变量名称要在python函数中分配为参数

Question

I am writing a function that builds and populates a tree-like data structure, using the treelib library. You create a tree as follows:

foo = Tree()

...and go from there, which is standard enough. Here is what I have, simplified:

def make_special_tree(tree, arg1, arg2):
    tree = Tree()
    other_stuff = things(arg1)
    modify_tree(tree, other_stuff, arg2)
    return tree

Here's the thing: let's say that I ultimately want a tree object called blah . If I do the following, the command runs without an error:

make_special_tree('blah', foo, bar)

...but when I type blah afterward, I get back NameError: name 'blah' is not defined . If I do the following, the command also runs without an error:

blah = make_special_tree('yoink', foo, bar)

...and when I type blah afterward, I get back <treelib.tree.Tree object at 0x10e60db10> , which is what I want to see. yoink meanwhile remains undefined, like blah in the previous version.

Hence my question--and I can tell this is basic, but I cannot untangle this, in part because I am not sure how precisely to ask the question. As you can see, right now I have to create an instance of class Tree() and I think I have to feed my function an argument to do so. I think that blah = make_special_tree(args) is the correct way to format this, but how can I pass the variable blah as the name of the tree structure I'd like returned?

Answer 1

The first argument in your function is redundant. Try this:

def make_special_tree(arg1, arg2):
    tree = Tree()
    other_stuff = things(arg1)
    modify_tree(tree, other_stuff, arg2)
    return tree

blah = make_special_tree(foo, bar)

Answer 2

Python functions generally don't "assign variable names", they just return values, and you can do whatever you want with those values.

A value doesn't have a variable name. ^* Because a value can be stored in five different variables, or it can be stored in things that aren't variables, like the elements of a list. It doesn't care.

So, what you almost certainly want to do is what @spectras suggested in a comment: Get rid of that first parameter, and then store the value in whatever variable you want:

def make_special_tree(arg1, arg2):
    tree = Tree()
    other_stuff = things(arg1)
    modify_tree(tree, other_stuff, arg2)
    return tree

# Two different trees with different names
blah = make_special_tree(foo, bar)
yoink = make_special_tree(foo, bar)

# One tree with two names
spam = make_special_tree(foo, bar)
eggs = spam

# A tree with no names at all
cheese = []
cheese.append(make_special_tree(foo, bar))

---

If you really do want to create a global variable with a specified name, you can do that, ^** but it's a really bad idea, unless you're trying to do something unusual, like the way collections.namedtuple acts like a class statement.

---

_{* A value can have a name by having, eg, a name attribute. You can even change the definition of the Tree class to use that name attribute as part of its output representation. But this still has nothing to do with what variables you store it in.}

_{** The least hacky way is globals()[name] = value .}