list.pop and list.pop()
I don't think I understand what the .pop() method does on a list....
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This feels like one of those python weird things. I am interested in explanations.
If I have a list=[1,2,3,4] and I do list.pop() the result is list=[1,2,3].
Perfect, just what I wanted.
However, if I am not careful and instead do list.pop--note there are no parentheses this time--I get no syntax error or warning and nothing happens, leading me to a strange debug session.
In the repl, if I do l.pop it just identifies it as a built-in method of list object at 0xwhatever. That's useful, but why is there not at least a runtime warning when I make this mistake in my code?
The pop method of dicts (like self.data, i.e. {'a':'aaa','b':'bbb','c':'ccc'}, here) takes two arguments -- see the docs
The second argument, default, is what pop returns if the first argument, key, is absent.
(If you call pop with just one argument, key, it raises an exception if that key's absent).
In your example, print b.pop('a',{'b':'bbb'}), this is irrelevant because 'a' is a key in b.data. But if you repeat that line...:
b=a()
print b.pop('a',{'b':'bbb'})
print b.pop('a',{'b':'bbb'})
print b.data
you'll see it makes a difference: the first pop removes the 'a' key, so in the second pop the default argument is actually returned (since 'a' is now absent from b.data).
So many questions here. I see at least two, maybe three:
- What does pop(a,b) do?/Why are there a second argument?
- What is
*argsbeing used for?
The first question is trivially answered in the Python Standard Library reference:
pop(key[, default])
If key is in the dictionary, remove it and return its value, else return default. If default is not given and key is not in the dictionary, a KeyError is raised.
The second question is covered in the Python Language Reference:
If the form “*identifier” is present, it is initialized to a tuple receiving any excess positional parameters, defaulting to the empty tuple. If the form “**identifier” is present, it is initialized to a new dictionary receiving any excess keyword arguments, defaulting to a new empty dictionary.
In other words, the pop function takes at least two arguments. The first two get assigned the names self and key; and the rest are stuffed into a tuple called args.
What's happening on the next line when *args is passed along in the call to self.data.pop is the inverse of this - the tuple *args is expanded to of positional parameters which get passed along. This is explained in the Python Language Reference:
If the syntax *expression appears in the function call, expression must evaluate to a sequence. Elements from this sequence are treated as if they were additional positional arguments
In short, a.pop() wants to be flexible and accept any number of positional parameters, so that it can pass this unknown number of positional parameters on to self.data.pop().
This gives you flexibility; data happens to be a dict right now, and so self.data.pop() takes either one or two parameters; but if you changed data to be a type which took 19 parameters for a call to self.data.pop() you wouldn't have to change class a at all. You'd still have to change any code that called a.pop() to pass the required 19 parameters though.
L=[1,2,3,4,5,6]
for element in L: print(L,element) L.pop(0)
I have the code above. This code returns the following:
[1, 2, 3, 4, 5, 6] 1 [2, 3, 4, 5, 6] 3 [3, 4, 5, 6] 5
I don't understand why the loop variable prints(1,3,5) here...