Why make lists unhashable? - python

A common issue on SO is removing duplicates from a list of lists. Since lists are unhashable, set([[1, 2], [3, 4], [1, 2]]) throws TypeError: unhashable type: 'list'. Answers to this kind of question usually involve using tuples, which are immutable and therefore hashable.
This answer to What makes lists unhashable? include the following:
If the hash value changes after it gets stored at a particular slot in the dictionary, it will lead to an inconsistent dictionary. For example, initially the list would have gotten stored at location A, which was determined based on the hash value. If the hash value changes, and if we look for the list we might not find it at location A, or as per the new hash value, we might find some other object.
but I don't quite understand because other types that can be used for dictionary keys can be changed without issue:
>>> d = {}
>>> a = 1234
>>> d[a] = 'foo'
>>> a += 1
>>> d[a] = 'bar'
>>> d
{1234: 'foo', 1235: 'bar'}
It is obvious that if the value of a changes, it will hash to a different location in the dictionary. Why is the same assumption dangerous for a list? Why is the following an unsafe method for hashing a list, since it is what we all use when we need to anyway?
>>> class my_list(list):
... def __hash__(self):
... return tuple(self).__hash__()
...
>>> a = my_list([1, 2])
>>> b = my_list([3, 4])
>>> c = my_list([1, 2])
>>> foo = [a, b, c]
>>> foo
[[1, 2], [3, 4], [1, 2]]
>>> set(foo)
set([[1, 2], [3, 4]])
It seems that this solves the set() problem, why is this an issue? Lists may be mutable, but they are ordered which seems like it would be all that's needed for hashing.

You seem to confuse mutability with rebinding. a += 1 assigns a new object, the int object with the numeric value 1235, to a. Under the hood, for immutable objects like int, a += 1 is just the same as a = a + 1.
The original 1234 object is not mutated. The dictionary is still using an int object with numeric value 1234 as the key. The dictionary still holds a reference to that object, even though a now references a different object. The two references are independent.
Try this instead:
>>> class BadKey:
... def __init__(self, value):
... self.value = value
... def __eq__(self, other):
... return other == self.value
... def __hash__(self):
... return hash(self.value)
... def __repr__(self):
... return 'BadKey({!r})'.format(self.value)
...
>>> badkey = BadKey('foo')
>>> d = {badkey: 42}
>>> badkey.value = 'bar'
>>> print(d)
{BadKey('bar'): 42}
Note that I altered the attribute value on the badkey instance. I didn't even touch the dictionary. The dictionary reflects the change; the actual key value itself was mutated, the object that both the name badkey and the dictionary reference.
However, you now can't access that key anymore:
>>> badkey in d
False
>>> BadKey('bar') in d
False
>>> for key in d:
... print(key, key in d)
...
BadKey('bar') False
I have thoroughly broken my dictionary, because I can no longer reliably locate the key.
That's because BadKey violates the principles of hashability; that the hash value must remain stable. You can only do that if you don't change anything about the object that the hash is based on. And the hash must be based on whatever makes two instances equal.
For lists, the contents make two list objects equal. And you can change those, so you can't produce a stable hash either.

Related

Hash function for collection of items that disregards ordering

I am using a the hash() function to get the hash value of my object which contains two integers and two Strings. Moreover, I have a dictionary where I store these objects; the process is that I check if the object exists with the hash value, if yes I update if not I insert the new one.
The thing is that when creating the objects, I do not know the order of the object variables and I want to treat the objects as same no matter the order of these variables.
Is there an alternative function to the hash() function that does not consider the order of the variables?
