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# beginner question about range with list and list

erik gartz
Guest
Posts: n/a

 11-24-2006
Hello,
I'm new to python and I'm having difficulty understanding the following
code. Why doesn't the variable a contain [[{}, {'x': 0}, {}], [{},
{'x': 1}, {}]] instead. Doesn't {} allocate new memory for the
dictionary each time? It almost appears as if the 2nd dictionary
created overwrites the first one. Thanks for your help,
Erik

>>>
>>> a = [[{}] * 3] * 2
>>> a

[[{}, {}, {}], [{}, {}, {}]]
>>> for i in range(2):

a[i][1] = {'x':i}
>>> a

[[{}, {'x': 1}, {}], [{}, {'x': 1}, {}]]
>>>

Ben Finney
Guest
Posts: n/a

 11-24-2006
"erik gartz" <(E-Mail Removed)> writes:

> Doesn't {} allocate new memory for the dictionary each time? It
> almost appears as if the 2nd dictionary created overwrites the first
> one.

<URL:http://effbot.org/pyfaq/how-do-i-create-a-multidimensional-list.htm>

--
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Ben Finney

Dan Lenski
Guest
Posts: n/a

 11-24-2006
erik gartz wrote:
> Doesn't {} allocate new memory for the
> dictionary each time? It almost appears as if the 2nd dictionary
> created overwrites the first one. Thanks for your help,
> Erik
>
> >>>
> >>> a = [[{}] * 3] * 2
> >>> a

> [[{}, {}, {}], [{}, {}, {}]]
> >>> for i in range(2):

> a[i][1] = {'x':i}
> >>> a

> [[{}, {'x': 1}, {}], [{}, {'x': 1}, {}]]
> >>>

You're right about both parts... sort of

Each time that {} is EVALUATED, it allocates a brand new dictionary.
However you're using the list repetition operator (* n). This operator
evaluates the list ONLY ONCE and then repeats that list n times.

The expression [{}]*3 will generate a list that contains the SAME
dictionary 3 times in a row. The expression [[{}]*3]*2 will generate a
list that contains the SAME list twice in a row. Thus, when you assign
to a[i][1], you are modifying a[0][1] and a[1][1], since a[0] and a[1]
are the same list.

The expression {'x':i} does indeed create a new dictionary each time it
is evaluated... however it gets assigned to the SAME list element each
time.

This is a fairly subtle point... if you want to make a new COPY of a
list (or other mutable object) rather than simply a new reference to
eat, you say a=b.copy() rather than just a=b. The ability to have
multiple references to one object is general considered a feature in
most programming languages, once you get used to it!

If you want to generate a 2x3 array with each element a DISTINCT
dictionary belonging to a DISTINCT list, try this:

a = [[{} for j in range(3)] for i in range(2)]
for i in range(2):
for j in range(3):
a[i][j] = {'row': i, 'column': j}

Dan

Fredrik Lundh
Guest
Posts: n/a

 11-24-2006
erik gartz wrote:

> I'm new to python and I'm having difficulty understanding the following
> code. Why doesn't the variable a contain [[{}, {'x': 0}, {}], [{},
> {'x': 1}, {}]] instead. Doesn't {} allocate new memory for the
> dictionary each time?

each time it's *executed*, yes. [{}]*3 doesn't execute [] or {} three
times; it executes them both once, and then creates a new list with
three references to the dictionary. see:

http://effbot.org/pyfaq/how-do-i-cre...ional-list.htm

</F>