List of lists surprising behaviour

C

candide

Let's the following code :
>>> t=[[0]*2]*3
>>> t [[0, 0], [0, 0], [0, 0]]
>>> t[0][0]=1
>>> t
[[1, 0], [1, 0], [1, 0]]

Rather surprising, isn't it ? So I suppose all the subarrays reférence
the same array :
>>> id(t[0]), id(t[1]), id(t[2]) (3077445996L, 3077445996L, 3077445996L)
>>>


So what is the right way to initialize to 0 a 2D array ? Is that way
correct :

>>> t=[[0 for _ in range(2)] for _ in range(3)]

It seems there is no more trouble now :
>>> t [[0, 0], [0, 0], [0, 0]]
>>> t[0][0]=1
>>> t [[1, 0], [0, 0], [0, 0]]
>>>

Correct ?
 
L

Lie Ryan

Let's the following code :
t=[[0]*2]*3
t [[0, 0], [0, 0], [0, 0]]
t[0][0]=1
t
[[1, 0], [1, 0], [1, 0]]

Rather surprising, isn't it ? So I suppose all the subarrays reférence
the same array :
id(t[0]), id(t[1]), id(t[2]) (3077445996L, 3077445996L, 3077445996L)

Yep, you're right. They share the same subarray if you uses
multiplication to build the array.
So what is the right way to initialize to 0 a 2D array ? Is that way
correct :
t=[[0 for _ in range(2)] for _ in range(3)]

Right again. That's the way to go. Although if the elements are
immutable, you can create the innermost array by multiplication:

t=[[0]*2 for _ in range(3)]
 
M

Matteo Landi

Yes you are. List comprehension makes you create list of lists without
reference-sharing. You should also find a recipe about that on the
python cookbook.

Let's the following code :
t=[[0]*2]*3
t [[0, 0], [0, 0], [0, 0]]
t[0][0]=1
t
[[1, 0], [1, 0], [1, 0]]

Rather surprising, isn't it ? So I suppose all the subarrays reférence the
same array :
id(t[0]), id(t[1]), id(t[2]) (3077445996L, 3077445996L, 3077445996L)


So what is the right way to initialize to 0 a 2D array ? Is that way correct
 :

t=[[0 for _ in range(2)] for _ in range(3)]

It seems there is no more trouble now :
t [[0, 0], [0, 0], [0, 0]]
t[0][0]=1
t [[1, 0], [0, 0], [0, 0]]

Correct ?
 
B

Boris Borcic

J

J Kenneth King

candide said:
Let's the following code :
t=[[0]*2]*3
t [[0, 0], [0, 0], [0, 0]]
t[0][0]=1
t
[[1, 0], [1, 0], [1, 0]]

Rather surprising, isn't it ?

Not at all, actually.

I'd be surprised if the multiplication operator was aware of object
constructors. Even arrays are "objects" in Python. Should the
multiplication operator know how to instantiate three arrays from a
single array instance? What about an instance of a user-defined class?
So I suppose all the subarrays reférence
the same array :
id(t[0]), id(t[1]), id(t[2]) (3077445996L, 3077445996L, 3077445996L)

As they should.
So what is the right way to initialize to 0 a 2D array ? Is that way
correct :

t=[[0 for _ in range(2)] for _ in range(3)]

It seems there is no more trouble now :
t [[0, 0], [0, 0], [0, 0]]
t[0][0]=1
t [[1, 0], [0, 0], [0, 0]]

Correct ?
2d_zero_vector = lambda len: [[0, 0] for _ in range(len)]
t = 2d_zero_vector(3)
print t [[0, 0], [0, 0], [0, 0]]
t[0][0] = 1
print t
[[1, 0], [0, 0], [0, 0], [0, 0]]

(Of course, if you're doing matrix math you'll probably want to work
with numpy which has a function for doing just this)
 
B

bart.c

J Kenneth King said:
candide said:
Let's the following code :
t=[[0]*2]*3
t [[0, 0], [0, 0], [0, 0]]
t[0][0]=1
t
[[1, 0], [1, 0], [1, 0]]

Rather surprising, isn't it ?

