CODEWITHSUNDEEP
pythonlist
snorkle213
snorkle213

Reputation: 5

Python appending a list to a list and then clearing it

I have this part of code isolated for testing purposes and this question

noTasks = int(input())
noOutput = int(input())
outputClist = []
outputCList = []

for i in range(0, noTasks):
    for w in range(0, noOutput):
        outputChecked = str(input())
        outputClist.append(outputChecked)
    outputCList.append(outputClist)
    outputClist[:] = []
print(outputCList)

I have this code here, and i get this output

[[], []]

I can't figure out how to get the following output, and i must clear that sublist or i get something completely wrong...

[["test lol", "here can be more stuff"], ["test 2 lol", "here can be more stuff"]]

Upvotes: 0

Views: 171

Answers (2)

Daniel
Daniel

Reputation: 42758

In Python everything is a object. A list is a object with elements. You only create one object outputclist filling and clearing its contents. In the end, you have one list multiple times in outputCList, and as your last thing is clearing the list, this list is empty.

Instead, you have to create a new list for every task:

noTasks = int(input())
noOutput = int(input())
output = []

for i in range(noTasks):
    checks = []
    for w in range(noOutput):
        checks.append(input())
    output.append(checks)
print(output)

Upvotes: 1

Blair
Blair

Reputation: 6693

Instead of passing the contained elements in outputClist to outputCList (not the greatest naming practice either to just have one capitalization partway through be the only difference in variable names), you are passing a reference to the list itself. To get around this important and useful feature of Python that you don't want to make use of, you can pretty easily just pass a new list containing the elements of outputClist by changing this line

 outputCList.append(outputClist)

to

 outputCList.append(list(outputClist))

or equivalently, as @jonrsharpe states in his comment

 outputCList.append(outputClist[:])

Upvotes: 1

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