easy name problem - Printable Version

easy name problem - Printable Version

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easy name problem - Steinsack - Jun-16-2021

Hello,

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

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

RE: easy name problem - snippsat - Jun-16-2021

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.

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.

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.

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

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

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