Not at all, actually.

The code is clearly trying to set only t[0][0] to 1, not t[1][0] and t[2][0]
as well.

This behaviour is quite scary actually, especially when t[0]=42 *does* work
as expected, while t[0][0]=42 is apparently duplicated. It appears
inconsistent.
I'd be surprised if the multiplication operator was aware of object
constructors. Even arrays are "objects" in Python. Should the
multiplication operator know how to instantiate three arrays from a
single array instance? What about an instance of a user-defined class?

Multiplication operators shouldn't need to be directly aware of any such
thing; it should just request that an object be duplicated without worrying
about how it's done.

I don't know how Python does things, but an object should either specify a
special way of duplicating itself, or lend itself to some standard way of
doing so. (So for a list, it's just a question of copying the data in the
list, then recursively duplicating each new element..)
 
B

Benjamin Kaplan

J Kenneth King said:
candide said:
Let's the following code :

t=[[0]*2]*3
t

[[0, 0], [0, 0], [0, 0]]

t[0][0]=1
t

[[1, 0], [1, 0], [1, 0]]

Rather surprising, isn't it ?

Not at all, actually.

The code is clearly trying to set only t[0][0] to 1, not t[1][0] and t[2][0]
as well.

This behaviour is quite scary actually, especially when t[0]=42 *does* work
as expected, while t[0][0]=42 is apparently duplicated. It appears
inconsistent.
I'd be surprised if the multiplication operator was aware of object
constructors.  Even arrays are "objects" in Python.  Should the
multiplication operator know how to instantiate three arrays from a
single array instance?  What about an instance of a user-defined class?

Multiplication operators shouldn't need to be directly aware of any such
thing; it should just request that an object be duplicated without worrying
about how it's done.

I don't know how Python does things, but an object should either specify a
special way of duplicating itself, or lend itself to some standard way of
doing so. (So for a list, it's just a question of copying the data in the
list, then recursively duplicating each new element..)

It's the recursively duplicating each element that's the problem. How
do you know when to stop?
 
L

Lie Ryan

J Kenneth King said:
candide said:
Let's the following code :

t=[[0]*2]*3
t
[[0, 0], [0, 0], [0, 0]]
t[0][0]=1
t
[[1, 0], [1, 0], [1, 0]]

Rather surprising, isn't it ?

Not at all, actually.

The code is clearly trying to set only t[0][0] to 1, not t[1][0] and
t[2][0]
as well.

This behaviour is quite scary actually, especially when t[0]=42 *does* work
as expected, while t[0][0]=42 is apparently duplicated. It appears
inconsistent.

I agree, the behavior is often quite inconvenient, but I disagree that
it is inconsistent. List multiplication behavior is consistent with the
tenet: "objects are never copied unless explicitly requested"

Peeking further:
t = [[0] * 2] * 3
print id(t[0]) == id(t[1]) # True
print id(t[0][0]) == id(t[1][0]) # True

so, it is consistent (though it is quite inconvenient).
Multiplication operators shouldn't need to be directly aware of any such
thing; it should just request that an object be duplicated without
worrying about how it's done.

I don't know how Python does things, but an object should either specify
a special way of duplicating itself, or lend itself to some standard way
of doing so. (So for a list, it's just a question of copying the data in
the list, then recursively duplicating each new element..)

That is inconsistent with the tenet. Moreover, the implicit copying
makes it quite difficult to reason about the program's behavior. How
would you propose this list should be copied:

class O(object):
def __init__(self):
self.attr = 0
def add(self):
self.attr += 1

b = [[O()] * 2] * 3
 
S

Steven D'Aprano

The code is clearly trying to set only t[0][0] to 1, not t[1][0] and
t[2][0] as well.

Trying to guess the motivation of the person writing code is tricky, but
in this case, that's a reasonable assumption. I can't think of any reason
why somebody would explicitly *want* that behaviour:

# Does it make sense to talk of anonymous aliases?
list_of_aliases = [[0]*2]*3
list_of_aliases[0][0] = 1
assert list_of_aliases[0][1] == 1

so it is a safe guess that anyone writing [[0]*2]*3 has probably made a
mistake.