#Consequently what I want is:
hash((int1,str1,int2,str2)) == hash((int2,str2,int1,str1))
You could use a frozenset instead of a tuple:
>>> hash(frozenset([1, 2, 'a', 'b']))
1190978740469805404
>>>
>>> hash(frozenset([1, 'a', 2, 'b']))
1190978740469805404
>>>
>>> hash(frozenset(['a', 2, 'b', 1]))
1190978740469805404
However, the removal of duplicates from the iterable presents a subtle problem:
>>> hash(frozenset([1,2,1])) == hash(frozenset([1,2,2]))
True
You can fix this by creating a counter from the iterable using collections.Counter, and calling frozenset on the counter's items, thus preserving the count of each item from the original iterable:
>>> from collections import Counter
>>>
>>> hash(frozenset(Counter([1,2,1]).items()))
-307001354391131208
>>> hash(frozenset(Counter([1,1,2]).items()))
-307001354391131208
>>>
>>> hash(frozenset(Counter([1,2,1]).items())) == hash(frozenset(Counter([1,2,2]).items()))
False
Usually for things like this it helps immeasurably if you post some sample code, but I'll assume you've got something like this:
class Foo():
def __init__(self, x, y):
self.x = x
self.y = y
def __hash__(self):
return hash((self.x, self.y))
You're taking a hash of a tuple there, which does care about order. If you want your hash to not care about the order of the ints, then just use a frozenset:
def __hash__(self):
return hash(frozenset([self.x, self.y]))
If the range of the values is not too great you could add them together, that way the order can be disregarded, however it does increase the possibility for 2 hashes to have the same value:
def hash_list(items):
value = 0
for item in items:
value+= hash(item)
return value
hash_list(['a', 'b', 'c'])
>>> 8409777985338339540
hash_list(['b', 'a', 'c'])
>>> 8409777985338339540

__getitem__ for a list vs a dict

The Dictionary __getitem__ method does not seem to work the same way as it does for List, and it is causing me headaches. Here is what I mean:
If I subclass list, I can overload __getitem__ as:
class myList(list):
def __getitem__(self,index):
if isinstance(index,int):
#do one thing
if isinstance(index,slice):
#do another thing
If I subclass dict, however, the __getitem__ does not expose index, but key instead as in:
class myDict(dict):
def __getitem__(self,key):
#Here I want to inspect the INDEX, but only have access to key!
So, my question is how can I intercept the index of a dict, instead of just the key?
Example use case:
a = myDict()
a['scalar'] = 1 # Create dictionary entry called 'scalar', and assign 1
a['vector_1'] = [1,2,3,4,5] # I want all subsequent vectors to be 5 long
a['vector_2'][[0,1,2]] = [1,2,3] # I want to intercept this and force vector_2 to be 5 long
print(a['vector_2'])
[1,2,3,0,0]
a['test'] # This should throw a KeyError
a['test'][[0,2,3]] # So should this
Dictionaries have no order; there is no index to pass in; this is why Python can use the same syntax ([..]) and the same magic method (__getitem__) for both lists and dictionaries.
When you index a dictionary on an integer like 0, the dictionary treats that like any other key:
>>> d = {'foo': 'bar', 0: 42}
>>> d.keys()
[0, 'foo']
>>> d[0]
42
>>> d['foo']
'bar'
Chained indexing applies to return values; the expression:
a['vector_2'][0, 1, 2]
is executed as:
_result = a['vector_2'] # via a.__getitem__('vector_2')
_result[0, 1, 2] # via _result.__getitem__((0, 1, 2))
so if you want values in your dictionary to behave in a certain way, you must return objects that support those operations.

python dictionary conundrum

On the console I typed in
>>> class S(str): pass
...
>>> a = 'hello'
>>> b = S('hello')
>>> d = {a:a, b:b}
>>> d
{'hello': 'hello'}
>>> type(d[a])
<class '__main__.S'>
>>> type(d[b])
<class '__main__.S'>
I thought at first that the reason that d only kept one pair was because hash(a) and hash(b) returned the same values, so I tried:
>>> class A(object):
... def __hash__(self):
... return 0
...
>>> class B(object):
... def __hash__(self):
... return 0
...
>>> d = {A():A(),B():B()}
>>> d
{<__main__.A object at 0x101808b90>: <__main__.A object at 0x101808b10>, <__main__.B object at 0x101808d10>: <__main__.B object at 0x101808cd0>}
Now I'm confused.
How come in the first code listing, d only kept one pair, but in the second listing d both keys were kept despite having same hash?
The two objects in your original example were collapsed not because they have the same hash, but because they compare equal. Dict keys are unique with respect to equality, not hash. Python requires that any two objects that compare equal must have the same hash (but not necessarily the reverse).
