Converting a String to a List of Words?

Question

I'm trying to convert a string to a list of words using python. I want to take something like the following:

string = 'This is a string, with words!'

Then convert to something like this :

list = ['This', 'is', 'a', 'string', 'with', 'words']

Notice the omission of punctuation and spaces. What would be the fastest way of going about this?

Answer 1

I think this is the simplest way for anyone else stumbling on this post given the late response:

>>> string = 'This is a string, with words!'
>>> string.split()
['This', 'is', 'a', 'string,', 'with', 'words!']

Answer 2

Try this:

import re

mystr = 'This is a string, with words!'
wordList = re.sub("[^\w]", " ",  mystr).split()

How it works:

From the docs :

re.sub(pattern, repl, string, count=0, flags=0)

Return the string obtained by replacing the leftmost non-overlapping occurrences of pattern in string by the replacement repl. If the pattern isn’t found, string is returned unchanged. repl can be a string or a function.

so in our case :

pattern is any non-alphanumeric character.

[\w] means any alphanumeric character and is equal to the character set [a-zA-Z0-9_]

a to z, A to Z , 0 to 9 and underscore.

so we match any non-alphanumeric character and replace it with a space .

and then we split() it which splits string by space and converts it to a list

so 'hello-world'

becomes 'hello world'

with re.sub

and then ['hello' , 'world']

after split()

let me know if any doubts come up.

Answer 3

To do this properly is quite complex. For your research, it is known as word tokenization. You should look at NLTK if you want to see what others have done, rather than starting from scratch:

>>> import nltk
>>> paragraph = u"Hi, this is my first sentence. And this is my second."
>>> sentences = nltk.sent_tokenize(paragraph)
>>> for sentence in sentences:
...     nltk.word_tokenize(sentence)
[u'Hi', u',', u'this', u'is', u'my', u'first', u'sentence', u'.']
[u'And', u'this', u'is', u'my', u'second', u'.']

Answer 4

The most simple way:

>>> import re
>>> string = 'This is a string, with words!'
>>> re.findall(r'\w+', string)
['This', 'is', 'a', 'string', 'with', 'words']

Answer 5

Using string.punctuation for completeness:

import re
import string
x = re.sub('['+string.punctuation+']', '', s).split()

This handles newlines as well.

Answer 6

Well, you could use

import re
list = re.sub(r'[.!,;?]', ' ', string).split()

Note that both string and list are names of builtin types, so you probably don't want to use those as your variable names.

Answer 7

Inspired by @mtrw's answer, but improved to strip out punctuation at word boundaries only:

import re
import string

def extract_words(s):
    return [re.sub('^[{0}]+|[{0}]+$'.format(string.punctuation), '', w) for w in s.split()]

>>> str = 'This is a string, with words!'
>>> extract_words(str)
['This', 'is', 'a', 'string', 'with', 'words']

>>> str = '''I'm a custom-built sentence with "tricky" words like https://stackoverflow.com/.'''
>>> extract_words(str)
["I'm", 'a', 'custom-built', 'sentence', 'with', 'tricky', 'words', 'like', 'https://stackoverflow.com']

Answer 8

Personally, I think this is slightly cleaner than the answers provided

def split_to_words(sentence):
    return list(filter(lambda w: len(w) > 0, re.split('\W+', sentence))) #Use sentence.lower(), if needed

Answer 9

A regular expression for words would give you the most control. You would want to carefully consider how to deal with words with dashes or apostrophes, like "I'm".

Answer 10

list=mystr.split(" ",mystr.count(" "))

Answer 11

This way you eliminate every special char outside of the alphabet:

def wordsToList(strn):
    L = strn.split()
    cleanL = []
    abc = 'abcdefghijklmnopqrstuvwxyz'
    ABC = abc.upper()
    letters = abc + ABC
    for e in L:
        word = ''
        for c in e:
            if c in letters:
                word += c
        if word != '':
            cleanL.append(word)
    return cleanL

s = 'She loves you, yea yea yea! '
L = wordsToList(s)
print(L)  # ['She', 'loves', 'you', 'yea', 'yea', 'yea']

I'm not sure if this is fast or optimal or even the right way to program.

Answer 12

def split_string(string):
    return string.split()

This function will return the list of words of a given string. In this case, if we call the function as follows,

string = 'This is a string, with words!'
split_string(string)

The return output of the function would be

['This', 'is', 'a', 'string,', 'with', 'words!']

Answer 13

This is from my attempt on a coding challenge that can't use regex,

outputList = "".join((c if c.isalnum() or c=="'" else ' ') for c in inputStr ).split(' ')

The role of apostrophe seems interesting.

Answer 14

Probably not very elegant, but at least you know what's going on.

my_str = "Simple sample, test! is, olny".lower()
my_lst =[]
temp=""
len_my_str = len(my_str)
number_letter_in_data=0
list_words_number=0
for number_letter_in_data in range(0, len_my_str, 1):
    if my_str[number_letter_in_data] in [',', '.', '!', '(', ')', ':', ';', '-']:
        pass
    else:
        if my_str[number_letter_in_data] in [' ']:
            #if you want longer than 3 char words
            if len(temp)>3:
                list_words_number +=1
                my_lst.append(temp)
                temp=""
            else:
                pass
        else:
            temp = temp+my_str[number_letter_in_data]
my_lst.append(temp)
print(my_lst)

Answer 15

You can try and do this:

tryTrans = string.maketrans(",!", "  ")
str = "This is a string, with words!"
str = str.translate(tryTrans)
listOfWords = str.split()