On 4/5/2015 7:12 AM, Narci Edson Venturini wrote:
The next code has an unexpected result:
a=3*[3*[0]]
a now contains three references to the same object, hence the results
you show below.
You can create three distinct objects as follows:
>>> a = [ [0,0,0] for i in (0,1,2) ]
>>> a[1][1]=1
>>> a
[[0, 0, 0], [0, 1, 0], [0, 0, 0]]
>>>
hth,
Emile
a
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
a[0][0]=1
a
[[1, 0, 0], [1, 0, 0], [1, 0, 0]]
The code assigned to "1" a(0,0), a(1,0) and a(2,0).
It was expected: [[1, 0, 0], [0, 0, 0], [0, 0, 0]]
When the followind code is ran, them the correct result is obtained:
a=[[0 for i in range(3)] for j in range(3)]
a
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
a[0][0]=1
a
[[1, 0, 0], [0, 0, 0], [0, 0, 0]]
So, what is wrong ?
Best Regars,
Narci
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Narci Edson Venturini
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