(k := next(iter(d)), d.pop(k))
will remove the leftmost (first) item (if it exists) from a dict object.
And if you want to remove the right most/recent value from the dict
d.popitem()
(k := next(iter(d)), d.pop(k))
will remove the leftmost (first) item (if it exists) from a dict object.
And if you want to remove the right most/recent value from the dict
d.popitem()
ex_dict.popitem()
it removes the last (most recently added) element from the dictionary
python 3.x - Pop element at the beginning or at the last from dict in python3 - Stack Overflow
python - Pop an element from a list in a dictionary - Stack Overflow
dictionary - In Python, what does dict.pop(a,b) mean? - Stack Overflow
Access an arbitrary element in a dictionary in Python - Stack Overflow
try it like this:
def get_item(item_name):
return d[item_name].pop()
d[item_name] gets you the list and then .pop() pops an item out
def epic(d, key, element):
if key in d and element in d[key]:
d[key].remove(element)
print(f"Deleted {element} from the list associated with key '{key}'")
print(f'deleted element: {element}')
else:
print(f"Element {element} not found in the list associated with key '{key}'")
epic(d, 'two', 4) ```
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.
On Python 3, non-destructively and iteratively:
next(iter(mydict.values()))
On Python 2, non-destructively and iteratively:
mydict.itervalues().next()
If you want it to work in both Python 2 and 3, you can use the six package:
six.next(six.itervalues(mydict))
though at this point it is quite cryptic and I'd rather prefer your code.
If you want to remove any item, do:
key, value = mydict.popitem()
Note that "first" may not be an appropriate term here because dict is not an ordered type in Python < 3.6. Python 3.6+ dicts are ordered.
If you only need to access one element (being the first by chance, since dicts do not guarantee ordering) you can simply do this in Python 2:
my_dict.keys()[0] # key of "first" element
my_dict.values()[0] # value of "first" element
my_dict.items()[0] # (key, value) tuple of "first" element
Please note that (at best of my knowledge) Python does not guarantee that 2 successive calls to any of these methods will return list with the same ordering. This is not supported with Python3.
in Python 3:
list(my_dict.keys())[0] # key of "first" element
list(my_dict.values())[0] # value of "first" element
list(my_dict.items())[0] # (key, value) tuple of "first" element