CODEWITHSUNDEEP
python
ASm
ASm

Reputation: 389

Create a dictionary where letters of the alphabet are the key?

I have a tuple that contains random words such as 

words = ("Apple","Charlie","Papa","Uniform","Tango","Echo","Romeo")

and I want to sort these into a dictionary like 

dict = {"A": "Apple","C": "Charlie", "P": "Papa", "U" : "Uniform"....}

I know I need to initialize the dictionary and create a for loop to look through the tuple, but I don't know how to sort by letter.

Upvotes: 1

Views: 1245

Answers (2)

mgilson
mgilson

Reputation: 310097

For the case you've mentioned, it's easy:

d = {}
for word in words:
    d[word[0]] = word

Of course, the normal caveats apply (dicts aren't ordered so there's no filtering involved, if "Adam" and "Apple" are both in your input, only the second will find its way into the output, etc). To take care of the latter problem, you might want to consider a collections.defaultdict:

from collections import defaultdict
d = defaultdict(list)
for word in words:
    d[word[0]].append(word)

which will leave you with a dict of characters mapped to a list. And, once you have this, if you really want it to be "sorted" by key, you can do that with a collections.OrderedDict after the fact:

odict = OrderedDict()
for key in sorted(d):  # d is the defaultdict from the last step.
    odict[key] = d[key]

Upvotes: 1

brunsgaard
brunsgaard

Reputation: 5168

You can use dict comprehensions. 

>>> words = ("Apple","Charlie","Papa","Uniform","Tango","Echo","Romeo")
>>> mydict = {value[0]: value for value in words}
>>> print(mydict)
{'A': 'Apple', 'C': 'Charlie', 'E': 'Echo', 'P': 'Papa', 'R': 'Romeo', 'U': 'Uniform', 'T': 'Tango'}

It will not be ordered, but if you need that pass it through

>>> OrderedDict(sorted(mydict.items()))

Upvotes: 2

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