In your first example, the two objects are equal, since they both have the str equality behavior. Since the two objects compare equal, they are collapsed to one. In the second example, they don't compare equal. By default user-defined classes use identity for equality --- that is, each object compares equal only to itself. So your two objects are not equal. It doesn't matter than they have the same hash.
The hash does not determine the unique key in a dictionary. In some ways, hash functions are an "implementation detail", in that they determine how the dictionary internally stores its entries. a == b implies hash(a) == hash(b), but the converse does not hold in general. Two keys also need to be equal to each other (when applying the == operator) to be treated as equivalent keys in a dictionary.
If you want a type to be hashable then you must also define __eq__(). str defines __eq__() properly, but A and B do not.
The keys are different for the first and second objects. As they are objects the key is some readable equivalent of the object not a string.

How to get Python to tell equal integers apart

Have a bit of a problem distinguishing between identical integers.
In the following (which is obviously a trivial case) a, b, c are integers. I wish to create a dicionary, diction, which will contain {a: 'foo', b: 'bar', c: 'baz'}
diction = {}
for i in (a, b, c):
j = ('foo', 'bar', 'baz')[(a, b, c).index(i)]
diction[i] = j
All runs very nicely until, for example, a and b are the same: the third line will give index 0 for both a and b, resulting in j = 'foo' for each case.
I know lists can be copied by
list_a = [1, 2, 3]
list_b = list(list_a)
or
list_b = list_a[:]
So, is there any way of maybe doing this with my identical integers?
(I tried making one a float, but the value remains the same , so that doesn't work.)
To create a dictionary from two different iterables, you can use the following code:
d = dict(zip((a, b, c), ('foo', 'bar', 'baz')))
where zip is used to combine both iterables in a list of tuples that can be passed to the dictionary constructor.
Note that if a==b, then the 'foo' will be overwritten with 'bar', since the values are added to the dictionary in the same order they are in the iterable as if you were using this code:
d[a] = 'foo'
d[b] = 'bar'
d[c] = 'baz'
This is just the standard behaviour of a dictionary, when a new value is assigned to a key that is already known, the value is overwritten.
If you prefer to keep all values in a list, then you can use a collections.defaultdict as follows:
from collections import defaultdict
d = defaultdict(list)
for key, value in zip((a, b, c), ('foo', 'bar', 'baz')):
d[key].append(value)
You can't distinguish between identical objects.
You can tell them apart if they do not fall between -5 and 256
See also "is" operator behaves unexpectedly with integers
http://docs.python.org/c-api/int.html
The current implementation keeps an array of integer objects for all
integers between -5 and 256, when you create an int in that range you
actually just get back a reference to the existing object. So it
should be possible to change the value of 1. I suspect the behaviour
of Python in this case is undefined. :-)
In [30]: a = 257
In [31]: a is 257
Out[31]: False
In [32]: a = 256
In [33]: a is 256
Out[33]: True
You may have to roll your own dictionary like object that implements this sort of behavior though... and it still wouldn't be able to do anything between -5 and 256. I'd need to do more digging to be sure though.
If a and b have the same value then you can't expect them to point to different positions in dictionary if used as keys. Key values in dictionaries must be unique.
Also if you have two sequences the simplest way to make a dictionary out of them is to zip them together:
tup = (a,b,c)
val = ('foo', 'bar', 'baz')
diction = dict(zip(tup, val))
All of the answers so far are correct - identical keys can't be re-used in a dictionary. If you absolutely have to try to do something like this, but can't ensure that a, b, and c have distinct values you could try something like this:
d = dict(zip((id(k) for k in (a,b,c)), ('foo', 'bar', 'baz')))
When you go to look up your values though, you'll have to remember to do so like this:
d[id(a)]
That might help, but I am not certain what you're actually after here.

