A Program That Identifies Individual Words In A Sentence, Stores These In A List And Replaces Each Word With The Position Of That Word In The List

I am developing a program that identifies individual words in a sentence, stores these in a list and replaces each word in the original sentence with the position of that word in t

Solution 1:

Here's a linear algorithm:

position = {} # word -> position
words = sentence.split()
for word in words:
    if word not in position: # new word
       position[word] = len(position) + 1# store its positionprint(*map(position.__getitem__, words), sep=",")
# -> 1,2,3,4,5,6,7,8,9,1,3,9,6,7,8,4,5

The print() call uses Python 3 * syntax to unpack the result returned by map() that returns positions for the corresponding words here. See What does ** (double star) and * (star) do for parameters?

Solution 2:

To get a list of word positions in sentence and recreate the original sentence from this list:

sentence = "ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU ASK WHAT YOU CAN DO FOR YOUR COUNTRY"
s = sentence.split()
positions = [s.index(x)+1 for x in s]
recreated = [s[i-1] for i in positions]
# the reconstructed sentenceprint(" ".join(recreated))
# the list of index numbers/word positionsprint(positions)
# the positions as a space separated string of numbersprint(" ".join(positions)

Lists are zero-indexed so the first element is index 0, not 1. You could, of course, add 1 to all indices in the list comprehension if you wanted it to start at 1.

To get exactly the same output as your script produces:

sentence = "ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU ASK WHAT YOU CAN DO FOR YOUR COUNTRY"
s = sentence.split()
positions = [s.index(x)+1 for x in s]
print(sentence)
print(positions)

Output:

ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU ASK WHAT YOU CAN DO FOR YOUR COUNTRY
[1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 3, 9, 6, 7, 8, 4, 5]

Solution 3:

sentence = 'ASK NOT WHAT YOUR COUNTRY CAN DO FOR YOU ASK WHAT YOU CAN DO FOR YOUR COUNTRY'
words = sentence.split()

# Counting backwards through the words means the last seen one will have # the lowest index without needing 'if' tests or incrementing counters.
positions = {word:indexforindex, word in reversed(list(enumerate(words, 1)))}

print(' '.join(str(positions.get(word)) for word in words))

Try it on repl.it here: https://repl.it/CHvy/0

Solution 4:

Not terribly efficient, but two lines.

words = sentence.split()
positions = [words.index(word) + 1 for word in words]

Note that list.index(entry) will return the index of the first occurrence of entry. If you're okay with 0-based indices, then the following is pretty concise:

positions = list(map(words.index, words))