Nested Lists Assignment Problem

[Licheng Fang]

I wanna use nested lists as an array, but here's the problem:

a = [[0]*3]*3
a [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
a[0][0] = 1
a

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

Could anybody please explain to me why three values were change? I'm
bewildered. Thanks!

 

[Dennis Lee Bieber]

Could anybody please explain to me why three values were change? I'm
bewildered. Thanks!

http://www.aifb.uni-karlsruhe.de/Le...hop/Python/Doku/The Whole Python FAQ.htm#4.50
--

 

[tiksin]

Le Mercredi 26 Avril 2006 10:13, Licheng Fang a écrit :

I wanna use nested lists as an array, but here's the problem:

a = [[0]*3]*3
a

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

a[0][0] = 1
a

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

Could anybody please explain to me why three values were change? I'm
bewildered. Thanks!

I guess is a probleme of reference. Write that instead:

l = [ [0]*3 for i in xrange(3) ]

cordialement,

 

[Licheng Fang]

Thank you very much!

But I still wonder why a nested assignment "a = [[0]*3]*3" generates 3
references to the same list, while the commands below apparently do
not.

a = [0] * 2
a [0, 0]
a[0] = 1
a

[1, 0]

 

[Fredrik Lundh]

I wanna use nested lists as an array, but here's the problem:

a = [[0]*3]*3
a [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
a[0][0] = 1
a

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

Could anybody please explain to me why three values were change? I'm
bewildered. Thanks!

http://pyfaq.infogami.com/how-do-i-create-a-multidimensional-list

</F>

 

[Fredrik Lundh]

But I still wonder why a nested assignment "a = [[0]*3]*3" generates 3
references to the same list, while the commands below apparently do
not.

that's because they're replacing a list item, rather than modifying it.

a = [0] * 2
a [0, 0] <- now you have two references to the same integer.
a[0] = 1 <- now you've *replaced* the first integer with another integer.
a

[1, 0]

if you do the same thing with lists, it behaves in exactly the same way:

a = [[0]*3]*3
a [[0, 0, 0], [0, 0, 0], [0, 0, 0]] <-- three references to the same list
a[0] = [1, 2, 3] <-- replace the first list
a

[[1, 2, 3], [0, 0, 0], [0, 0, 0]]

however, if you modify the *shared* object, the modification will of course
be visible everywhere that object is used:

a[1][0] = "hello" <-- modified the first item in the shared list
a

[[1, 2, 3], ['hello', 0, 0], ['hello', 0, 0]]

</F>

 

[Deepan Chakravarthy]

Hello,
I had used pyumlgraph to generate dot files for single python script. Is
it possible to use pyumlgraph to generate dot files for multiple python
scripts ??? Or is it automatically done if i generate a dot file for the
final python script that imports all other files?

Thanks
Deepan Chakravarthy N
www.codeshepherd.com

 

[Gary Herron]

Dennis Lee Bieber wrote:


Thank you very much!

But I still wonder why a nested assignment "a = [[0]*3]*3" generates 3
references to the same list, while the commands below apparently do
not.

It's got nothing to do with "nested"-ness. If X is ANY object in
Python, then [X]*3 will make a list with three references to X. If
you don't want three references to the single object X (and you don't),
then you have to find a different way to create three separate objects.
The copy module helps with this in some cases, but for your simple
example, you just want to create the three inner objects by evaluating
the expression [0]*3 three times. Here's several ways:

a = []
for i in range(3):
a.append([0]*3)

or

a = [ [0]*3 for i in range(3)]

Clear?

Gary Herron