Immutable vs Mutable types

I'm confused on what an immutable type is. I know the float object is considered to be immutable, with this type of example from my book:
class RoundFloat(float):
def __new__(cls, val):
return float.__new__(cls, round(val, 2))
Is this considered to be immutable because of the class structure / hierarchy?, meaning float is at the top of the class and is its own method call. Similar to this type of example (even though my book says dict is mutable):
class SortedKeyDict(dict):
def __new__(cls, val):
return dict.__new__(cls, val.clear())
Whereas something mutable has methods inside the class, with this type of example:
class SortedKeyDict_a(dict):
def example(self):
return self.keys()
Also, for the last class(SortedKeyDict_a), if I pass this type of set to it:
d = (('zheng-cai', 67), ('hui-jun', 68),('xin-yi', 2))
without calling the example method, it returns a dictionary. The SortedKeyDict with __new__ flags it as an error. I tried passing integers to the RoundFloat class with __new__ and it flagged no errors.
What? Floats are immutable? But can't I do
x = 5.0
x += 7.0
print x # 12.0
Doesn't that "mut" x?
Well you agree strings are immutable right? But you can do the same thing.
s = 'foo'
s += 'bar'
print s # foobar
The value of the variable changes, but it changes by changing what the variable refers to. A mutable type can change that way, and it can also change "in place".
Here is the difference.
x = something # immutable type
print x
func(x)
print x # prints the same thing
x = something # mutable type
print x
func(x)
print x # might print something different
x = something # immutable type
y = x
print x
# some statement that operates on y
print x # prints the same thing
x = something # mutable type
y = x
print x
# some statement that operates on y
print x # might print something different
Concrete examples
x = 'foo'
y = x
print x # foo
y += 'bar'
print x # foo
x = [1, 2, 3]
y = x
print x # [1, 2, 3]
y += [3, 2, 1]
print x # [1, 2, 3, 3, 2, 1]
def func(val):
val += 'bar'
x = 'foo'
print x # foo
func(x)
print x # foo
def func(val):
val += [3, 2, 1]
x = [1, 2, 3]
print x # [1, 2, 3]
func(x)
print x # [1, 2, 3, 3, 2, 1]
You have to understand that Python represents all its data as objects. Some of these objects like lists and dictionaries are mutable, meaning you can change their content without changing their identity. Other objects like integers, floats, strings and tuples are objects that can not be changed.
An easy way to understand that is if you have a look at an objects ID.
Below you see a string that is immutable. You can not change its content. It will raise a TypeError if you try to change it. Also, if we assign new content, a new object is created instead of the contents being modified.
>>> s = "abc"
>>> id(s)
4702124
>>> s[0]
'a'
>>> s[0] = "o"
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'str' object does not support item assignment
>>> s = "xyz"
>>> id(s)
4800100
>>> s += "uvw"
>>> id(s)
4800500
You can do that with a list and it will not change the objects identity
>>> i = [1,2,3]
>>> id(i)
2146718700
>>> i[0]
1
>>> i[0] = 7
>>> id(i)
2146718700
To read more about Python's data model you could have a look at the Python language reference:
Python 2 datamodel
Python 3 datamodel
Common immutable type:
numbers: int(), float(), complex()
immutable sequences: str(), tuple(), frozenset(), bytes()
Common mutable type (almost everything else):
mutable sequences: list(), bytearray()
set type: set()
mapping type: dict()
classes, class instances
etc.
One trick to quickly test if a type is mutable or not, is to use id() built-in function.
Examples, using on integer,
>>> i = 1
>>> id(i)
***704
>>> i += 1
>>> i
2
>>> id(i)
***736 (different from ***704)
using on list,
>>> a = [1]
>>> id(a)
***416
>>> a.append(2)
>>> a
[1, 2]
>>> id(a)
***416 (same with the above id)
First of all, whether a class has methods or what it's class structure is has nothing to do with mutability.
ints and floats are immutable. If I do
a = 1
a += 5
It points the name a at a 1 somewhere in memory on the first line. On the second line, it looks up that 1, adds 5, gets 6, then points a at that 6 in memory -- it didn't change the 1 to a 6 in any way. The same logic applies to the following examples, using other immutable types:
b = 'some string'
b += 'some other string'
c = ('some', 'tuple')
c += ('some', 'other', 'tuple')
For mutable types, I can do thing that actallly change the value where it's stored in memory. With:
d = [1, 2, 3]
I've created a list of the locations of 1, 2, and 3 in memory. If I then do
e = d
I just point e to the same list d points at. I can then do:
e += [4, 5]
And the list that both e and d points at will be updated to also have the locations of 4 and 5 in memory.
