Python - implement CUSTOM "lshift" method - but not for bitarray elements - python

I have a class which has as parameter list of integer values.
I must implement custom lshift method which will do the adding to the list (I must override lshift method for my use case). This is custom lshift method so it does not have anything with bitarrays but only working with integers in the list (adding it to the list ) and it receives argument element. I should do this without import of additional python classes. I have defined also tests for this functionality so you can see how result should look like. (I have already implemented custom methods for add,len and iter but I suppose this is not relevant)
class CustomShift:
def __init__(self, iterator=None):
self.iterator = iterator
if (self.iterator):
self.iterator = list(dict.fromkeys(self.iterator))
def __lshift__(self, element):
"""Add an element to the list.
>>> shiftInstance = CustomShift()
>>> _ = shiftInstance << 4
>>> sorted(shiftInstance << 5 << 6 << 4)
[4, 5, 6]
"""
if self.iterator is None:
self.iterator = []
if element:
self.iterator.append(element)
if __name__ == "__main__":
import doctest
from pprint import pprint
doctest.testmod()
first two tests pass, but third fails!
TypeError: unsupported operand type(s) for <<: 'NoneType' and 'int'
Not sure what I am doing wrong, any hint will be appreciated.
Thanks in advance

First off, __lshift__ need to return a value - even if it mutates the instance it is working on. In order for chaining to work the way you expect to, it must return itself. Since the first << operation doesn't return anything, any subsequent ones will result in the TypeError exception you noted.
Second, the __lshift__ method has a logic error - the element is only appended if the iterator argument is None. Thus in the most general use case that you intend, nothing would actually happen.
Finally, in your test case, you wanted to call sort on the object which implies that this object need to provide some kind of iterator (via __iter__), otherwise it will simply fail with TypeError: 'CustomShift' object is not iterable. Putting this together, your class will look like so:
class CustomShift:
def __init__(self, iterator=None):
self.iterator = iterator
if (self.iterator):
self.iterator = list(dict.fromkeys(self.iterator))
def __lshift__(self, element):
"""Add an element to the list.
>>> shiftInstance = CustomShift()
>>> _ = shiftInstance << 4
>>> sorted(shiftInstance << 5 << 6 << 4)
[4, 5, 6]
"""
if self.iterator is None:
self.iterator = []
if element:
self.iterator.append(element)
return self
def __iter__(self):
return iter(self.iterator)
However, the test will still fail, because as it is structured, shiftInstance already has a 4 appended which was assigned to _.
Failed example:
sorted(shiftInstance << 5 << 6 << 4)
Expected:
[4, 5, 6]
Got:
[4, 4, 5, 6]
This should however put you towards a direction on how you might want to proceed from what you got.

Related

Using the map() function within a class

I've been trying to get the map() function to work within a class but have had trouble because I'm not sure if I should be passing self into it. If so, I'm unsure how to make self into a list to go with my other iterables. Here is my code so far:
from itertools import repeat
class test:
def __init__(self):
self.nums = [1, 4, 8]
self.empty_list = []
map(self.fxn, repeat(self, len(self.nums)), self.nums)
print(self.empty_list)
def fxn(self, num):
self.empty_list.append(num ** num)
instance = test()
Even after trying to append to the empty list, the list still seems to be blank, what am I doing wrong in that example?
map doesn't mutate its argument. It returns a new iterable.
self.nums = list(map(...))
test.fxn is a function of two arguments: self and num. self.fxn is a bound method of one argument: num. Since you're just repeatedly applying it on self, you can bind it and save yourself the extra argument.
self.nums = list(map(self.fxn, self.nums))
You can change map(self.fxn, repeat(self, len(self.nums)), self.nums) to self.empty_list=list(map(self.fxn,self.nums)) which is equivalent to self.empty_list=list(map(lambda x: self.fxn(x),self.nums)) and self.empty_list=[self.fxn(i) for i in self.nums] and remember to change fxn() from self.empty_list.append(num ** num) to return num ** num (and since it is list(map(...))-list it will be a list), so try the below:
class test:
def __init__(self):
self.nums = [1, 4, 8]
self.empty_list=list(map(self.fxn,self.nums))
print(self.empty_list)
def fxn(self, num):
return num ** num
instance = test()
Output:
[1, 256, 16777216]

Python - append reference in a list [duplicate]

