>>> a.number
2
I'm asking because I am interested in writing a static method which works with the object on which it was called, but would like to avoid using the object as an argument as it would be much nicer to call a method like my_object.prop.static_method() instead of my_object.prop.static_method(my_object).
python
asked Nov 16 '18 at 10:57
pesekon2pesekon2
254
add a comment |
0
Is there a way in Python to get a reference to an object on which a method was called?
And in case it is, is it possible even in a nested way?
my_class.py:
from modules import math_ops
class A():
def __init__(self):
self.math_ops = math_ops.B()
self.number = 1
modules/math_ops.py:
class B():
def add_1():
where_to_add = # Get instance of A() object
where_to_add.number += 1
To execute this:
>>> a = A()
>>> a.math_ops.add_1()
And get this:
>>> a.number
2
I'm asking because I am interested in writing a static method which works with the object on which it was called, but would like to avoid using the object as an argument as it would be much nicer to call a method like my_object.prop.static_method() instead of my_object.prop.static_method(my_object).
python
asked Nov 16 '18 at 10:57
pesekon2pesekon2
254
2
If your static method needs the object on which it is called, why not just use a regular instance method?
– merlyn
Nov 16 '18 at 10:59
If you mean usingselfinB.add_1(), then it would refer to the instance ofB(e.g.A.math_ops) and not to the instance ofA, I think.
– pesekon2
Nov 16 '18 at 12:27
Then I guess the statement you have a problem with is notmy_object.static_method(my_object), but rathermy_object.prop.static_method(my_object). Is that correct?
– merlyn
Nov 16 '18 at 12:40
Ouch. Right, I forgot to include it there. Sorry. Edited.
– pesekon2
Nov 16 '18 at 12:43
add a comment |
0
0
0
Is there a way in Python to get a reference to an object on which a method was called?
And in case it is, is it possible even in a nested way?
my_class.py:
from modules import math_ops
class A():
def __init__(self):
self.math_ops = math_ops.B()
self.number = 1
modules/math_ops.py:
class B():
def add_1():
where_to_add = # Get instance of A() object
where_to_add.number += 1
To execute this:
>>> a = A()
>>> a.math_ops.add_1()
And get this:
>>> a.number
2
I'm asking because I am interested in writing a static method which works with the object on which it was called, but would like to avoid using the object as an argument as it would be much nicer to call a method like my_object.prop.static_method() instead of my_object.prop.static_method(my_object).
python
asked Nov 16 '18 at 10:57
pesekon2pesekon2
254
Is there a way in Python to get a reference to an object on which a method was called?
And in case it is, is it possible even in a nested way?
my_class.py:
from modules import math_ops
class A():
def __init__(self):
self.math_ops = math_ops.B()
self.number = 1
modules/math_ops.py:
class B():
def add_1():
where_to_add = # Get instance of A() object
where_to_add.number += 1
To execute this:
>>> a = A()
>>> a.math_ops.add_1()
And get this:
>>> a.number
2
I'm asking because I am interested in writing a static method which works with the object on which it was called, but would like to avoid using the object as an argument as it would be much nicer to call a method like my_object.prop.static_method() instead of my_object.prop.static_method(my_object).
python
python
asked Nov 16 '18 at 10:57
pesekon2pesekon2
254
asked Nov 16 '18 at 10:57
pesekon2pesekon2
254
edited Nov 16 '18 at 13:20
pesekon2
asked Nov 16 '18 at 10:57
pesekon2pesekon2
254
asked Nov 16 '18 at 10:57
pesekon2pesekon2
254
asked Nov 16 '18 at 10:57
pesekon2pesekon2
254
254
2
If your static method needs the object on which it is called, why not just use a regular instance method?
– merlyn
Nov 16 '18 at 10:59
If you mean usingselfinB.add_1(), then it would refer to the instance ofB(e.g.A.math_ops) and not to the instance ofA, I think.
– pesekon2
Nov 16 '18 at 12:27
Then I guess the statement you have a problem with is notmy_object.static_method(my_object), but rathermy_object.prop.static_method(my_object). Is that correct?
– merlyn
Nov 16 '18 at 12:40
Ouch. Right, I forgot to include it there. Sorry. Edited.
