Keeping a value the same despite changing the variable it was equated to

Keeping a value the same despite changing the variable it was equated to - Printable Version

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Keeping a value the same despite changing the variable it was equated to - TheTypicalDoge - Mar-13-2022

I'm beginning to learn Python and I've come across a problem that is foreign to me as a past Lua programmer. I've noticed that after appending a variable to a list, if that variable changes value so does the value in the list as well, even though the list was never directly redefined.
Here's what I mean,

currentPath = ["EWR", "JFK", "SFO"]
completedPaths = []

completedPaths.append(currentPath)
print(completedPaths)

currentPath.clear()
print(completedPaths)

Output:["EWR", "JFK", "SFO"]
[[]]

This confuses me because in Lua the list in completedPaths would stay the same until directly redefined, even if the variable it was once equated to has changed. How would I solidify the list inside of completedPaths while also changing the currentPath variable to input more values later on?

RE: Keeping a value the same despite changing the variable it was equated to - menator01 - Mar-13-2022

Might can do something like this

currentPath = ["EWR", "JFK", "SFO"]
completedPaths = []

# completedPaths.append(currentPath.copy())
# or maybe
completedPaths = completedPaths + currentPath
print(completedPaths)

currentPath.clear()
print(currentPath)
print(completedPaths)

Output:[['EWR', 'JFK', 'SFO']]
[]
[['EWR', 'JFK', 'SFO']]

RE: Keeping a value the same despite changing the variable it was equated to - Yoriz - Mar-13-2022

The list that is referenced by currentPath that is appended to the list referenced by completedPaths is not a copy of the list it is the same list.

currentPath = ["EWR", "JFK", "SFO"]
completedPaths = []
 
completedPaths.append(currentPath)
print(id(currentPath))
print(id(completedPaths[0]))

Output:2794339074624
2794339074624

You can view this here this visualizer will show you that the same list object is being referenced by currentPath and index 0 of completedPaths.

If you don't want this behaviour rather than passing the reference to the list pass a copy by using a slice of all items

currentPath = ["EWR", "JFK", "SFO"]
completedPaths = []
 
completedPaths.append(currentPath[:])
print(completedPaths)
 
currentPath.clear()
print(completedPaths)

Output:[['EWR', 'JFK', 'SFO']]
[['EWR', 'JFK', 'SFO']]

or by using the copy module

import copy

currentPath = ["EWR", "JFK", "SFO"]
completedPaths = []
 
completedPaths.append(copy.copy(currentPath))
print(completedPaths)
 
currentPath.clear()
print(completedPaths)

Output:[['EWR', 'JFK', 'SFO']]
[['EWR', 'JFK', 'SFO']]

or the list itself's copy method

completedPaths.append(currentPath.copy())

Both of these are shallow copies, the copy module also has copy.deepcopy() for list within list.