I know Python doesn't have pointers, but is there a way to have this yield 2 instead
>>> a = 1
>>> b = a # modify this line somehow so that b "points to" a
>>> a = 2
>>> b
1
?
Here's an example: I want form.data['field'] and form.field.value to always have the same value. It's not completely necessary, but I think it would be nice.
In PHP, for example, I can do this:
<?php
class Form {
public $data = [];
public $fields;
function __construct($fields) {
$this->fields = $fields;
foreach($this->fields as &$field) {
$this->data[$field['id']] = &$field['value'];
}
}
}
$f = new Form([
[
'id' => 'fname',
'value' => 'George'
],
[
'id' => 'lname',
'value' => 'Lucas'
]
]);
echo $f->data['fname'], $f->fields[0]['value']; # George George
$f->data['fname'] = 'Ralph';
echo $f->data['fname'], $f->fields[0]['value']; # Ralph Ralph
Output:
GeorgeGeorgeRalphRalph
ideone
Or like this in C++ (I think this is right, but my C++ is rusty):
#include <iostream>
using namespace std;
int main() {
int* a;
int* b = a;
*a = 1;
cout << *a << endl << *b << endl; # 1 1
return 0;
}
There's no way you can do that changing only that line. You can do:
a = [1]
b = a
a[0] = 2
b[0]
That creates a list, assigns the reference to a, then b also, uses the a reference to set the first element to 2, then accesses using the b reference variable.
I want form.data['field'] and
form.field.value to always have the
same value
This is feasible, because it involves decorated names and indexing -- i.e., completely different constructs from the barenames a and b that you're asking about, and for with your request is utterly impossible. Why ask for something impossible and totally different from the (possible) thing you actually want?!
Maybe you don't realize how drastically different barenames and decorated names are. When you refer to a barename a, you're getting exactly the object a was last bound to in this scope (or an exception if it wasn't bound in this scope) -- this is such a deep and fundamental aspect of Python that it can't possibly be subverted. When you refer to a decorated name x.y, you're asking an object (the object x refers to) to please supply "the y attribute" -- and in response to that request, the object can perform totally arbitrary computations (and indexing is quite similar: it also allows arbitrary computations to be performed in response).
Now, your "actual desiderata" example is mysterious because in each case two levels of indexing or attribute-getting are involved, so the subtlety you crave could be introduced in many ways. What other attributes is form.field suppose to have, for example, besides value? Without that further .value computations, possibilities would include:
class Form(object):
...
def __getattr__(self, name):
return self.data[name]
and
class Form(object):
...
#property
def data(self):
return self.__dict__
The presence of .value suggests picking the first form, plus a kind-of-useless wrapper:
class KouWrap(object):
def __init__(self, value):
self.value = value
class Form(object):
...
def __getattr__(self, name):
return KouWrap(self.data[name])
If assignments such form.field.value = 23 is also supposed to set the entry in form.data, then the wrapper must become more complex indeed, and not all that useless:
class MciWrap(object):
def __init__(self, data, k):
self._data = data
self._k = k
#property
def value(self):
return self._data[self._k]
#value.setter
def value(self, v)
self._data[self._k] = v
class Form(object):
...
def __getattr__(self, name):
return MciWrap(self.data, name)
The latter example is roughly as close as it gets, in Python, to the sense of "a pointer" as you seem to want -- but it's crucial to understand that such subtleties can ever only work with indexing and/or decorated names, never with barenames as you originally asked!
It's not a bug, it's a feature :-)
When you look at the '=' operator in Python, don't think in terms of assignment. You don't assign things, you bind them. = is a binding operator.
So in your code, you are giving the value 1 a name: a. Then, you are giving the value in 'a' a name: b. Then you are binding the value 2 to the name 'a'. The value bound to b doesn't change in this operation.
Coming from C-like languages, this can be confusing, but once you become accustomed to it, you find that it helps you to read and reason about your code more clearly: the value which has the name 'b' will not change unless you explicitly change it. And if you do an 'import this', you'll find that the Zen of Python states that Explicit is better than implicit.
Note as well that functional languages such as Haskell also use this paradigm, with great value in terms of robustness.
Yes! there is a way to use a variable as a pointer in python!
I am sorry to say that many of answers were partially wrong. In principle every equal(=) assignation shares the memory address (check the id(obj) function), but in practice it is not such. There are variables whose equal("=") behaviour works in last term as a copy of memory space, mostly in simple objects (e.g. "int" object), and others in which not (e.g. "list","dict" objects).
Here is an example of pointer assignation
dict1 = {'first':'hello', 'second':'world'}
dict2 = dict1 # pointer assignation mechanism
dict2['first'] = 'bye'
dict1
>>> {'first':'bye', 'second':'world'}
Here is an example of copy assignation
a = 1
b = a # copy of memory mechanism. up to here id(a) == id(b)
b = 2 # new address generation. therefore without pointer behaviour
a
>>> 1
Pointer assignation is a pretty useful tool for aliasing without the waste of extra memory, in certain situations for performing comfy code,
class cls_X():
...
def method_1():
pd1 = self.obj_clsY.dict_vars_for_clsX['meth1'] # pointer dict 1: aliasing
pd1['var4'] = self.method2(pd1['var1'], pd1['var2'], pd1['var3'])
#enddef method_1
...
#endclass cls_X
but one have to be aware of this use in order to prevent code mistakes.
To conclude, by default some variables are barenames (simple objects like int, float, str,...), and some are pointers when assigned between them (e.g. dict1 = dict2). How to recognize them? just try this experiment with them. In IDEs with variable explorer panel usually appears to be the memory address ("#axbbbbbb...") in the definition of pointer-mechanism objects.
I suggest investigate in the topic. There are many people who know much more about this topic for sure. (see "ctypes" module). I hope it is helpful. Enjoy the good use of the objects! Regards, José Crespo
>> id(1)
1923344848 # identity of the location in memory where 1 is stored
>> id(1)