However, I've certainly done something like this:

a = [0]*2
my_instance.items = a
a[0] = 1
assert my_instance.items[0] = 1

This is the same fundamental behaviour with the same cause: Python does
not copy objects unless you explicitly tell it to.

I cheerfully accept that the behaviour of [[0]*2]*3 is a Gotcha, but it
follows logically from Python's object model and assignment rules. If you
are surprised by it, it just goes to show that your understanding of
Python has at least one hole in it.

This behaviour is quite scary actually, especially when t[0]=42 *does*
work as expected, while t[0][0]=42 is apparently duplicated. It appears
inconsistent.

Emphasis on the word "appears". It actually displays a deep consistency
with the language fundamentals.

If you're ever interviewing somebody for a position as Python developer,
this is a quick test to distinguish those who know the language from
those who know the language *well*.

Multiplication operators shouldn't need to be directly aware of any such
thing; it should just request that an object be duplicated without
worrying about how it's done.

The multiplication operator is not a duplicator (copier). It is a
*repetition* operator: repeat the object N times, not make N copies.

I don't know how Python does things,

Well there you go :)

but an object should either specify
a special way of duplicating itself, or lend itself to some standard way
of doing so.

import copy
copy.copy(obj)

Dicts have a copy() method as a shortcut, and for lists you can use
slicing:

L = [1,2,3]
Lcopy = L[:]

(So for a list, it's just a question of copying the data in
the list, then recursively duplicating each new element..)

There's nothing "just" about that. Consider:

L = [1, 2]
L.append(L)

How would you copy that?

The correct answer is:

x = copy.deepcopy(L)
 
B

bart.c

It's the recursively duplicating each element that's the problem. How
do you know when to stop?

When you reach a primitive object (one not comprising other objects). (I
don't know if Python allows circular references, but that would give
problems anyway: how would you even print out such a list?)
 
B

bart.c

Lie said:
(I
don't know if Python allows circular references, but that would give
problems anyway: how would you even print out such a list?)


Python uses ellipsis to indicate recursive list:
a = [1, 2, 3]
a.append(a)
a
[1, 2, 3, [...]]

Ok, perhaps whatever logic print uses to know when to stop and show "...",
can also be used by copy handlers...

(Although I have an issue with the way that that append works. I tried it in
another, simpler language (which always does deep copies):

L:=(1,2,3)
L append:= L
print L

output: (1,2,3,(1,2,3))

which is exactly what I'd expect, and not (1,2,3,(1,2,3,(1,2,3,...))) )
 
S

Steven D'Aprano

(Although I have an issue with the way that that append works. I tried
it in another, simpler language (which always does deep copies):

L:=(1,2,3)
L append:= L
print L

output: (1,2,3,(1,2,3))

which is exactly what I'd expect,
and not (1,2,3,(1,2,3,(1,2,3,...))) )

I find that behaviour a big surprise. You asked to append the list L, not
a copy of the list L. So why is this "simpler" language making a copy
without being asked?

If you asked for:

L:=(1,2,3)
M:=(0,1)
M append:= L

does it also append a copy of L instead of L? If so, how do you append
the original rather than wastefully making a copy? If L is huge, making a
copy before appending will be slow, and potentially fail.
 
B

bart.c

Steven D'Aprano said:
I find that behaviour a big surprise. You asked to append the list L, not
a copy of the list L. So why is this "simpler" language making a copy
without being asked?

If you asked for:

L:=(1,2,3)
M:=(0,1)
M append:= L

does it also append a copy of L instead of L?

It make a copy.
If so, how do you append
the original rather than wastefully making a copy?

I don't think it can, without perhaps doing something with explicit
pointers.
If L is huge, making a
copy before appending will be slow, and potentially fail.

I don't know whether L append:=L requires 3 times the space of L, or 2
times, during the operation. But it should be doable using just twice the
space.

