Dynamic variable in Python [duplicate]

2019-01-02 21:59发布

问题:

This question already has an answer here:

  • How do I create a variable number of variables? 15 answers

How can I create lists with dynamic names in python, for example

for i in range(len(myself)):
   list(i) = []

what should I use instead of list(i) ? it means that i want some names as below:

list1
list2
list3
...

回答1:

I'd advise you to just use a list or dictionary instead of dynamic variable names. All the versions below result in lists[0], lists[1] etc being [], which seems close enough to what you want, and will be more readable/maintainable in the long term. (Note: I'm using lists instead of list as a variable name because the latter would overwrite the builtin list function, which you probably don't want).

1) Version with lists being a list of lists (the numbers are just the order of the lists):

lists = [[] for i in range(len(myself))]

2) Same but with a for loop instead of a list comprehension:

lists = []
for i in range(len(myself)):
   lists.append([])

3) Version with lists being a dictionary of lists with numbers as keys (a bit more flexible if you want to remove some of the values later or such):

lists = {}
for i in range(len(myself)):
   lists[i] = []

About dynamic variable names, i.e. variables like list1 instead of lists[1]... Seriously, you probably shouldn't do that. It's unnecessarily complicated and hard to maintain. Think about it - next month you'll want to modify the script, and you'll try to figure out where the variable list1 was defined, and you won't be able to do that with a plain text search. It's a pain.
But if you really want to for some reason, it's possible with exec - here are some reasons not to use it - or with modifying locals() - bad idea according to documentation. Also see comments for more discussion on why these things are a bad idea and how confusing it gets even talking about them.



回答2:

Dynamic variable names are very rarely a good idea; it is almost always better to use a dictionary:

myLists = {"list{}".format(i):[] for i in range(len_i)}

Do not use list as a variable name - it hides the existing keyword.



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