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