How to index into a dictionary?


Question

I have a Dictionary below:

colors = {
    "blue" : "5",
    "red" : "6",
    "yellow" : "8",
}

How do I index the first entry in the dictionary?

colors[0] will return a KeyError for obvious reasons.

1
68
6/7/2017 1:41:51 AM

Accepted Answer

Dictionaries are unordered in Python versions up to and including Python 3.6. If you do not care about the order of the entries and want to access the keys or values by index anyway, you can use d.keys()[i] and d.values()[i] or d.items()[i]. (Note that these methods create a list of all keys, values or items in Python 2.x. So if you need them more then once, store the list in a variable to improve performance.)

If you do care about the order of the entries, starting with Python 2.7 you can use collections.OrderedDict. Or use a list of pairs

l = [("blue", "5"), ("red", "6"), ("yellow", "8")]

if you don't need access by key. (Why are your numbers strings by the way?)

In Python 3.7, normal dictionaries are ordered, so you don't need to use OrderedDict anymore (but you still can – it's basically the same type). The CPython implementation of Python 3.6 already included that change, but since it's not part of the language specification, you can't rely on it in Python 3.6.

86
8/3/2018 10:34:37 AM

If anybody still looking at this question, the currently accepted answer is now outdated:

Since Python 3.7* the dictionaries are order-preserving, that is they now behave exactly as collections.OrderedDicts used to. Unfortunately, there is still no dedicated method to index into keys() / values() of the dictionary, so getting the first key / value in the dictionary can be done as

first_key = list(colors)[0]
first_val = list(colors.values())[0]

or alternatively (this avoids instantiating the keys view into a list):

def get_first_key(dictionary):
    for key in dictionary:
        return key
    raise IndexError

first_key = get_first_key(colors)
first_val = colors[first_key]

If you need an n-th key, then similarly

def get_nth_key(dictionary, n=0):
    if n < 0:
        n += len(dictionary)
    for i, key in enumerate(dictionary.keys()):
        if i == n:
            return key
    raise IndexError("dictionary index out of range") 

(*CPython 3.6 already included ordered dicts, but this was only an implementation detail. The language specification includes ordered dicts from 3.7 onwards.)


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