1923344848 # always the same
>> a = 1
>> b = a # or equivalently b = 1, because 1 is immutable
>> id(a)
1923344848
>> id(b) # equal to id(a)
1923344848
As you can see a and b are just two different names that reference to the same immutable object (int) 1. If later you write a = 2, you reassign the name a to a different object (int) 2, but the b continues referencing to 1:
>> id(2)
1923344880
>> a = 2
>> id(a)
1923344880 # equal to id(2)
>> b
1 # b hasn't changed
>> id(b)
1923344848 # equal to id(1)
What would happen if you had a mutable object instead, such as a list [1]?
>> id([1])
328817608
>> id([1])
328664968 # different from the previous id, because each time a new list is created
>> a = [1]
>> id(a)
328817800
>> id(a)
328817800 # now same as before
>> b = a
>> id(b)
328817800 # same as id(a)
Again, we are referencing to the same object (list) [1] by two different names a and b. However now we can mutate this list while it remains the same object, and a, b will both continue referencing to it
>> a[0] = 2
>> a
[2]
>> b
[2]
>> id(a)
328817800 # same as before
>> id(b)
328817800 # same as before
From one point of view, everything is a pointer in Python. Your example works a lot like the C++ code.
int* a = new int(1);
int* b = a;
a = new int(2);
cout << *b << endl; // prints 1
(A closer equivalent would use some type of shared_ptr<Object> instead of int*.)
Here's an example: I want
form.data['field'] and
form.field.value to always have the
same value. It's not completely
necessary, but I think it would be
nice.
You can do this by overloading __getitem__ in form.data's class.
This is a python pointer (different of c/c++)
>>> a = lambda : print('Hello')
>>> a
<function <lambda> at 0x0000018D192B9DC0>
>>> id(a) == int(0x0000018D192B9DC0)
True
>>> from ctypes import cast, py_object
>>> cast(id(a), py_object).value == cast(int(0x0000018D192B9DC0), py_object).value
True
>>> cast(id(a), py_object).value
<function <lambda> at 0x0000018D192B9DC0>
>>> cast(id(a), py_object).value()
Hello
I wrote the following simple class as, effectively, a way to emulate a pointer in python:
class Parameter:
"""Syntactic sugar for getter/setter pair
Usage:
p = Parameter(getter, setter)
Set parameter value:
p(value)
p.val = value
p.set(value)
Retrieve parameter value:
p()
p.val
p.get()
"""
def __init__(self, getter, setter):
"""Create parameter
Required positional parameters:
getter: called with no arguments, retrieves the parameter value.
setter: called with value, sets the parameter.
"""
self._get = getter
self._set = setter
def __call__(self, val=None):
if val is not None:
self._set(val)
return self._get()
def get(self):
return self._get()
def set(self, val):
self._set(val)
#property
def val(self):
return self._get()
#val.setter
def val(self, val):
self._set(val)
Here's an example of use (from a jupyter notebook page):
l1 = list(range(10))
def l1_5_getter(lst=l1, number=5):
return lst[number]
def l1_5_setter(val, lst=l1, number=5):
lst[number] = val
[
l1_5_getter(),
l1_5_setter(12),
l1,
l1_5_getter()
]
Out = [5, None, [0, 1, 2, 3, 4, 12, 6, 7, 8, 9], 12]
p = Parameter(l1_5_getter, l1_5_setter)
print([
p(),
p.get(),
p.val,
p(13),
p(),
p.set(14),
p.get()
])
p.val = 15
print(p.val, l1)
[12, 12, 12, 13, 13, None, 14]
15 [0, 1, 2, 3, 4, 15, 6, 7, 8, 9]
Of course, it is also easy to make this work for dict items or attributes of an object. There is even a way to do what the OP asked for, using globals():
def setter(val, dict=globals(), key='a'):
dict[key] = val
def getter(dict=globals(), key='a'):
return dict[key]
pa = Parameter(getter, setter)
pa(2)
print(a)
pa(3)
print(a)
This will print out 2, followed by 3.
Messing with the global namespace in this way is kind of transparently a terrible idea, but it shows that it is possible (if inadvisable) to do what the OP asked for.
The example is, of course, fairly pointless. But I have found this class to be useful in the application for which I developed it: a mathematical model whose behavior is governed by numerous user-settable mathematical parameters, of diverse types (which, because they depend on command line arguments, are not known at compile time). And once access to something has been encapsulated in a Parameter object, all such objects can be manipulated in a uniform way.
Although it doesn't look much like a C or C++ pointer, this is solving a problem that I would have solved with pointers if I were writing in C++.
The following code emulates exactly the behavior of pointers in C:
from collections import deque # more efficient than list for appending things
pointer_storage = deque()
pointer_address = 0
class new:
def __init__(self):
global pointer_storage
global pointer_address
self.address = pointer_address
self.val = None
pointer_storage.append(self)
pointer_address += 1
def get_pointer(address):
return pointer_storage[address]
def get_address(p):
return p.address
null = new() # create a null pointer, whose address is 0
Here are examples of use:
p = new()
p.val = 'hello'
q = new()
q.val = p
r = new()
r.val = 33
p = get_pointer(3)
print(p.val, flush = True)
p.val = 43
print(get_pointer(3).val, flush = True)
But it's now time to give a more professional code, including the option of deleting pointers, that I've just found in my personal library:
# C pointer emulation:
from collections import deque # more efficient than list for appending things
from sortedcontainers import SortedList #perform add and discard in log(n) times
class new:
# C pointer emulation:
# use as : p = new()
# p.val
# p.val = something
# p.address
# get_address(p)
# del_pointer(p)
# null (a null pointer)
__pointer_storage__ = SortedList(key = lambda p: p.address)
__to_delete_pointers__ = deque()
__pointer_address__ = 0
def __init__(self):
self.val = None
if new.__to_delete_pointers__:
p = new.__to_delete_pointers__.pop()
self.address = p.address
new.__pointer_storage__.discard(p) # performed in log(n) time thanks to sortedcontainers
new.__pointer_storage__.add(self) # idem
else:
self.address = new.__pointer_address__
new.__pointer_storage__.add(self)
new.__pointer_address__ += 1
def get_pointer(address):
return new.__pointer_storage__[address]
def get_address(p):
return p.address
def del_pointer(p):
new.__to_delete_pointers__.append(p)
null = new() # create a null pointer, whose address is 0
I don't know if my comment will help or not but if you want to use pointers in python, you can use dictionaries instead of variables
Let's say in your example will be
a = {'value': 1}
b = {'value': 2}
then you changed a to 2
a['value'] = 2
print(a) #{'value': 2}

