easy name problem

[Steinsack]

Hello,

I am a beginner in Phyton. It is a very basic question I guess, but pls help me anyway :)

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listA=[] a=input() globals()[a]=[]    #generates a new list with the name of the input string listA.append(????????)    #question

--> I would like to add the new list to listA. I know the name is in a, but if I put a in, the string gets added to listA and not the new list.
What do I have to put inside the brackets?

Cheers

Larz60+ write Jun-16-2021, 03:13 PM:
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[snippsat]

You should not use globals() that's the internal dictionary that Python use.
Can make visible dictionary that do the same,then it's much more understandable and readable for all.
Here a look a two versions.

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list_a = [] animal = input('Enter a animal: ') # cat more_animals = ['Sheep', 'dog'] list_a.append(animal) list_a.append(more_animals) print(list_a)

Output:
['cat', ['Sheep', 'dog']]

So this is normal way to make a list and append to it,will lose refence to variable names.
If need also names for what is the data structure then can use a dictionary.

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list_a = [] animal_dict = {} animal_dict['animal'] = input('Enter a animal: ') more_animals = ['Sheep', 'dog'] animal_dict['more_animals'] = ['Sheep', 'dog'] # Can also append a dicionray to a list list_a.append(animal_dict) print(animal_dict) print(list_a)

Output:
{'animal': 'cat', 'more_animals': ['Sheep', 'dog']}
[{'animal': 'cat', 'more_animals': ['Sheep', 'dog']}]

So now we have a visible dictionary that really dos the same as globals().
If first code i could to this which is a bad💀 way of doing this.

1 2	>>> globals()['more_animals'] ['Sheep', 'dog']

In second code it make more sense as can do this.

1 2	>>> animal_dict['more_animals'] ['Sheep', 'dog']