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.

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.

If anybody still looking at this question, the currently accepted answer is now outdated: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

Addressing an element of dictionary is like sitting on donkey and enjoy the ride.

As rule of Python DICTIONARY is orderless

If there is

dic = {1: "a", 2: "aa", 3: "aaa"}

Now suppose if I go like dic[10] = "b", then it will not add like this always

dic = {1:"a",2:"aa",3:"aaa",10:"b"}

It may be like

dic = {1: "a", 2: "aa", 3: "aaa", 10: "b"}

Or

dic = {1: "a", 2: "aa", 10: "b", 3: "aaa"}

Or

dic = {1: "a", 10: "b", 2: "aa", 3: "aaa"}

Or any such combination.

So thumb rule is DICTIONARY is orderless!

If you need an ordered dictionary, you can use odict.

actually I found a novel solution that really helped me out, If you are especially concerned with the index of a certain value in a list or data set, you can just set the value of dictionary to that Index!:

Just watch:

list = ['a', 'b', 'c']
dictionary = {}
counter = 0
for i in list:
   dictionary[i] = counter
   counter += 1

print(dictionary) # dictionary = {'a':0, 'b':1, 'c':2}

Now through the power of hashmaps you can pull the index your entries in constant time (aka a whole lot faster)

oh, that’s a tough one. What you have here, basically, is two values for each item. Then you are trying to call them with a number as the key. Unfortunately, one of your values is already set as the key!

Try this:

colors = {1: ["blue", "5"], 2: ["red", "6"], 3: ["yellow", "8"]}

Now you can call the keys by number as if they are indexed like a list. You can also reference the color and number by their position within the list.

For example,

colors[1][0]
// returns 'blue'

colors[3][1]
// returns '8'

Of course, you will have to come up with another way of keeping track of what location each color is in. Maybe you can have another dictionary that stores each color’s key as it’s value.

colors_key = {‘blue’: 1, ‘red’: 6, ‘yllow’: 8}

Then, you will be able to also look up the colors key if you need to.

colors[colors_key[‘blue’]][0] will return ‘blue’

Something like that.

And then, while you’re at it, you can make a dict with the number values as keys so that you can always use them to look up your colors, you know, if you need.

values = {5: [1, ‘blue’], 6: [2, ‘red’], 8: [3, ‘yellow’]}

Then, (colors[colors_key[values[5][1]]][0]) will return ‘blue’.

Or you could use a list of lists.

Good luck!

You can’t, since dict is unordered. you can use .popitem() to get an arbitrary item, but that will remove it from the dict.