– pesekon2
Nov 16 '18 at 12:43
add a comment |
2
If your static method needs the object on which it is called, why not just use a regular instance method?
– merlyn
Nov 16 '18 at 10:59
If you mean usingselfinB.add_1(), then it would refer to the instance ofB(e.g.A.math_ops) and not to the instance ofA, I think.
– pesekon2
Nov 16 '18 at 12:27
Then I guess the statement you have a problem with is notmy_object.static_method(my_object), but rathermy_object.prop.static_method(my_object). Is that correct?
– merlyn
Nov 16 '18 at 12:40
Ouch. Right, I forgot to include it there. Sorry. Edited.
– pesekon2
Nov 16 '18 at 12:43
2
2
If your static method needs the object on which it is called, why not just use a regular instance method?
– merlyn
Nov 16 '18 at 10:59
If your static method needs the object on which it is called, why not just use a regular instance method?
– merlyn
Nov 16 '18 at 10:59
If you mean using
self in B.add_1(), then it would refer to the instance of B (e.g. A.math_ops) and not to the instance of A, I think.– pesekon2
Nov 16 '18 at 12:27
If you mean using
self in B.add_1(), then it would refer to the instance of B (e.g. A.math_ops) and not to the instance of A, I think.– pesekon2
Nov 16 '18 at 12:27
Then I guess the statement you have a problem with is not
my_object.static_method(my_object), but rather my_object.prop.static_method(my_object). Is that correct?– merlyn
Nov 16 '18 at 12:40
Then I guess the statement you have a problem with is not
my_object.static_method(my_object), but rather my_object.prop.static_method(my_object). Is that correct?– merlyn
Nov 16 '18 at 12:40
Ouch. Right, I forgot to include it there. Sorry. Edited.
– pesekon2
Nov 16 '18 at 12:43
Ouch. Right, I forgot to include it there. Sorry. Edited.
– pesekon2
Nov 16 '18 at 12:43
add a comment |
2 Answers
2
active
oldest
votes
1
If you never plan on reassigning math_ops outside A, this is fairly simple to do.
from modules import math_ops
class A():
def __init__():
self.math_ops = math_ops.B(self)
self.number = 1
modules/math_ops.py:
class B():
def __init__(self, creator):
self.creator = creator
def add_1():
creator.number += 1
I will mention it again in case you skimmed the first line, the following will generate unexpected results since B is tracking the creator of the object rather than the caller.
a1 = A()
a2 = A()
a1.math_ops = a2.math_ops
a1.math_ops.add_1() # a2 is updated
If that looks like something you might wanna do, the answer is a tad more complicated. Here's my attempt:
from modules import math_ops
class A():
def __init__(self):
self._math_ops = math_ops.B(self)
self.number = 1
@property
def math_ops(self):
self._math_ops.set_caller(self)
return self._math_ops
@math_ops.setter
def math_ops(self, new_math_ops):
self._math_ops = new_math_ops
modules/math_ops.py:
class B():
def __init__(self, caller):
self.caller = caller
def set_caller(self, caller):
self.caller = caller
def add_1(self):
self.caller.number += 1
answered Nov 16 '18 at 13:04
merlynmerlyn
1,77011323
Looks good, but one question: Wouldn't it create a circular link and therefore also a possibility for memory leaks to keep the reference to theAobject inBas we are also keeping a reference toBfromA?
– pesekon2
Nov 16 '18 at 13:52
@pesekon2 Not if they don't have delete methods. See this answer.
– merlyn
Nov 16 '18 at 14:01
Oh, I see. This is for sure the correct answer, just one more question before accepting it: The memory is not released immediately and then when I am creating in a loop a lot of big objects reqriting each other (e.g. in each loop isa1 = A()), I can run out of memory when not usinggc.collect(). Is there another way to avoid it besides including it in the property definition?
– pesekon2
Nov 19 '18 at 14:37
@pesekon2 You don't have to usegc.collect(). The gc in python is enabled by default. It will collect unclaimed objects automatically before you run out of memory.
