I tried to use multiple assignment as show below to initialize variables, but I got confused by the behavior, I expect to reassign the values list separately, I mean b[0] and c[0] equal 0 as before.
a=b=c=[0,3,5]
a[0]=1
print(a)
print(b)
print(c)
Result is:
[1, 3, 5]
[1, 3, 5]
[1, 3, 5]
Is that correct? what should I use for multiple assignment?
what is different from this?
d=e=f=3
e=4
print('f:',f)
print('e:',e)
result:
('f:', 3)
('e:', 4)
If you're coming to Python from a language in the C/Java/etc. family, it may help you to stop thinking about a as a "variable", and start thinking of it as a "name".
a, b, and c aren't different variables with equal values; they're different names for the same identical value. Variables have types, identities, addresses, and all kinds of stuff like that.
Names don't have any of that. Values do, of course, and you can have lots of names for the same value.
If you give Notorious B.I.G. a hot dog,* Biggie Smalls and Chris Wallace have a hot dog. If you change the first element of a to 1, the first elements of b and c are 1.
If you want to know if two names are naming the same object, use the is operator:
>>> a=b=c=[0,3,5]
>>> a is b
True
You then ask:
what is different from this?
d=e=f=3
e=4
print('f:',f)
print('e:',e)
Here, you're rebinding the name e to the value 4. That doesn't affect the names d and f in any way.
In your previous version, you were assigning to a[0], not to a. So, from the point of view of a[0], you're rebinding a[0], but from the point of view of a, you're changing it in-place.
You can use the id function, which gives you some unique number representing the identity of an object, to see exactly which object is which even when is can't help:
>>> a=b=c=[0,3,5]
>>> id(a)
4473392520
>>> id(b)
4473392520
>>> id(a[0])
4297261120
>>> id(b[0])
4297261120
>>> a[0] = 1
>>> id(a)
4473392520
>>> id(b)
4473392520
>>> id(a[0])
4297261216
>>> id(b[0])
4297261216
Notice that a[0] has changed from 4297261120 to 4297261216—it's now a name for a different value. And b[0] is also now a name for that same new value. That's because a and b are still naming the same object.
Under the covers, a[0]=1 is actually calling a method on the list object. (It's equivalent to a.__setitem__(0, 1).) So, it's not really rebinding anything at all. It's like calling my_object.set_something(1). Sure, likely the object is rebinding an instance attribute in order to implement this method, but that's not what's important; what's important is that you're not assigning anything, you're just mutating the object. And it's the same with a[0]=1.
user570826 asked:
What if we have, a = b = c = 10
That's exactly the same situation as a = b = c = [1, 2, 3]: you have three names for the same value.
But in this case, the value is an int, and ints are immutable. In either case, you can rebind a to a different value (e.g., a = "Now I'm a string!"), but the won't affect the original value, which b and c will still be names for. The difference is that with a list, you can change the value [1, 2, 3] into [1, 2, 3, 4] by doing, e.g., a.append(4); since that's actually changing the value that b and c are names for, b will now b [1, 2, 3, 4]. There's no way to change the value 10 into anything else. 10 is 10 forever, just like Claudia the vampire is 5 forever (at least until she's replaced by Kirsten Dunst).
* Warning: Do not give Notorious B.I.G. a hot dog. Gangsta rap zombies should never be fed after midnight.
Cough cough
>>> a,b,c = (1,2,3)
>>> a
1
>>> b
2
>>> c
3
>>> a,b,c = ({'test':'a'},{'test':'b'},{'test':'c'})
>>> a
{'test': 'a'}
>>> b
{'test': 'b'}
>>> c
{'test': 'c'}
>>>
In python, everything is an object, also "simple" variables types (int, float, etc..).
When you changes a variable value, you actually changes it's pointer, and if you compares between two variables it's compares their pointers.
(To be clear, pointer is the address in physical computer memory where a variable is stored).
As a result, when you changes an inner variable value, you changes it's value in the memory and it's affects all the variables that point to this address.
For your example, when you do:
a = b = 5
This means that a and b points to the same address in memory that contains the value 5, but when you do:
a = 6
It's not affect b because a is now points to another memory location that contains 6 and b still points to the memory address that contains 5.