Annoying generator bug

The original context of this bug is a piece of code too large to post in a question like this. I had to whittle this code down to a minimal snippet that still exhibits the bug. This is why the code shown below is somewhat bizarre-looking.
In the code below, the class Foo may thought of as a convoluted way to get something like xrange.
class Foo(object):
def __init__(self, n):
self.generator = (x for x in range(n))
def __iter__(self):
for e in self.generator:
yield e
Indeed, Foo seems to behave very much like xrange:
for c in Foo(3):
print c
# 0
# 1
# 2
print list(Foo(3))
# [0, 1, 2]
Now, the subclass Bar of Foo adds only a __len__ method:
class Bar(Foo):
def __len__(self):
return sum(1 for _ in self.generator)
Bar behaves just like Foo when used in a for-loop:
for c in Bar(3):
print c
# 0
# 1
# 2
BUT:
print list(Bar(3))
# []
My guess is that, in the evaluation of list(Bar(3)), the __len__ method of Bar(3) is getting called, thereby using up the generator.
(If this guess is correct, the call to Bar(3).__len__ is unnecessary; after all, list(Foo(3)) produces the correct result even though Foo has no __len__ method.)
This situation is annoying: there's no good reason for list(Foo(3)) and list(Bar(3)) to produce different results.
Is it possible to fix Bar (without, of course, getting rid of its __len__ method) such that list(Bar(3)) returns [0, 1, 2]?
Your problem is that Foo does not behave the same as xrange: xrange gives you a new iterator each time you asks its iter method, while Foo gives you always the same, meaning that once it is exhausted the object is too:
>>> a = Foo(3)
>>> list(a)
[0, 1, 2]
>>> list(a)
[]
>>> a = range(3)
>>> list(a)
[0, 1, 2]
>>> list(a)
[0, 1, 2]
I could easily confirm that the __len__ method is called by list by adding spys to your methods:
class Bar(Foo):
def __len__(self):
print "LEN"
return sum(1 for _ in self.generator)
(and I added a print "ITERATOR" in Foo.__iter__). It yields:
>>> list(Bar(3))
LEN
ITERATOR
[]
I can only imagine two workarounds:
my preferred one: return a new iterator on each call to __iter__ at Foo level to mimic xrange:
class Foo(object):
def __init__(self, n):
self.n = n
def __iter__(self):
print "ITERATOR"
return ( x for x in range(self.n))
class Bar(Foo):
def __len__(self):
print "LEN"
return sum(1 for _ in self.generator)
we get correctly:
>>> list(Bar(3))
ITERATOR
LEN
ITERATOR
[0, 1, 2]
the alternative: change len to not call the iterator and let Foo untouched:
class Bar(Foo):
def __init__(self, n):
self.len = n
super(Bar, self).__init__(n)
def __len__(self):
print "LEN"
return self.len
Here again we get:
>>> list(Bar(3))
LEN
ITERATOR
[0, 1, 2]
but Foo and Bar objects are exhausted once first iterator reaches its end.
But I must admit that I do not know the context of your real classes...
This behaviour might be annoying but it's actually quite understandable. Internally a list is simply an array and an array is a fixed size datastructure. The result of this is that if you have a list that has size n and you want to add an extra item to reach n+1 it will have to create a whole new array and completely copy the old one to the new one. Effectively your list.append(x) is now a O(n) operation instead of the regular O(1).
To prevent this, list() tries to get the size of your input so it can guess what size the array needs to be.
So one solution for this problem is to force it to guess by using iter:
list(iter(Bar(3)))