– merlyn
Nov 19 '18 at 15:05
add a comment |
-1
class A():
number = 1
class B():
def add_1():
where_to_add = A
where_to_add.number += 1
B.add_1()
print(A.number)
B.add_1()
print(A.number)
B.add_1()
print(A.number)
answered Nov 16 '18 at 11:07
KarlKarl
2,50943055
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
If you never plan on reassigning math_ops outside A, this is fairly simple to do.
from modules import math_ops
class A():
def __init__():
self.math_ops = math_ops.B(self)
self.number = 1
modules/math_ops.py:
class B():
def __init__(self, creator):
self.creator = creator
def add_1():
creator.number += 1
I will mention it again in case you skimmed the first line, the following will generate unexpected results since B is tracking the creator of the object rather than the caller.
a1 = A()
a2 = A()
a1.math_ops = a2.math_ops
a1.math_ops.add_1() # a2 is updated
If that looks like something you might wanna do, the answer is a tad more complicated. Here's my attempt:
from modules import math_ops
class A():
def __init__(self):
self._math_ops = math_ops.B(self)
self.number = 1
@property
def math_ops(self):
self._math_ops.set_caller(self)
return self._math_ops
@math_ops.setter
def math_ops(self, new_math_ops):
self._math_ops = new_math_ops
modules/math_ops.py:
class B():
def __init__(self, caller):
self.caller = caller
def set_caller(self, caller):
self.caller = caller
def add_1(self):
self.caller.number += 1
answered Nov 16 '18 at 13:04
merlynmerlyn
1,77011323
Looks good, but one question: Wouldn't it create a circular link and therefore also a possibility for memory leaks to keep the reference to theAobject inBas we are also keeping a reference toBfromA?
– pesekon2
Nov 16 '18 at 13:52
@pesekon2 Not if they don't have delete methods. See this answer.
– merlyn
Nov 16 '18 at 14:01
Oh, I see. This is for sure the correct answer, just one more question before accepting it: The memory is not released immediately and then when I am creating in a loop a lot of big objects reqriting each other (e.g. in each loop isa1 = A()), I can run out of memory when not usinggc.collect(). Is there another way to avoid it besides including it in the property definition?
– pesekon2
Nov 19 '18 at 14:37
@pesekon2 You don't have to usegc.collect(). The gc in python is enabled by default. It will collect unclaimed objects automatically before you run out of memory.
– merlyn
Nov 19 '18 at 15:05
add a comment |
1
If you never plan on reassigning math_ops outside A, this is fairly simple to do.
from modules import math_ops
class A():
def __init__():
self.math_ops = math_ops.B(self)
self.number = 1
modules/math_ops.py:
class B():
def __init__(self, creator):
self.creator = creator
def add_1():
creator.number += 1
I will mention it again in case you skimmed the first line, the following will generate unexpected results since B is tracking the creator of the object rather than the caller.
a1 = A()
a2 = A()
a1.math_ops = a2.math_ops
a1.math_ops.add_1() # a2 is updated
If that looks like something you might wanna do, the answer is a tad more complicated. Here's my attempt:
from modules import math_ops
class A():
def __init__(self):
self._math_ops = math_ops.B(self)
self.number = 1
@property
def math_ops(self):
self._math_ops.set_caller(self)
return self._math_ops
@math_ops.setter
def math_ops(self, new_math_ops):
self._math_ops = new_math_ops
modules/math_ops.py:
class B():
def __init__(self, caller):
self.caller = caller
def set_caller(self, caller):
self.caller = caller
def add_1(self):
self.caller.number += 1
answered Nov 16 '18 at 13:04
merlynmerlyn
1,77011323
Looks good, but one question: Wouldn't it create a circular link and therefore also a possibility for memory leaks to keep the reference to theAobject inBas we are also keeping a reference toBfromA?
– pesekon2
Nov 16 '18 at 13:52
@pesekon2 Not if they don't have delete methods. See this answer.
– merlyn
Nov 16 '18 at 14:01
Oh, I see. This is for sure the correct answer, just one more question before accepting it: The memory is not released immediately and then when I am creating in a loop a lot of big objects reqriting each other (e.g. in each loop isa1 = A()), I can run out of memory when not usinggc.collect(). Is there another way to avoid it besides including it in the property definition?
– pesekon2
Nov 19 '18 at 14:37
@pesekon2 You don't have to usegc.collect(). The gc in python is enabled by default. It will collect unclaimed objects automatically before you run out of memory.