But, when you do:
a = b = [1,2,3]
a and b, again, points to the same location but the difference is that if you change the one of the list values:
a[0] = 2
It's changes the value of the memory that a is points on, but a is still points to the same address as b, and as a result, b changes as well.
Yes, that's the expected behavior. a, b and c are all set as labels for the same list. If you want three different lists, you need to assign them individually. You can either repeat the explicit list, or use one of the numerous ways to copy a list:
b = a[:] # this does a shallow copy, which is good enough for this case
import copy
c = copy.deepcopy(a) # this does a deep copy, which matters if the list contains mutable objects
Assignment statements in Python do not copy objects - they bind the name to an object, and an object can have as many labels as you set. In your first edit, changing a[0], you're updating one element of the single list that a, b, and c all refer to. In your second, changing e, you're switching e to be a label for a different object (4 instead of 3).
You can use id(name) to check if two names represent the same object:
>>> a = b = c = [0, 3, 5]
>>> print(id(a), id(b), id(c))
46268488 46268488 46268488
Lists are mutable; it means you can change the value in place without creating a new object. However, it depends on how you change the value:
>>> a[0] = 1
>>> print(id(a), id(b), id(c))
46268488 46268488 46268488
>>> print(a, b, c)
[1, 3, 5] [1, 3, 5] [1, 3, 5]
If you assign a new list to a, then its id will change, so it won't affect b and c's values:
>>> a = [1, 8, 5]
>>> print(id(a), id(b), id(c))
139423880 46268488 46268488
>>> print(a, b, c)
[1, 8, 5] [1, 3, 5] [1, 3, 5]
Integers are immutable, so you cannot change the value without creating a new object:
>>> x = y = z = 1
>>> print(id(x), id(y), id(z))
507081216 507081216 507081216
>>> x = 2
>>> print(id(x), id(y), id(z))
507081248 507081216 507081216
>>> print(x, y, z)
2 1 1
in your first example a = b = c = [1, 2, 3] you are really saying:
'a' is the same as 'b', is the same as 'c' and they are all [1, 2, 3]
If you want to set 'a' equal to 1, 'b' equal to '2' and 'c' equal to 3, try this:
a, b, c = [1, 2, 3]
print(a)
--> 1
print(b)
--> 2
print(c)
--> 3
Hope this helps!
What you need is this:
a, b, c = [0,3,5] # Unpack the list, now a, b, and c are ints
a = 1 # `a` did equal 0, not [0,3,5]
print(a)
print(b)
print(c)
Simply put, in the first case, you are assigning multiple names to a list. Only one copy of list is created in memory and all names refer to that location. So changing the list using any of the names will actually modify the list in memory.
In the second case, multiple copies of same value are created in memory. So each copy is independent of one another.
The code that does what I need could be this:
# test
aux=[[0 for n in range(3)] for i in range(4)]
print('aux:',aux)
# initialization
a,b,c,d=[[0 for n in range(3)] for i in range(4)]
# changing values
a[0]=1
d[2]=5
print('a:',a)
print('b:',b)
print('c:',c)
print('d:',d)
Result:
('aux:', [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]])
('a:', [1, 0, 0])
('b:', [0, 0, 0])
('c:', [0, 0, 0])
('d:', [0, 0, 5])
To assign multiple variables same value I prefer list
a, b, c = [10]*3#multiplying 3 because we have 3 variables
print(a, type(a), b, type(b), c, type(c))
output:
10 <class 'int'> 10 <class 'int'> 10 <class 'int'>
Initialize multiple objects:
import datetime
time1, time2, time3 = [datetime.datetime.now()]*3
print(time1)
print(time2)
print(time3)
output:
2022-02-25 11:52:59.064487
2022-02-25 11:52:59.064487
2022-02-25 11:52:59.064487
E.g: basically a = b = 10 means both a and b are pointing to 10 in the memory, you can test by id(a) and id(b) which comes out exactly equal to a is b as True.
is matches the memory location but not its value, however == matches the value.
let's suppose, you want to update the value of a from 10 to 5, since the memory location was pointing to the same memory location you will experience the value of b will also be pointing to 5 because of the initial declaration.
The conclusion is to use this only if you know the consequences otherwise simply use , separated assignment like a, b = 10, 10 and won't face the above-explained consequences on updating any of the values because of different memory locations.
