CODEWITHSUNDEEP
python
Tom F
Tom F

Reputation: 453

Appending to a list inside of a dictionary

I'm going through a list of individual words and creating a dictionary where the word is the key, and the index of the word is the value.

dictionary = {}
for x in wordlist:
    dictionary[x] = wordlist.index(x)

This works fine at the moment, but I want more indexes to be added for when the same word is found a second, or third time etc. So if the phrase was "I am going to go to town", I would be looking to create a dictionary like this:

{'I': 0, 'am' : 1, 'going' : 2, 'to': (3, 5), 'go' : 4, 'town' : 6}

So I suppose I need lists inside the dictionary? And then to append more indexes to them? Any advice on how to accomplish this would be great!

Upvotes: 1

Views: 2578

Answers (5)

Mario Rossi
Mario Rossi

Reputation: 7799

A possible solution:

dictionary= {}
for i, x in enumerate(wordlist):
    if not x in dictionary : dictionary[x]= []
    dictionary[x].append( i )

Upvotes: 0

dansalmo
dansalmo

Reputation: 11704

>>> wl = ['I', 'am', 'going', 'to', 'go', 'to', 'town']
>>> {w: [i for i, x in enumerate(wl) if x == w] for w in wl}
{'town': [6], 'I': [0], 'am': [1], 'to': [3, 5], 'going': [2], 'go': [4]}

Upvotes: 2

Mario Rossi
Mario Rossi

Reputation: 7799

import collections
dictionary= collections.defaultdict(list)
for i, x in enumerate( wordlist ) : 
    dictionary[x].append( i )

Upvotes: 0

rodrigo
rodrigo

Reputation: 98516

You can do this way:

dictionary = {}
for i, x in enumerate(wordlist):
    dictionary.setdefault(x, []).append(i)

Explanation:

  • You do not need the call to index(). It is more efficient and cooler to use enumerate().
  • dict.setdefault() uses the first argument as key. If it is not found, inserts the second argument, else it ignores it. Then it returns the (possibly newly inserted) value.
  • list.append() appends the item to the list.

You will get something like this:

{'I': [0], 'am' : [1], 'going' : [2], 'to': [3, 5], 'go' : [4], 'town' : [6]}

With lists instead of tuples, and using lists even if it is only one element. I really think it is better this way.

UPDATE:

Inspired shamelessly by the comment by @millimoose to the OP (thanks!), this code is nicer and faster, because it does not build a lot of [] that are never inserted in the dictionary:

import collections
dictionary = collections.defaultdict(list)
for i, x in enumerate(wordlist):
    dictionary[x].append(i)

Upvotes: 7

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 799430

Objects are objects, regardless of where they are.

dictionary[x] = []
 ...
dictionary[x].append(y)

Upvotes: 0

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