如何迭代列表的元组？

Question

alright this might be a stupid one but I am kinda stuck right now. I am dealing with arguments of type tuple, containing lists of lists and just can't read them properly.

Here is a basic example:

def test(*args):
    for arg in args:
        print(arg)
        print(arg[0])
        print(arg[1])

testTuple1 = ([["x_1", "x_2", "x_3"], []])
testTuple2 = ([['x_1', 'x_2', 'x_3'], []], [['x_2', 'x_3'], ['x_1']])

test(testTuple1)
test(testTuple2)

Which results in the following outputs:

[['x_1', 'x_2', 'x_3'], []]
['x_1', 'x_2', 'x_3']
[]

([['x_1', 'x_2', 'x_3'], []], [['x_2', 'x_3'], ['x_1']])
[['x_1', 'x_2', 'x_3'], []]
[['x_2', 'x_3'], ['x_1']]

The output of the first example is the output I would have expected. However I don't get the first output of the second example. Why am I getting a tuple and not the first of the lists contained in the tuple: [['x_1', 'x_2', 'x_3'], []]?

Answer 1

Because in python there's some ambiguity created by the fact that parens are used both for creating literal tuples and for general grouping of expressions. Here's your lists, with some added formatting for clarity:

testTuple1 = (
               [
                 ["x_1", "x_2", "x_3"], 
                 []
               ]
             )

testTuple2 = (
               [
                 ['x_1', 'x_2', 'x_3'], 
                 []
               ], 
               [
                 ['x_2', 'x_3'], 
                 ['x_1']
               ]
             )

testTuple1 , since the outer parens contain only a single item (the outermost list), will evaluate to just that outer list. The parens just sort of go away. But in testTuple2 , since there are multiple comma-separated lists, it will evaluate to a tuple.

If you want the first one to be a tuple as well, you need to add a trailing comma after the outer list to explicitly let python know that those parens are meant to denote a 1-tuple, and aren't just for grouping:

testTuple1 = ([["x_1", "x_2", "x_3"], []],)

Answer 2

When you use the asterisk as modifier for your argument, it will enable you to pass arbitrarily many arguments when calling the function.

def test(*args):
    # Do anything

test(v1)
test(v1, arg, another, something)
test()

It will end up concatenate all arguments from the call into one tuple.

When you simply pass one argument to the function call, the tupled argument will end consisting only of this one argument, no matter what class this object actually has. Just do the printing right before the iteration. Then you will see the actual value of args , which resolves to this:

([['x_1', 'x_2', 'x_3'], []],)

---

If you want to pass an iterable to a function and have the element on the first layer spread out to the arguments just use the asterisk at the call again.

>> test(*testTuple1)
['x_1', 'x_2', 'x_3']
'x_1'
'x_2'
[]

IndexError: ....

---

The IndexError occurrs, because an empty list does not have a zero-th or even a first element to get.