The behavior is correct. However, all the variables will share the same reference. Please note the behavior below:
>>> a = b = c = [0,1,2]
>>> a
[0, 1, 2]
>>> b
[0, 1, 2]
>>> c
[0, 1, 2]
>>> a[0]=1000
>>> a
[1000, 1, 2]
>>> b
[1000, 1, 2]
>>> c
[1000, 1, 2]
So, yes, it is different in the sense that if you assign a, b and c differently on a separate line, changing one will not change the others.
Here are two codes for you to choose one:
a = b = c = [0, 3, 5]
a = [1, 3, 5]
print(a)
print(b)
print(c)
or
a = b = c = [0, 3, 5]
a = [1] + a[1:]
print(a)
print(b)
print(c)
>>> a = [1,2,3]
>>> b = []
>>> b.append(a)
>>> print(b)
[[1, 2, 3]]
>>> num = a.pop(0)
>>> a.append(num)
>>> print(a)
[2, 3, 1]
>>> b.append(a)
>>> print(b)
[[2, 3, 1], [2, 3, 1]]
>>>
Why is this happening and how to fix it? I need the list like
[[1, 2, 3], [2, 3, 1]]
Thank you.
Edit:
Also, why is this working?
>>> a = []
>>> b = []
>>> a = [1,2,3]
>>> b.append(a)
>>> a = [1,2,3,4]
>>> b.append(a)
>>> print(b)
[[1, 2, 3], [1, 2, 3, 4]]
>>>
'''
Append a copy of your list a, at least the first time. Otherwise, you've appended the same list both times.
b.append(a[:])
When you append the list a, python creates a reference to that variable inside the list b. So when you edit the list a, it is reflected again in the list b. You need to create a copy of your variable and then append it to get the desired result.
Every variable name in Python should be thought of as a reference to a piece of data. In your first listing, b contains two references to the same underlying object that is also referenced by the name a. That object gets changed in-place by the operations you’re using to rotate its members. The effect of that change is seen when you look at either of the two references to the object found in b, or indeed when you look at the reference associated with the name a.
Their identicality can be seen by using the id() function: id(a), id(b[0]) and id(b[1]) all return the same number, which is the unique identifier of the underlying list object that they all refer to. Or you can use the is operator: b[0] is b[1] evaluates to True.
By contrast, in the second listing, you reassign a—in other words, by using the assignment operator = you cause that name to become associated with a different object: in this case, a new list object that you just created with your square-bracketed literal expression. b still contains one reference to the old list, and now you append a new reference that points to this different piece of underlying data. So the two elements of b now look different from each other—and indeed they are different objects and accordingly have different id() numbers, only one of which is the same as the current id(a). b[0] is b[1] now evaluates to False
How to fix it? Reassign the name a before changing it: for example, create a copy:
a = list(a)
or:
import copy
a = copy.copy(a)
(or you could even use copy.deepcopy()—study the difference). Alternatively, rotate the members a using methods that entail reassignment rather than in-place changes—e.g.:
a = a[1:] + a[:1]
(NB immutable objects such as the tuple avoid this whole confusion —not because they behave fundamentally differently but because they lack methods that produce in-place changes and therefore force you to use reassignment strategies.)
In addition to making the copy of a by doing a[:] and assigning it to b.
You can also use collections.deque.rotate to rotate your list
from collections import deque
a = [1,2,3]
#Make a deque of copy of a
b = deque(a[:])
#Rotate the deque
b.rotate(len(a)-1)
#Create the list and print it
print([a,list(b)])
#[[1, 2, 3], [2, 3, 1]]
>>> a = [3, 2]
>>> a[0:1][0] = 1
>>> a
[3, 2]
>>> a[0:1] = [1]
>>> a
[1, 2]
What does a[0:1] mean?
If it's a pointer to the range of a, then a[0:1][0] = 1 should change the value of a.
If it's a copy of the range of a, then a[0:1] = [1] shouldn't change the value of a.
I think the result of the two is inconsistent with each other. Could you please help me work out the problem?
Internally, this is a big difference:
>>> a = [3, 2]
>>> a[0:1][0] = 1
is a shorthand for
temp = a[0:1]
temp[0] = 1
and is internally expressed as
a.__getitem__(slice(0, 1)).__setitem__(0, 1)
resp.
temp = a.__getitem__(slice(0, 1))
temp.__setitem__(0, 1)
so it accesses a part of the list, making a separate object, and doing an assignment on this object, which is then dropped.
