Python assigning multiple variables to same value? list behavior

Question

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)

Answer 1

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.

Answer 2

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'}
>>> 

Answer 3

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.

Answer 4

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).

Answer 5

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

Answer 6

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!

Answer 7

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)

Answer 8

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.

Answer 9

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])

Answer 10

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

Answer 11

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.

Answer 12

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.

Answer 13

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)

Answer 14

The reason for observed behavior was already described in other comments, but there was no solution given, which could help to achieve the required result in a "similar way" as it was given in the topic-description.
I would give a try:

import copy

s = [0,3,5] # Define a value we would like to assign
a,b,c=(copy.deepcopy(s) for _ in range(3)) # assign same value to variables a,b,c as its "deep copy"

# Same as a single line:
a,b,c=(copy.deepcopy([0,3,5]) for _ in range(3))

In my example - changing of a would neither affect b nor c .