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name

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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)

      • 2
    • Do you want a, b, and c, to all point to the same value (in this case a list), or do you want a=0, b=3, and c=5. In that case, you want a,b,c = [0,3,5] or just a,b,c = 0,3,5.

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.

  • 271
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    • What if we have, a = b = c = 10; and when we try to update the value of b, it does effect any other? although i checked their ids are the same.?
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    • @user570826: 10 is immutable—that means there is no way to update the value, so your question doesn't make sense. You can point b at a different value, but doing so has no effect on a and c, which are still pointing at the original value. The difference that lists make is that they're mutable—e.g., you can append to a list, or lst[0] = 3, and that will update the value, which will be visible through all names for that value.

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'}
>>> 
  • 72
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      • 1
    • IMHO, this actually answers OP first key question of what should I use for multiple assignment, whereas the higher rated and more cerebral answer above doesn't.

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 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).

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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.

  • 13
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    • No. In one implementation of Python (CPython), every variable is a pointer to a PyObject. That's not true in other implementations like PyPy or Jython. (In fact, it's not even clear how it could be true, because the languages those implementations are written in don't even have pointers.)
    • I think the use of "pointer" in a conceptual sense is ok (perhaps with a disclaimer that implementations may vary), esp if the goal is to convey behavior.
    • @abarnert When people say Python they mean CPython, not other rarely used implementations. Just like when people say Kleenex they mean facial tissue. Playing the semantics game in these comments is really unnecessary. As for what he wrote, is the behavior of what he described wrong?
      • 2
    • This is highly misleading. Pointers are certainly not visible at the Python level, and at least two of the four major implementations (PyPy and Jython) don't use them even inside the implementation.
    • You welcome to read and explore python internals and you'll discover that every variable in python is actually pointer.

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
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      • 1
    • id isn't necessarily a memory location. As the docs say, this returns the "identity… an integer… which is guaranteed to be unique and constant for this object during its lifetime." CPython happens to use the memory address as the id, but other Python implementations may not. PyPy, for example, doesn't. And saying "two vars point to the same memory location" is misleading to anyone who understands it C-style. "Two names for the same object" is both more accurate and less misleading.

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!

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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.

  • 4
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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])
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