On the other hand,
>>> a[0:1] = [1]
does
a.__setitem__(slice(0, 1), [1])
which just operates on the original object.
So, while looking similar, these expressions are distinct on what they mean.
Let's test that:
class Itemtest(object):
def __init__(self, name):
self.name = name
def __repr__(self):
return self.name
def __setitem__(self, item, value):
print "__setitem__", self, item, value
def __getitem__(self, item):
print "__getitem__", self, item
return Itemtest("inner")
a = Itemtest("outer")
a[0:1] = [4]
temp = a[0:1]
temp[0] = 4
a[0:1][0] = 4
outputs
__setitem__ outer slice(0, 1, None) [4]
__getitem__ outer slice(0, 1, None)
__setitem__ inner 0 4
__getitem__ outer slice(0, 1, None)
__setitem__ inner 0 4
Slicing a list creates a shallow copy- it is not a reference to the original. So when you get that slice, it is not bound to the original list a. Therefore, you can try and change a single element of it, but it is not stored in a variable so no changes will be made to any original list.
To clarify, with the former to you doing __getitem__- accessing part of the list (a copy):
a[0:1][0] = 1
You are editing the slice [0:1], which is a only shallow copy of a, so will not edit a itself.
But with the latter, one is calling __setitem__, which will of course edit the object in-place.:
a[0:1] = [1]
You are directly referring to and editing part of a, so it changes in real-time.
The following statement:
>>> a[0:1] = [1]
assigns the list [1] as a subset of the list a from 0 to 1.
By doing a[0:1][0] you are getting the first element of [3] which is 3. Then if you try to assign it a value of 1, it simply wont work because 3 cannot be 1. However, if you stick to a[0:1], you are getting [3], which can be changed to [1]. Hope that helps
Examples
>>> a = [1,2,3,4]
>>> a[1:4]
[2,3,4]
>>> a[1:4] = [6,5,4,3,2]
>>> a
[1,6,5,4,3,2]
Say I have a list:
L = [1,2,3]
and I assigned L[0] to a variable a
a = L[0]
then if I change a, it won't affect L.
a = a + 1
print L # [1,2,3] not affected
Why is this happening? isn't python passing everything around with references? I thought that a is pointing to L[0]
The problem is that a and L[0] are references to an immutable object, so changing any one of them won't affect the other references:
>>> L = [1, 2, [3]]
>>> a = L[0]
>>> a = a + 1
a now points to a new object, while L[0] still points to the same object.
>>> a, L[0]
(2, 1)
Now in this case b and L[2] are references to a mutable object(list), any in-place operation on them will affect all the references:
>>> b = L[2]
>>> b.append(4) #list.append is an in-place operation
>>> b, L[2]
([3, 4], [3, 4])
>>> b = b + [5] #mutable object, but not an in-place operation
>>> b #b is assigned to new list object
[3, 4, 5]
>>> L[2] #L[2] is unchanged
[3, 4]
L[0] is a name, and when you create the list L, you assign an object to that name, the integer 1. a is also a name, and when you assign a as in a = L[0], you make a to point to the same object that L[0] points to.
But when you later do a = a + 1, this is another assignment. You are not modifying the object that a points to -- the = sign can't do that. You are creating a new object, the integer 2, and assigning that to a.
So in the end, you have two objects in memory; one is referred to by L[0] and the other is referred to by a.
Integers are immutable, which means that there is no possible way to change the properties of the objects in this example; however, that's not salient in this example exactly, because even if the object was mutable it wouldn't change the fact that you're doing assignment (with the = sign). In a case where the object in question was mutable, you could theoretically change the properties of the object when it is still referenced by L[0] and a, instead of doing any additional assignment with = as you are doing. At that point, you would see the properties change regardless of which name you used to inspect the object.
Since L[0] in your case is immutable, changing a doesn't affect the value of L[0]. When you change a, the new object is created and a starts to pointing to it.
See what happens if L[0] is of a mutable type:
>>> L = [[1],2,3]
>>> a = L[0]
>>> a.append(2)
>>> L
[[1, 2], 2, 3]
In this case a and L[0] both point to the same object.
Also see Raymond Hettinger's answer in the relevant thread.
Change the assignment to:
a = L
then when you change L as:
L[0] += 1
you will see that a also changes. This is the reference magic.