Pointers in Python?

I know Python doesn't have pointers, but is there a way to have this yield 2 instead
>>> a = 1
>>> b = a # modify this line somehow so that b "points to" a
>>> a = 2
>>> b
1
?
Here's an example: I want form.data['field'] and form.field.value to always have the same value. It's not completely necessary, but I think it would be nice.
In PHP, for example, I can do this:
<?php
class Form {
public $data = [];
public $fields;
function __construct($fields) {
$this->fields = $fields;
foreach($this->fields as &$field) {
$this->data[$field['id']] = &$field['value'];
}
}
}
$f = new Form([
[
'id' => 'fname',
'value' => 'George'
],
[
'id' => 'lname',
'value' => 'Lucas'
]
]);
echo $f->data['fname'], $f->fields[0]['value']; # George George
$f->data['fname'] = 'Ralph';
echo $f->data['fname'], $f->fields[0]['value']; # Ralph Ralph
Output:
GeorgeGeorgeRalphRalph
ideone
Or like this in C++ (I think this is right, but my C++ is rusty):
#include <iostream>
using namespace std;
int main() {
int* a;
int* b = a;
*a = 1;
cout << *a << endl << *b << endl; # 1 1
return 0;
}
There's no way you can do that changing only that line. You can do:
a = [1]
b = a
a[0] = 2
b[0]
That creates a list, assigns the reference to a, then b also, uses the a reference to set the first element to 2, then accesses using the b reference variable.
I want form.data['field'] and
form.field.value to always have the
same value
This is feasible, because it involves decorated names and indexing -- i.e., completely different constructs from the barenames a and b that you're asking about, and for with your request is utterly impossible. Why ask for something impossible and totally different from the (possible) thing you actually want?!
Maybe you don't realize how drastically different barenames and decorated names are. When you refer to a barename a, you're getting exactly the object a was last bound to in this scope (or an exception if it wasn't bound in this scope) -- this is such a deep and fundamental aspect of Python that it can't possibly be subverted. When you refer to a decorated name x.y, you're asking an object (the object x refers to) to please supply "the y attribute" -- and in response to that request, the object can perform totally arbitrary computations (and indexing is quite similar: it also allows arbitrary computations to be performed in response).
Now, your "actual desiderata" example is mysterious because in each case two levels of indexing or attribute-getting are involved, so the subtlety you crave could be introduced in many ways. What other attributes is form.field suppose to have, for example, besides value? Without that further .value computations, possibilities would include:
class Form(object):
...
def __getattr__(self, name):
return self.data[name]
and
class Form(object):
...
#property
def data(self):
return self.__dict__
The presence of .value suggests picking the first form, plus a kind-of-useless wrapper:
class KouWrap(object):
def __init__(self, value):
self.value = value
class Form(object):
...
def __getattr__(self, name):
return KouWrap(self.data[name])
If assignments such form.field.value = 23 is also supposed to set the entry in form.data, then the wrapper must become more complex indeed, and not all that useless:
class MciWrap(object):
def __init__(self, data, k):
self._data = data
self._k = k
#property
def value(self):
return self._data[self._k]
#value.setter
def value(self, v)
self._data[self._k] = v
class Form(object):
...
def __getattr__(self, name):
return MciWrap(self.data, name)
The latter example is roughly as close as it gets, in Python, to the sense of "a pointer" as you seem to want -- but it's crucial to understand that such subtleties can ever only work with indexing and/or decorated names, never with barenames as you originally asked!
It's not a bug, it's a feature :-)
When you look at the '=' operator in Python, don't think in terms of assignment. You don't assign things, you bind them. = is a binding operator.
So in your code, you are giving the value 1 a name: a. Then, you are giving the value in 'a' a name: b. Then you are binding the value 2 to the name 'a'. The value bound to b doesn't change in this operation.