– merlyn
Nov 19 '18 at 15:05
add a comment |
1
1
1
If you never plan on reassigning math_ops outside A, this is fairly simple to do.
from modules import math_ops
class A():
def __init__():
self.math_ops = math_ops.B(self)
self.number = 1
modules/math_ops.py:
class B():
def __init__(self, creator):
self.creator = creator
def add_1():
creator.number += 1
I will mention it again in case you skimmed the first line, the following will generate unexpected results since B is tracking the creator of the object rather than the caller.
a1 = A()
a2 = A()
a1.math_ops = a2.math_ops
a1.math_ops.add_1() # a2 is updated
If that looks like something you might wanna do, the answer is a tad more complicated. Here's my attempt:
from modules import math_ops
class A():
def __init__(self):
self._math_ops = math_ops.B(self)
self.number = 1
@property
def math_ops(self):
self._math_ops.set_caller(self)
return self._math_ops
@math_ops.setter
def math_ops(self, new_math_ops):
self._math_ops = new_math_ops
modules/math_ops.py:
class B():
def __init__(self, caller):
self.caller = caller
def set_caller(self, caller):
self.caller = caller
def add_1(self):
self.caller.number += 1
answered Nov 16 '18 at 13:04
merlynmerlyn
1,77011323
If you never plan on reassigning math_ops outside A, this is fairly simple to do.
from modules import math_ops
class A():
def __init__():
self.math_ops = math_ops.B(self)
self.number = 1
modules/math_ops.py:
class B():
def __init__(self, creator):
self.creator = creator
def add_1():
creator.number += 1
I will mention it again in case you skimmed the first line, the following will generate unexpected results since B is tracking the creator of the object rather than the caller.
a1 = A()
a2 = A()
a1.math_ops = a2.math_ops
a1.math_ops.add_1() # a2 is updated
If that looks like something you might wanna do, the answer is a tad more complicated. Here's my attempt:
from modules import math_ops
class A():
def __init__(self):
self._math_ops = math_ops.B(self)
self.number = 1
@property
def math_ops(self):
self._math_ops.set_caller(self)
return self._math_ops
@math_ops.setter
def math_ops(self, new_math_ops):
self._math_ops = new_math_ops
modules/math_ops.py:
class B():
def __init__(self, caller):
self.caller = caller
def set_caller(self, caller):
self.caller = caller
def add_1(self):
self.caller.number += 1
answered Nov 16 '18 at 13:04
merlynmerlyn
1,77011323
answered Nov 16 '18 at 13:04
merlynmerlyn
1,77011323
answered Nov 16 '18 at 13:04
merlynmerlyn
1,77011323
answered Nov 16 '18 at 13:04
merlynmerlyn
1,77011323
1,77011323
Looks good, but one question: Wouldn't it create a circular link and therefore also a possibility for memory leaks to keep the reference to theAobject inBas we are also keeping a reference toBfromA?
– pesekon2
Nov 16 '18 at 13:52
@pesekon2 Not if they don't have delete methods. See this answer.
– merlyn
Nov 16 '18 at 14:01
Oh, I see. This is for sure the correct answer, just one more question before accepting it: The memory is not released immediately and then when I am creating in a loop a lot of big objects reqriting each other (e.g. in each loop isa1 = A()), I can run out of memory when not usinggc.collect(). Is there another way to avoid it besides including it in the property definition?
– pesekon2
Nov 19 '18 at 14:37
@pesekon2 You don't have to usegc.collect(). The gc in python is enabled by default. It will collect unclaimed objects automatically before you run out of memory.
– merlyn
Nov 19 '18 at 15:05
add a comment |
Looks good, but one question: Wouldn't it create a circular link and therefore also a possibility for memory leaks to keep the reference to theAobject inBas we are also keeping a reference toBfromA?
– pesekon2
Nov 16 '18 at 13:52
@pesekon2 Not if they don't have delete methods. See this answer.
– merlyn
Nov 16 '18 at 14:01
Oh, I see. This is for sure the correct answer, just one more question before accepting it: The memory is not released immediately and then when I am creating in a loop a lot of big objects reqriting each other (e.g. in each loop isa1 = A()), I can run out of memory when not usinggc.collect(). Is there another way to avoid it besides including it in the property definition?
