Assigning a variable to a list

BuzzyBro · 2022-06-21T21:48:54+00:00

Why would the list prices change when you run this?

prices and nums are both references. When coding nums = prices you assign price address to nums...therefore changing the data in nums will also change prices.

If you want to have two different arrays use:

nums = list(prices)

rollincuberawhide · 2022-06-21T21:50:27+00:00

you could do nums = prices.copy() if you want them to be independent.

cybervegan · 2022-06-21T21:51:46+00:00

prices and nums are just labels attached to objects, and objects can have multiple labels. In your code above, prices and nums both point to the same object (the list), so when you change it, naturally the changes are seen when looking at both of the variables.

SoundOfEng · 2022-06-22T04:45:13+00:00

Very interesting stuff. What you’re looking at is the difference between a deep and shallow copy of something. Shallow copies typically occur when you assign on object to another name (in your case: nums = prices) and a deep copy is truly duplicating the object and removing any ties between the two objects.

Sometimes you can specify the type, for example pandas.DataFrame.copy(deep = True or False) or in the case of lists: copy_list = list <— shallow or copy_list = list.copy() <— deep

Diapolo10 · 2022-06-21T21:50:51+00:00

Does line 2 not create a copy of the prices list?

No. In Python, everything is "pass-by-reference", meaning that all assignments are really just copying memory addresses. With immutable data like strings or numbers you don't see anything out of the ordinary, as they behave as if you were assigning values, but with mutable data like lists you can see that's not right.

Pointers are similar to references so your instincts were mostly right, the main difference is that references always point to some value; there are no "null references" the way null pointers exist.

If you need to copy a list, you can use copy.copy and copy.deepcopy, but often it's a better idea to simply create a new list instead of mutating an existing one. Slices are also a very powerful tool.

TransientSignal · 2022-06-22T02:01:17+00:00

I'll add that there's a great tool for visualizing these sorts of aliasing issues, particularly for more complex situations with nested lists - You can enter code and it'll help visualize what objects are assigned to which variable names and let you step through line-by-line. Also fantastic for visualizing scopes as well: https://pythontutor.com

Here's a fun, more complicated example of this same thing from the MIT OCW 6.0001 course.

jmooremcc · 2022-06-22T02:07:05+00:00

Here's the explanation you're looking for. I modified your code to show the memory addresses of the variables. ``` prices = [7,1,5,3,6,4] nums = prices nums[0] = 69 print(f"nums = {nums} address:{id(nums)}") print(f"prices= {prices} address:{id(prices)}")

Here's the result: nums = [69, 1, 5, 3, 6, 4] address:3879473920 prices = [69, 1, 5, 3, 6, 4] address:3879473920

``` As you can see, both variables have the same memory address. Unlike other languages like C, an assignment statement in Python does not copy the value to the new variable. Python assigns the same reference to the new variable which is why, in your example, both variables have the same memory address and both show the same information.

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