Coming from C-like languages, this can be confusing, but once you become accustomed to it, you find that it helps you to read and reason about your code more clearly: the value which has the name 'b' will not change unless you explicitly change it. And if you do an 'import this', you'll find that the Zen of Python states that Explicit is better than implicit.
Note as well that functional languages such as Haskell also use this paradigm, with great value in terms of robustness.
Yes! there is a way to use a variable as a pointer in python!
I am sorry to say that many of answers were partially wrong. In principle every equal(=) assignation shares the memory address (check the id(obj) function), but in practice it is not such. There are variables whose equal("=") behaviour works in last term as a copy of memory space, mostly in simple objects (e.g. "int" object), and others in which not (e.g. "list","dict" objects).
Here is an example of pointer assignation
dict1 = {'first':'hello', 'second':'world'}
dict2 = dict1 # pointer assignation mechanism
dict2['first'] = 'bye'
dict1
>>> {'first':'bye', 'second':'world'}
Here is an example of copy assignation
a = 1
b = a # copy of memory mechanism. up to here id(a) == id(b)
b = 2 # new address generation. therefore without pointer behaviour
a
>>> 1
Pointer assignation is a pretty useful tool for aliasing without the waste of extra memory, in certain situations for performing comfy code,
class cls_X():
...
def method_1():
pd1 = self.obj_clsY.dict_vars_for_clsX['meth1'] # pointer dict 1: aliasing
pd1['var4'] = self.method2(pd1['var1'], pd1['var2'], pd1['var3'])
#enddef method_1
...
#endclass cls_X
but one have to be aware of this use in order to prevent code mistakes.
To conclude, by default some variables are barenames (simple objects like int, float, str,...), and some are pointers when assigned between them (e.g. dict1 = dict2). How to recognize them? just try this experiment with them. In IDEs with variable explorer panel usually appears to be the memory address ("#axbbbbbb...") in the definition of pointer-mechanism objects.
I suggest investigate in the topic. There are many people who know much more about this topic for sure. (see "ctypes" module). I hope it is helpful. Enjoy the good use of the objects! Regards, José Crespo
>> id(1)
1923344848 # identity of the location in memory where 1 is stored
>> id(1)
1923344848 # always the same
>> a = 1
>> b = a # or equivalently b = 1, because 1 is immutable
>> id(a)
1923344848
>> id(b) # equal to id(a)
1923344848
As you can see a and b are just two different names that reference to the same immutable object (int) 1. If later you write a = 2, you reassign the name a to a different object (int) 2, but the b continues referencing to 1:
>> id(2)
1923344880
>> a = 2
>> id(a)
1923344880 # equal to id(2)
>> b
1 # b hasn't changed
>> id(b)
1923344848 # equal to id(1)
What would happen if you had a mutable object instead, such as a list [1]?
>> id([1])
328817608
>> id([1])
328664968 # different from the previous id, because each time a new list is created
>> a = [1]
>> id(a)
328817800
>> id(a)
328817800 # now same as before
>> b = a
>> id(b)
328817800 # same as id(a)
Again, we are referencing to the same object (list) [1] by two different names a and b. However now we can mutate this list while it remains the same object, and a, b will both continue referencing to it
>> a[0] = 2
>> a
[2]
>> b
[2]
>> id(a)
328817800 # same as before
>> id(b)
328817800 # same as before
From one point of view, everything is a pointer in Python. Your example works a lot like the C++ code.
int* a = new int(1);
int* b = a;
a = new int(2);
cout << *b << endl; // prints 1
(A closer equivalent would use some type of shared_ptr<Object> instead of int*.)
Here's an example: I want
form.data['field'] and
form.field.value to always have the
same value. It's not completely
necessary, but I think it would be
nice.
You can do this by overloading __getitem__ in form.data's class.
This is a python pointer (different of c/c++)
>>> a = lambda : print('Hello')
>>> a
<function <lambda> at 0x0000018D192B9DC0>
>>> id(a) == int(0x0000018D192B9DC0)
True
>>> from ctypes import cast, py_object
>>> cast(id(a), py_object).value == cast(int(0x0000018D192B9DC0), py_object).value
True
>>> cast(id(a), py_object).value
<function <lambda> at 0x0000018D192B9DC0>
>>> cast(id(a), py_object).value()
Hello
I wrote the following simple class as, effectively, a way to emulate a pointer in python:
class Parameter:
"""Syntactic sugar for getter/setter pair
Usage:
p = Parameter(getter, setter)
Set parameter value:
p(value)
p.val = value
p.set(value)
Retrieve parameter value:
p()
p.val
p.get()
"""
def __init__(self, getter, setter):
"""Create parameter
Required positional parameters:
getter: called with no arguments, retrieves the parameter value.
setter: called with value, sets the parameter.
"""
self._get = getter
self._set = setter
def __call__(self, val=None):
if val is not None:
self._set(val)
return self._get()
def get(self):
return self._get()
def set(self, val):
self._set(val)
#property
def val(self):
return self._get()
#val.setter
def val(self, val):
self._set(val)
Here's an example of use (from a jupyter notebook page):
l1 = list(range(10))
def l1_5_getter(lst=l1, number=5):
return lst[number]
def l1_5_setter(val, lst=l1, number=5):
lst[number] = val
[
l1_5_getter(),
l1_5_setter(12),
l1,
l1_5_getter()
]
Out = [5, None, [0, 1, 2, 3, 4, 12, 6, 7, 8, 9], 12]
p = Parameter(l1_5_getter, l1_5_setter)
print([
p(),
p.get(),
p.val,
p(13),
p(),
p.set(14),
p.get()
])
p.val = 15
print(p.val, l1)
[12, 12, 12, 13, 13, None, 14]
15 [0, 1, 2, 3, 4, 15, 6, 7, 8, 9]
Of course, it is also easy to make this work for dict items or attributes of an object. There is even a way to do what the OP asked for, using globals():
def setter(val, dict=globals(), key='a'):
dict[key] = val
def getter(dict=globals(), key='a'):
return dict[key]
pa = Parameter(getter, setter)
pa(2)
print(a)
pa(3)
print(a)
This will print out 2, followed by 3.
Messing with the global namespace in this way is kind of transparently a terrible idea, but it shows that it is possible (if inadvisable) to do what the OP asked for.
The example is, of course, fairly pointless. But I have found this class to be useful in the application for which I developed it: a mathematical model whose behavior is governed by numerous user-settable mathematical parameters, of diverse types (which, because they depend on command line arguments, are not known at compile time). And once access to something has been encapsulated in a Parameter object, all such objects can be manipulated in a uniform way.
Although it doesn't look much like a C or C++ pointer, this is solving a problem that I would have solved with pointers if I were writing in C++.
The following code emulates exactly the behavior of pointers in C:
from collections import deque # more efficient than list for appending things
pointer_storage = deque()
pointer_address = 0
class new:
def __init__(self):
global pointer_storage
global pointer_address
self.address = pointer_address
self.val = None
pointer_storage.append(self)
pointer_address += 1
def get_pointer(address):
return pointer_storage[address]
def get_address(p):
return p.address
null = new() # create a null pointer, whose address is 0
Here are examples of use:
p = new()
p.val = 'hello'
q = new()
q.val = p
r = new()
r.val = 33
p = get_pointer(3)
print(p.val, flush = True)
p.val = 43
print(get_pointer(3).val, flush = True)
But it's now time to give a more professional code, including the option of deleting pointers, that I've just found in my personal library:
# C pointer emulation:
from collections import deque # more efficient than list for appending things
from sortedcontainers import SortedList #perform add and discard in log(n) times
class new:
# C pointer emulation:
# use as : p = new()
# p.val
# p.val = something
# p.address
# get_address(p)
# del_pointer(p)
# null (a null pointer)
__pointer_storage__ = SortedList(key = lambda p: p.address)
__to_delete_pointers__ = deque()
__pointer_address__ = 0
def __init__(self):
self.val = None
if new.__to_delete_pointers__:
p = new.__to_delete_pointers__.pop()
self.address = p.address
new.__pointer_storage__.discard(p) # performed in log(n) time thanks to sortedcontainers
new.__pointer_storage__.add(self) # idem
else:
self.address = new.__pointer_address__
new.__pointer_storage__.add(self)
new.__pointer_address__ += 1
def get_pointer(address):
return new.__pointer_storage__[address]
def get_address(p):
return p.address
def del_pointer(p):
new.__to_delete_pointers__.append(p)
null = new() # create a null pointer, whose address is 0
I don't know if my comment will help or not but if you want to use pointers in python, you can use dictionaries instead of variables
Let's say in your example will be
a = {'value': 1}
b = {'value': 2}
then you changed a to 2
a['value'] = 2
print(a) #{'value': 2}