If I go back to an immutable type and do that with a tuple:
f = (1, 2, 3)
g = f
g += (4, 5)
Then f still only points to the original tuple -- you've pointed g at an entirely new tuple.
Now, with your example of
class SortedKeyDict(dict):
def __new__(cls, val):
return dict.__new__(cls, val.clear())
Where you pass
d = (('zheng-cai', 67), ('hui-jun', 68),('xin-yi', 2))
(which is a tuple of tuples) as val, you're getting an error because tuples don't have a .clear() method -- you'd have to pass dict(d) as val for it to work, in which case you'll get an empty SortedKeyDict as a result.
Difference between Mutable and Immutable objects
Definitions
Mutable object: Object that can be changed after creating it.
Immutable object: Object that cannot be changed after creating it.
In Python if you change the value of the immutable object it will create a new object.
Mutable Objects
Here are the objects in Python that are of mutable type:
list
Dictionary
Set
bytearray
user defined classes
Immutable Objects
Here are the objects in Python that are of immutable type:
int
float
decimal
complex
bool
string
tuple
range
frozenset
bytes
Some Unanswered Questions
Question: Is string an immutable type?
Answer: yes it is, but can you explain this:
Proof 1:
a = "Hello"
a +=" World"
print a
Output
"Hello World"
In the above example the string got once created as "Hello" then changed to "Hello World". This implies that the string is of the mutable type. But it is not when we check its identity to see whether it is of a mutable type or not.
a = "Hello"
identity_a = id(a)
a += " World"
new_identity_a = id(a)
if identity_a != new_identity_a:
print "String is Immutable"
Output
String is Immutable
Proof 2:
a = "Hello World"
a[0] = "M"
Output
TypeError 'str' object does not support item assignment
Question: Is Tuple an immutable type?
Answer: yes, it is.
Proof 1:
tuple_a = (1,)
tuple_a[0] = (2,)
print a
Output
'tuple' object does not support item assignment
If you're coming to Python from another language (except one that's a lot like Python, like Ruby), and insist on understanding it in terms of that other language, here's where people usually get confused:
>>> a = 1
>>> a = 2 # I thought int was immutable, but I just changed it?!
In Python, assignment is not mutation in Python.
In C++, if you write a = 2, you're calling a.operator=(2), which will mutate the object stored in a. (And if there was no object stored in a, that's an error.)
In Python, a = 2 does nothing to whatever was stored in a; it just means that 2 is now stored in a instead. (And if there was no object stored in a, that's fine.)
Ultimately, this is part of an even deeper distinction.
A variable in a language like C++ is a typed location in memory. If a is an int, that means it's 4 bytes somewhere that the compiler knows is supposed to be interpreted as an int. So, when you do a = 2, it changes what's stored in those 4 bytes of memory from 0, 0, 0, 1 to 0, 0, 0, 2. If there's another int variable somewhere else, it has its own 4 bytes.
A variable in a language like Python is a name for an object that has a life of its own. There's an object for the number 1, and another object for the number 2. And a isn't 4 bytes of memory that are represented as an int, it's just a name that points at the 1 object. It doesn't make sense for a = 2 to turn the number 1 into the number 2 (that would give any Python programmer way too much power to change the fundamental workings of the universe); what it does instead is just make a forget the 1 object and point at the 2 object instead.
So, if assignment isn't a mutation, what is a mutation?
Calling a method that's documented to mutate, like a.append(b). (Note that these methods almost always return None). Immutable types do not have any such methods, mutable types usually do.
Assigning to a part of the object, like a.spam = b or a[0] = b. Immutable types do not allow assignment to attributes or elements, mutable types usually allow one or the other.
Sometimes using augmented assignment, like a += b, sometimes not. Mutable types usually mutate the value; immutable types never do, and give you a copy instead (they calculate a + b, then assign the result to a).