– pesekon2
Nov 19 '18 at 14:37
@pesekon2 You don't have to usegc.collect(). The gc in python is enabled by default. It will collect unclaimed objects automatically before you run out of memory.
– merlyn
Nov 19 '18 at 15:05
Looks good, but one question: Wouldn't it create a circular link and therefore also a possibility for memory leaks to keep the reference to the
A object in B as we are also keeping a reference to B from A?– pesekon2
Nov 16 '18 at 13:52
Looks good, but one question: Wouldn't it create a circular link and therefore also a possibility for memory leaks to keep the reference to the
A object in B as we are also keeping a reference to B from A?– pesekon2
Nov 16 '18 at 13:52
@pesekon2 Not if they don't have delete methods. See this answer.
– merlyn
Nov 16 '18 at 14:01
@pesekon2 Not if they don't have delete methods. See this answer.
– merlyn
Nov 16 '18 at 14:01
Oh, I see. This is for sure the correct answer, just one more question before accepting it: The memory is not released immediately and then when I am creating in a loop a lot of big objects reqriting each other (e.g. in each loop is
a1 = A()), I can run out of memory when not using gc.collect(). Is there another way to avoid it besides including it in the property definition?– pesekon2
Nov 19 '18 at 14:37
Oh, I see. This is for sure the correct answer, just one more question before accepting it: The memory is not released immediately and then when I am creating in a loop a lot of big objects reqriting each other (e.g. in each loop is
a1 = A()), I can run out of memory when not using gc.collect(). Is there another way to avoid it besides including it in the property definition?– pesekon2
Nov 19 '18 at 14:37
@pesekon2 You don't have to use
gc.collect(). The gc in python is enabled by default. It will collect unclaimed objects automatically before you run out of memory.– merlyn
Nov 19 '18 at 15:05
@pesekon2 You don't have to use
gc.collect(). The gc in python is enabled by default. It will collect unclaimed objects automatically before you run out of memory.– merlyn
Nov 19 '18 at 15:05
add a comment |
-1
class A():
number = 1
class B():
def add_1():
where_to_add = A
where_to_add.number += 1
B.add_1()
print(A.number)
B.add_1()
print(A.number)
B.add_1()
print(A.number)
answered Nov 16 '18 at 11:07
KarlKarl
2,50943055
add a comment |
-1
class A():
number = 1
class B():
def add_1():
where_to_add = A
where_to_add.number += 1
B.add_1()
print(A.number)
B.add_1()
print(A.number)
B.add_1()
print(A.number)
answered Nov 16 '18 at 11:07
KarlKarl
2,50943055
add a comment |
-1
-1
-1
class A():
number = 1
class B():
def add_1():
where_to_add = A
where_to_add.number += 1
B.add_1()
print(A.number)
B.add_1()
print(A.number)
B.add_1()
print(A.number)
answered Nov 16 '18 at 11:07
KarlKarl
2,50943055
class A():
number = 1
class B():
def add_1():
where_to_add = A
where_to_add.number += 1
B.add_1()
print(A.number)
B.add_1()
print(A.number)
B.add_1()
print(A.number)
answered Nov 16 '18 at 11:07
KarlKarl
2,50943055
answered Nov 16 '18 at 11:07
KarlKarl
2,50943055
answered Nov 16 '18 at 11:07
KarlKarl
2,50943055
answered Nov 16 '18 at 11:07
KarlKarl
2,50943055
2,50943055
add a comment |
add a comment |
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2
If your static method needs the object on which it is called, why not just use a regular instance method?
– merlyn
Nov 16 '18 at 10:59
If you mean using
selfinB.add_1(), then it would refer to the instance ofB(e.g.A.math_ops) and not to the instance ofA, I think.– pesekon2
Nov 16 '18 at 12:27
Then I guess the statement you have a problem with is not
my_object.static_method(my_object), but rathermy_object.prop.static_method(my_object). Is that correct?– merlyn
Nov 16 '18 at 12:40
Ouch. Right, I forgot to include it there. Sorry. Edited.
– pesekon2
Nov 16 '18 at 12:43