Ignore python multiple return value

Say I have a Python function that returns multiple values in a tuple:
def func():
return 1, 2
Is there a nice way to ignore one of the results rather than just assigning to a temporary variable? Say if I was only interested in the first value, is there a better way than this:
x, temp = func()
You can use x = func()[0] to return the first value, x = func()[1] to return the second, and so on.
If you want to get multiple values at a time, use something like x, y = func()[2:4].
One common convention is to use a "_" as a variable name for the elements of the tuple you wish to ignore. For instance:
def f():
return 1, 2, 3
_, _, x = f()
If you're using Python 3, you can you use the star before a variable (on the left side of an assignment) to have it be a list in unpacking.
# Example 1: a is 1 and b is [2, 3]
a, *b = [1, 2, 3]
# Example 2: a is 1, b is [2, 3], and c is 4
a, *b, c = [1, 2, 3, 4]
# Example 3: b is [1, 2] and c is 3
*b, c = [1, 2, 3]
# Example 4: a is 1 and b is []
a, *b = [1]
The common practice is to use the dummy variable _ (single underscore), as many have indicated here before.
However, to avoid collisions with other uses of that variable name (see this response) it might be a better practice to use __ (double underscore) instead as a throwaway variable, as pointed by ncoghlan. E.g.:
x, __ = func()
Remember, when you return more than one item, you're really returning a tuple. So you can do things like this:
def func():
return 1, 2
print func()[0] # prints 1
print func()[1] # prints 2
The best solution probably is to name things instead of returning meaningless tuples (unless there is some logic behind the order of the returned items). You can for example use a dictionary:
def func():
return {'lat': 1, 'lng': 2}
latitude = func()['lat']
You could even use namedtuple if you want to add extra information about what you are returning (it's not just a dictionary, it's a pair of coordinates):
from collections import namedtuple
Coordinates = namedtuple('Coordinates', ['lat', 'lng'])
def func():
return Coordinates(lat=1, lng=2)
latitude = func().lat
If the objects within your dictionary/tuple are strongly tied together then it may be a good idea to even define a class for it. That way you'll also be able to define more complex operations. A natural question that follows is: When should I be using classes in Python?
Most recent versions of python (≥ 3.7) have dataclasses which you can use to define classes with very few lines of code:
from dataclasses import dataclass
#dataclass
class Coordinates:
lat: float = 0
lng: float = 0
def func():
return Coordinates(lat=1, lng=2)
latitude = func().lat
The primary advantage of dataclasses over namedtuple is that its easier to extend, but there are other differences. Note that by default, dataclasses are mutable, but you can use #dataclass(frozen=True) instead of #dataclass to force them being immutable.
Here is a video that might help you pick the right data class for your use case.
Three simple choices.
Obvious
x, _ = func()
x, junk = func()
Hideous
x = func()[0]
And there are ways to do this with a decorator.
def val0( aFunc ):
def pick0( *args, **kw ):
return aFunc(*args,**kw)[0]
return pick0
func0= val0(func)
This seems like the best choice to me:
val1, val2, ignored1, ignored2 = some_function()
It's not cryptic or ugly (like the func()[index] method), and clearly states your purpose.
If this is a function that you use all the time but always discard the second argument, I would argue that it is less messy to create an alias for the function without the second return value using lambda.
def func():
return 1, 2
func_ = lambda: func()[0]
func_() # Prints 1
This is not a direct answer to the question. Rather it answers this question: "How do I choose a specific function output from many possible options?".
If you are able to write the function (ie, it is not in a library you cannot modify), then add an input argument that indicates what you want out of the function. Make it a named argument with a default value so in the "common case" you don't even have to specify it.
def fancy_function( arg1, arg2, return_type=1 ):
ret_val = None
if( 1 == return_type ):
ret_val = arg1 + arg2
elif( 2 == return_type ):
ret_val = [ arg1, arg2, arg1 * arg2 ]
else:
ret_val = ( arg1, arg2, arg1 + arg2, arg1 * arg2 )
return( ret_val )
This method gives the function "advanced warning" regarding the desired output. Consequently it can skip unneeded processing and only do the work necessary to get your desired output. Also because Python does dynamic typing, the return type can change. Notice how the example returns a scalar, a list or a tuple... whatever you like!
When you have many output from a function and you don't want to call it multiple times, I think the clearest way for selecting the results would be :
results = fct()
a,b = [results[i] for i in list_of_index]
As a minimum working example, also demonstrating that the function is called only once :
def fct(a):
b=a*2
c=a+2
d=a+b
e=b*2
f=a*a
print("fct called")
return[a,b,c,d,e,f]
results=fct(3)
> fct called
x,y = [results[i] for i in [1,4]]
And the values are as expected :
results
> [3,6,5,9,12,9]
x
> 6
y
> 12
For convenience, Python list indexes can also be used :
x,y = [results[i] for i in [0,-2]]
Returns : a = 3 and b = 12
It is possible to ignore every variable except the first with less syntax if you like. If we take your example,
# The function you are calling.
def func():
return 1, 2
# You seem to only be interested in the first output.
x, temp = func()
I have found the following to works,
x, *_ = func()
This approach "unpacks" with * all other variables into a "throwaway" variable _. This has the benefit of assigning the one variable you want and ignoring all variables behind it.
However, in many cases you may want an output that is not the first output of the function. In these cases, it is probably best to indicate this by using the func()[i] where i is the index location of the output you desire. In your case,
# i == 0 because of zero-index.
x = func()[0]
As a side note, if you want to get fancy in Python 3, you could do something like this,
# This works the other way around.
*_, y = func()
Your function only outputs two potential variables, so this does not look too powerful until you have a case like this,
def func():
return 1, 2, 3, 4
# I only want the first and last.
x, *_, d = func()

Categories

Resources