But if assignment isn't mutation, how is assigning to part of the object mutation? That's where it gets tricky. a[0] = b does not mutate a[0] (again, unlike C++), but it does mutate a (unlike C++, except indirectly).
All of this is why it's probably better not to try to put Python's semantics in terms of a language you're used to, and instead learn Python's semantics on their own terms.
Whether an object is mutable or not depends on its type. This doesn't depend on whether or not it has certain methods, nor on the structure of the class hierarchy.
User-defined types (i.e. classes) are generally mutable. There are some exceptions, such as simple sub-classes of an immutable type. Other immutable types include some built-in types such as int, float, tuple and str, as well as some Python classes implemented in C.
A general explanation from the "Data Model" chapter in the Python Language Reference":
The value of some objects can change. Objects whose value can change
are said to be mutable; objects whose value is unchangeable once they
are created are called immutable.
(The value of an immutable container
object that contains a reference to a mutable object can change when
the latter’s value is changed; however the container is still
considered immutable, because the collection of objects it contains
cannot be changed. So, immutability is not strictly the same as having
an unchangeable value, it is more subtle.)
An object’s mutability is
determined by its type; for instance, numbers, strings and tuples are
immutable, while dictionaries and lists are mutable.
A mutable object has to have at least a method able to mutate the object. For example, the list object has the append method, which will actually mutate the object:
>>> a = [1,2,3]
>>> a.append('hello') # `a` has mutated but is still the same object
>>> a
[1, 2, 3, 'hello']
but the class float has no method to mutate a float object. You can do:
>>> b = 5.0
>>> b = b + 0.1
>>> b
5.1
but the = operand is not a method. It just make a bind between the variable and whatever is to the right of it, nothing else. It never changes or creates objects. It is a declaration of what the variable will point to, since now on.
When you do b = b + 0.1 the = operand binds the variable to a new float, wich is created with te result of 5 + 0.1.
When you assign a variable to an existent object, mutable or not, the = operand binds the variable to that object. And nothing more happens
In either case, the = just make the bind. It doesn't change or create objects.
When you do a = 1.0, the = operand is not wich create the float, but the 1.0 part of the line. Actually when you write 1.0 it is a shorthand for float(1.0) a constructor call returning a float object. (That is the reason why if you type 1.0 and press enter you get the "echo" 1.0 printed below; that is the return value of the constructor function you called)
Now, if b is a float and you assign a = b, both variables are pointing to the same object, but actually the variables can't comunicate betweem themselves, because the object is inmutable, and if you do b += 1, now b point to a new object, and a is still pointing to the oldone and cannot know what b is pointing to.
but if c is, let's say, a list, and you assign a = c, now a and c can "comunicate", because list is mutable, and if you do c.append('msg'), then just checking a you get the message.
(By the way, every object has an unique id number asociated to, wich you can get with id(x). So you can check if an object is the same or not checking if its unique id has changed.)
A class is immutable if each object of that class has a fixed value upon instantiation that cannot SUBSEQUENTLY be changed
In another word change the entire value of that variable (name) or leave it alone.
Example:
my_string = "Hello world"
my_string[0] = "h"
print my_string
you expected this to work and print hello world but this will throw the following error:
Traceback (most recent call last):
File "test.py", line 4, in <module>
my_string[0] = "h"
TypeError: 'str' object does not support item assignment
The interpreter is saying : i can't change the first character of this string
you will have to change the whole string in order to make it works:
my_string = "Hello World"
my_string = "hello world"
print my_string #hello world
check this table:
source
It would seem to me that you are fighting with the question what mutable/immutable actually means. So here is a simple explenation:
First we need a foundation to base the explenation on.
So think of anything that you program as a virtual object, something that is saved in a computers memory as a sequence of binary numbers. (Don't try to imagine this too hard, though.^^) Now in most computer languages you will not work with these binary numbers directly, but rather more you use an interpretation of binary numbers.
E.g. you do not think about numbers like 0x110, 0xaf0278297319 or similar, but instead you think about numbers like 6 or Strings like "Hello, world". Never the less theses numbers or Strings are an interpretation of a binary number in the computers memory. The same is true for any value of a variable.