I suppose there are pros and cons to both approaches; copying all the time
at least avoids some of the odd effects and inconsistencies you get using
Python:

a1=[1,2,3]
a1.append(a1)

a2=[1,2,3]
b=[1,2,3]
a2.append(b)

a3=[1,2,3]
a3.append([1,2,3])

print ("a1 = ",a1)
print ("a2 = ",a2)
print ("a3 = ",a3)

Here, a1 ends up with a different result from a2, a3, even though the same
value is being appended to the same list of numbers in each case. And it
might sometimes bite you in the arse as the OP demonstrated:

L=[1,2,3]
M=[0,1]
M.append(L)

print (M) # output: [0, 1, [1, 2, 3]]

L[1]=31416

print (M) # output: [0, 1, [1, 31416, 3]], yikes!
 
S

Stephen Hansen

I suppose there are pros and cons to both approaches; copying all the time
at least avoids some of the odd effects and inconsistencies you get using
Python:

What inconsistencies? All your examples are perfectly consistent. Its
just consistent to different ideals. Python never copies implicitly; and
every object you create has a concrete identity in and of itself,
utterly separate from its potential equality.
a1=[1,2,3]
a1.append(a1)

The "a1" object is a distinct object unto itself; you are appending said
distinct object onto the end of itself. Entirely doable, even if you
don't usually want to. Recursive, but doable if that's what you want. If
you wished to append a copy, you must-- as always, consistently--
explicitly copy it.

I.e.,

a1.append(a1[:])
a2=[1,2,3]
b=[1,2,3]
a2.append(b)

a2 is a distinct object from b; that the two objects are equal means
nothing. So you're appending an object to the end of a2 which happens to
be equal to it.
a3=[1,2,3]
a3.append([1,2,3])

This is just another way of writing the previous example; that in one
you are naming the object [1,2,3] and in the other you are not, doesn't
mean anything. A named vs unnamed object in Python behaves exactly the same.
print ("a1 = ",a1)
print ("a2 = ",a2)
print ("a3 = ",a3)

Here, a1 ends up with a different result from a2, a3, even though the same
value is being appended to the same list of numbers in each case.

There's the rub: the VALUE is not being appended. The *actual object* is.

Consider:
False

And it
might sometimes bite you in the arse as the OP demonstrated:

L=[1,2,3]
M=[0,1]
M.append(L)

print (M) # output: [0, 1, [1, 2, 3]]

L[1]=31416

print (M) # output: [0, 1, [1, 31416, 3]], yikes!

That might bite you on your arse if you think Python implicitly copies;
but since the rule is very simple -- it /never/ implicitly copies -- and
that it objects are discrete entities and not just names of certain
values, it more likely then not will be beneficial to you quite often
down the road. You'll -want- that to be how things work. Eventually.
When you learn more Python.

--

Stephen Hansen
... Also: Ixokai
... Mail: me+list/python (AT) ixokai (DOT) io
... Blog: http://meh.ixokai.io/


-----BEGIN PGP SIGNATURE-----
Version: GnuPG v2.0.10 (Darwin)

iQEcBAEBAgAGBQJMG6MqAAoJEKcbwptVWx/ll/YIALfJMaQuxVQGeF0EXGllNGl4
zAH5akcpA3nFr+X6E2I1Xc9MxN162dIbBuRJ6GrLwej2wyklkgm3snMDCQAUA0QU
xGEzyIma0Mi6KotydyEkmpgxM6b3vQj8SSlmM6mex+zKgR+HlIXvl/zapyPc9v+8
Kfc8glJgoFXmiv7W9fbwCu8g/w6KeIiAaaC7JepMwm6IoVsrP888p65Add8eEtKb
PUQKuU1bSEnysmCVstzUnY4lIsByOW3ygpnnA6++uDeR3AwZ4sLRN0J33UsUiGJM
kPV6c6LkJ9+jY3dObbnSusVPcdfvOIyUlN8i4nxKb/JddPQ9aTptLO/kupZm8xA=
=18Pu
-----END PGP SIGNATURE-----
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Members online

No members online now.

Forum statistics

Threads
473,744
Messages
2,569,483
Members
44,903
Latest member
orderPeak8CBDGummies

Latest Threads

Top