CODEWITHSUNDEEP
python
qiaop
qiaop

Reputation: 169

Adding key-value pairs to a dictionary in an ascending order

Say I have a list of words with repeats:

["Apple", "Orange", "Grape", "Orange", "Watermelon", "Apple", "Grape"]

I need to put them into a dictionary so that each word has its own key in an ascending order:

{1 : "Apple", 2 : "Orange", 3 : "Grape", 4: "Watermelon"}

The order is based on which word appears first. If "Apple" appears first, it will have the key 1 and if "Apple" appears in the future, it will be ignored, since "Apple" is already included in 1. If "Orange" appears after "Apple", it will have the key 2 . Thus, the key-value pairs are added in an ascending order.

I don't have a problem with adding and checking repeats, but I'm struggling with making the keys in an ascending order. Any clues?

Upvotes: 0

Views: 233

Answers (5)

domoarigato
domoarigato

Reputation: 2912

fruits = ["Apple", "Orange", "Grape", "Orange", "Watermelon", "Apple", "Grape"] 
basket = []
for i in range(len(fruits)):
    if fruit[i] not in basket:
        basket.append(fruit)

This doesn't get you a dictionary, but it shouldn't matter, because the elements are accessed the same way: basket[2] == "Grape" either way, assuming you correct for arrays being 0-indexed while your desired dict layout is 1-indexed.

Upvotes: 0

celeritas
celeritas

Reputation: 2281

Edited after the discussion in the comments:

A pythonic way to check if a key exists in a dictionary:

fruits = ['Apple', 'Orange', 'Grape', 'Orange', 'Watermelon', 'Apple', 'Grape']
index = 1
basket = {}
for fruit in fruits:
    if fruit not in basket:
        basket[fruit] = index
        index = index + 1

Upvotes: 0

clintval
clintval

Reputation: 369

You can convert your list into a set which eliminates redundant entries but you lose your initial list order:

fruit = set(["Apple", "Orange", "Grape", "Orange", "Watermelon", "Apple", "Grape"])

The object fruit becomes:

{'Apple', 'Grape', 'Orange', 'Watermelon'}

I find it easier to have the iteration index starting at 1 for cases like these. If you would like to sort your fruits before assigning them to numerical keys in a dictionary you will need to to apply the sorted() function to your fruit set for an alphabetical sort (you can also setup a parameter to sort in other ways using the sorted() function).

basket = {}
for i, item in enumerate(sorted(fruit), start=1):
    basket[i] = item
print basket

Printed result:

{1: 'Apple', 2: 'Grape', 3: 'Orange', 4: 'Watermelon'}

Upvotes: 1

DSM
DSM

Reputation: 353059

If you don't mind an import:

>>> from collections import OrderedDict
>>> s = ["Apple", "Orange", "Grape", "Orange", "Watermelon", "Apple", "Grape"] 
>>> dict(enumerate(OrderedDict.fromkeys(s), 1))
{1: 'Apple', 2: 'Orange', 3: 'Grape', 4: 'Watermelon'}

This works because OrderedDict.fromkeys makes an ordered dictionary where the order is insertion order:

>>> OrderedDict.fromkeys(s)
OrderedDict([('Apple', None), ('Orange', None), ('Grape', None), ('Watermelon', None)])

Upvotes: 5

jh314
jh314

Reputation: 27792

Just maintain an index and check each fruit against the values of the dictionary.

fruits = ["Apple", "Orange", "Grape", "Orange", "Watermelon", "Apple", "Grape"] 
d = {}
index = 1
for fruit in fruits:
    if fruit not in d.values():
        d[index] = fruit
        index += 1

Will give you:

>>> d
{1: 'Apple', 2: 'Orange', 3: 'Grape', 4: 'Watermelon'}

Upvotes: 3

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