In short: We do not program with actual values but with interpretations of actual binary values.
Now we do have interpretations that must not be changed for the sake of logic and other "neat stuff" while there are interpretations that may well be changed. For example think of the simulation of a city, in other words a program where there are many virtual objects and some of these are houses. Now may these virtual objects (the houses) be changed and can they still be considered to be the same houses? Well of course they can. Thus they are mutable: They can be changed without becoming a "completely" different object.
Now think of integers: These also are virtual objects (sequences of binary numbers in a computers memory). So if we change one of them, like incrementing the value six by one, is it still a six? Well of course not. Thus any integer is immutable.
So: If any change in a virtual object means that it actually becomes another virtual object, then it is called immutable.
Final remarks:
(1) Never mix up your real-world experience of mutable and immutable with programming in a certain language:
Every programming language has a definition of its own on which objects may be muted and which ones may not.
So while you may now understand the difference in meaning, you still have to learn the actual implementation for each programming language. ... Indeed there might be a purpose of a language where a 6 may be muted to become a 7. Then again this would be quite some crazy or interesting stuff, like simulations of parallel universes.^^
(2) This explenation is certainly not scientific, it is meant to help you to grasp the difference between mutable and immutable.
The goal of this answer is to create a single place to find all the good ideas about how to tell if you are dealing with mutating/nonmutating (immutable/mutable), and where possible, what to do about it? There are times when mutation is undesirable and python's behavior in this regard can feel counter-intuitive to coders coming into it from other languages.
As per a useful post by #mina-gabriel:
Books to read that might help: "Data Structures and Algorithms in Python"
Excerpt from that book that lists mutable/immutable types:
mutable/imutable types image
Analyzing the above and combining w/ a post by #arrakëën:
What cannot change unexpectedly?
scalars (variable types storing a single value) do not change unexpectedly
numeric examples: int(), float(), complex()
there are some "mutable sequences":
str(), tuple(), frozenset(), bytes()
What can?
list like objects (lists, dictionaries, sets, bytearray())
a post on here also says classes and class instances but this may depend on what the class inherits from and/or how its built.
by "unexpectedly" I mean that programmers from other languages might not expect this behavior (with the exception or Ruby, and maybe a few other "Python like" languages).
Adding to this discussion:
This behavior is an advantage when it prevents you from accidentally populating your code with mutliple copies of memory-eating large data structures. But when this is undesirable, how do we get around it?
With lists, the simple solution is to build a new one like so:
list2 = list(list1)
with other structures ... the solution can be trickier. One way is to loop through the elements and add them to a new empty data structure (of the same type).
functions can mutate the original when you pass in mutable structures. How to tell?
There are some tests given on other comments on this thread but then there are comments indicating these tests are not full proof
object.function() is a method of the original object but only some of these mutate. If they return nothing, they probably do. One would expect .append() to mutate without testing it given its name. .union() returns the union of set1.union(set2) and does not mutate. When in doubt, the function can be checked for a return value. If return = None, it does not mutate.
sorted() might be a workaround in some cases. Since it returns a sorted version of the original, it can allow you to store a non-mutated copy before you start working on the original in other ways. However, this option assumes you don't care about the order of the original elements (if you do, you need to find another way). In contrast .sort() mutates the original (as one might expect).
Non-standard Approaches (in case helpful):
Found this on github published under an MIT license:
github repository under: tobgu named: pyrsistent
What it is: Python persistent data structure code written to be used in place of core data structures when mutation is undesirable
For custom classes, #semicolon suggests checking if there is a __hash__ function because mutable objects should generally not have a __hash__() function.
This is all I have amassed on this topic for now. Other ideas, corrections, etc. are welcome. Thanks.
One way of thinking of the difference:
Assignments to immutable objects in python can be thought of as deep copies,
whereas assignments to mutable objects are shallow
The simplest answer:
A mutable variable is one whose value may change in place, whereas in an immutable variable change of value will not happen in place. Modifying an immutable variable will rebuild the same variable.
