Question

Sorry if this is already mentioned somewhere(I couldn’t find it).

I basically want to list an item from a list but its including quotes and brackets(which I don’t want).
Here’s my data:

inputData = {'red':3, 'blue':1, 'green':2, 'organge':5}

Here’s my class to find items either based on key or value.

class Lookup(dict):
    """
    a dictionary which can lookup value by key, or keys by value
    """
    def __init__(self, items=[]):
        """items can be a list of pair_lists or a dictionary"""
        dict.__init__(self, items)

    def get_key(self, value):
        """find the key(s) as a list given a value"""
        return [item[0] for item in self.items() if item[1] == value]

    def get_value(self, key):
        """find the value given a key"""
        return self[key]

it works fine except for the brackets.

print Lookup().get_key(2) # ['blue']  but I want it to just output blue

I know I can do this via replacing the bracket/quotes( LookupVariable.replace("'", "") ) but I was wondering if there was a more pythonic way of doing this.

Thanks.

Answer 1

Change

return [item[0] for item in self.items() if item[1] == value]

to

return next(item[0] for item in self.items() if item[1] == value)

Right now you’re returning the result of a list comprehension — a list. Instead, you want to return the first item returned by the equivalent generator expression — that’s what next does.

Edit: If you actually want multiple items, use Greg’s answer — but it sounds to me like you’re only thinking about getting a single key — this is a good way to do that.

If you want it to raise a StopIteration error if the value doesn’t exist, leave it as above. If you want it to return something else instead (like None) do:

return next((item[0] for item in self.items() if item[1] == value), None)