How does the logic behind this piece of code work?
Question:
I have typed out the following code.
la = [1,[2,[3,[4]]]]
lb = [la[1], la[1][1]]
print(la)
lb[0][1]=9
print(la)
I was expecting la to remain as in the original first line, but it changed as shown below.
[1, [2, [3, [4]]]]
[1, [2, 9]]
Does this have to do with shallow and deep copy? I can’t seem to wrap my head around what’s going on. Apologies for the formatting, I’m trying to fix it.
Answers:
You should use copy() in your case:
lb = [la[1].copy(), la[1][1].copy()]
Assignment statements in Python do not copy objects, they create bindings between a target and an object.
See doc.
So the answer gives you the solution to how to properly copy data in your case, but it does not explains what happens. I commented your script to get an insight into what you are doing:
A comment on notation:
ptr(X) means a reference to somewhere in memory which I call X.
X is a un-named address, if you will.
Don’t look to much into the words "pointer" and "reference" as I have only used them to explain a concept, not as strict keywords like one might think if coming from C++ or similar.
la = [1,[2,[3,[4]]]]
# la = [ 1 , ptr(X) ]
# X = [ 2 , ptr(Y) ]
# Y = [ 3 , ptr(Z) ]
# Z = [ 4 ]
print(la)
lb = [la[1], la[1][1]]
# lb = [ ptr(X) , ptr(Y) ]
lb[0][1]=9
# lb[0] is ptr(X)
# lb[0][1] is X[1] is ptr(Y)
# lb[0][1] = 9 --> X = [2,9]
# now if we look at how la is defined
# la = [ 1 , ptr(X) ]
# X = [ 2 , ptr(Y) ]
# meaning that now X is [2,9] and ptr(X) points to [2,9],
# so la is
# la = [ 1 , ptr(X) ]
# X = [ 2 , 9 ]
# so if we print la we get
# [1,[2,9]]
# et voila
print(la)
I have typed out the following code.
la = [1,[2,[3,[4]]]]
lb = [la[1], la[1][1]]
print(la)
lb[0][1]=9
print(la)
I was expecting la to remain as in the original first line, but it changed as shown below.
[1, [2, [3, [4]]]]
[1, [2, 9]]
Does this have to do with shallow and deep copy? I can’t seem to wrap my head around what’s going on. Apologies for the formatting, I’m trying to fix it.
You should use copy() in your case:
lb = [la[1].copy(), la[1][1].copy()]
Assignment statements in Python do not copy objects, they create bindings between a target and an object.
See doc.
So the answer gives you the solution to how to properly copy data in your case, but it does not explains what happens. I commented your script to get an insight into what you are doing:
A comment on notation:
ptr(X) means a reference to somewhere in memory which I call X.
X is a un-named address, if you will.
Don’t look to much into the words "pointer" and "reference" as I have only used them to explain a concept, not as strict keywords like one might think if coming from C++ or similar.
la = [1,[2,[3,[4]]]]
# la = [ 1 , ptr(X) ]
# X = [ 2 , ptr(Y) ]
# Y = [ 3 , ptr(Z) ]
# Z = [ 4 ]
print(la)
lb = [la[1], la[1][1]]
# lb = [ ptr(X) , ptr(Y) ]
lb[0][1]=9
# lb[0] is ptr(X)
# lb[0][1] is X[1] is ptr(Y)
# lb[0][1] = 9 --> X = [2,9]
# now if we look at how la is defined
# la = [ 1 , ptr(X) ]
# X = [ 2 , ptr(Y) ]
# meaning that now X is [2,9] and ptr(X) points to [2,9],
# so la is
# la = [ 1 , ptr(X) ]
# X = [ 2 , 9 ]
# so if we print la we get
# [1,[2,9]]
# et voila
print(la)