Example:
>>>x = 5
Will create a value 5 referenced by x
x -> 5
>>>y = x
This statement will make y refer to 5 of x
x -------------> 5 <-----------y
>>>x = x + y
As x being an integer (immutable type) has been rebuild.
In the statement, the expression on RHS will result into value 10 and when this is assigned to LHS (x), x will rebuild to 10. So now
x--------->10
y--------->5
Mutable means that it can change/mutate. Immutable the opposite.
Some Python data types are mutable, others not.
Let's find what are the types that fit in each category and see some examples.
Mutable
In Python there are various mutable types:
lists
dict
set
Let's see the following example for lists.
list = [1, 2, 3, 4, 5]
If I do the following to change the first element
list[0] = '!'
#['!', '2', '3', '4', '5']
It works just fine, as lists are mutable.
If we consider that list, that was changed, and assign a variable to it
y = list
And if we change an element from the list such as
list[0] = 'Hello'
#['Hello', '2', '3', '4', '5']
And if one prints y it will give
['Hello', '2', '3', '4', '5']
As both list and y are referring to the same list, and we have changed the list.
Immutable
In some programming languages one can define a constant such as the following
const a = 10
And if one calls, it would give an error
a = 20
However, that doesn't exist in Python.
In Python, however, there are various immutable types:
None
bool
int
float
str
tuple
Let's see the following example for strings.
Taking the string a
a = 'abcd'
We can get the first element with
a[0]
#'a'
If one tries to assign a new value to the element in the first position
a[0] = '!'
It will give an error
'str' object does not support item assignment
When one says += to a string, such as
a += 'e'
#'abcde'
It doesn't give an error, because it is pointing a to a different string.
It would be the same as the following
a = a + 'f'
And not changing the string.
Some Pros and Cons of being immutable
• The space in memory is known from the start. It would not require extra space.
• Usually, it makes things more efficiently. Finding, for example, the len() of a string is much faster, as it is part of the string object.
Every time we change value of a immutable variable it basically destroy the previous instance and create a new instance of variable class
var = 2 #Immutable data
print(id(var))
var += 4
print(id(var))
list_a = [1,2,3] #Mutable data
print(id(list_a))
list_a[0]= 4
print(id(list_a))
Output:
9789024
9789088
140010877705856
140010877705856
Note:Mutable variable memory_location is change when we change the value
I haven't read all the answers, but the selected answer is not correct and I think the author has an idea that being able to reassign a variable means that whatever datatype is mutable. That is not the case. Mutability has to do with passing by reference rather than passing by value.
Lets say you created a List
a = [1,2]
If you were to say:
b = a
b[1] = 3
Even though you reassigned a value on B, it will also reassign the value on a. Its because when you assign "b = a". You are passing the "Reference" to the object rather than a copy of the value. This is not the case with strings, floats etc. This makes list, dictionaries and the likes mutable, but booleans, floats etc immutable.
In Python, there's a easy way to know:
Immutable:
>>> s='asd'
>>> s is 'asd'
True
>>> s=None
>>> s is None
True
>>> s=123
>>> s is 123
True
Mutable:
>>> s={}
>>> s is {}
False
>>> {} is {}
Flase
>>> s=[1,2]
>>> s is [1,2]
False
>>> s=(1,2)
>>> s is (1,2)
False
And:
>>> s=abs
>>> s is abs
True
So I think built-in function is also immutable in Python.
But I really don't understand how float works:
>>> s=12.3
>>> s is 12.3
False
>>> 12.3 is 12.3
True
>>> s == 12.3
True
>>> id(12.3)
140241478380112
>>> id(s)
140241478380256
>>> s=12.3
>>> id(s)
140241478380112
>>> id(12.3)
140241478380256
>>> id(12.3)
140241478380256
It's so weird.
For immutable objects, assignment creates a new copy of values, for example.
x=7
y=x
print(x,y)
x=10 # so for immutable objects this creates a new copy so that it doesnot
#effect the value of y
print(x,y)
For mutable objects, the assignment doesn't create another copy of values. For example,
x=[1,2,3,4]
print(x)
y=x #for immutable objects assignment doesn't create new copy
x[2]=5
print(x,y) # both x&y holds the same list

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