Python for loop, iterate through a string: 'int' object is not iterable

Question

I've been having trouble with this for loop that is supposed to iterate through and save a list with all the words in the book.

The error I get is: 'int' object is not iterable.

def create_word_dict ()
    word_list = open("mobydick.txt", "r")
    all_list = word_list.read()
    all_list = all_list.split()
    word_list.close()
    for index in len(all_list):
        all_list[index] = parseString(all_list[index])
    return all_list

# Removes punctuation marks from a string
def parseString (st):
    s = ""

    for ch in st:
        if ch.isalpha() or ch.isspace():
            s += ch
        else:
            s += ""
    return s #return was outside code block

I edited your post to fix all the indentation. What part are you having issues with exactly? — abe
– abe, Commented Feb 12, 2016 at 4:24

Joran Beasley · Accepted Answer · 2016-02-12 04:24:04Z

2

I guess you want

for index in range(len(all_list)):
   all_list[index]=parseString(all_list[index])

since for i in 5: means nothing in python(since an int cannot be iterated), however for i in range(5) is indeed a valid statement, since a range can be iterated...

however its probably better to just iterate over the objects directly

new_list = []
for word in all_list:
    new_list.append(parseString(word))

or even better just do a list comprehension

new_list = [parseString(word) for word in all_list]

answered Feb 12, 2016 at 4:24

Joran Beasley

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1 Comment

Joran Beasley Over a year ago

well I doubt its all of the examples ;P but most of the common ones at least

Hugh Bothwell · Accepted Answer · 2016-02-12 04:51:10Z

0

You can speed things up a lot by using some of Python's built-in methods:

from string import ascii_lowercase as lower, ascii_uppercase as upper
from string import digits, punctuation

# create a translation table which
#   makes the string lowercase and
#   replaces all digits and punctuation with spaces
TRANS = str.maketrans(
    lower + upper +           digits + punctuation,
    lower + lower + " " * len(digits + punctuation)
)

def get_word_list(filename):
    with open(filename) as inf:
        return inf.read().translate(TRANS).split()

words = get_word_list("mobydick.txt")

For sake of comparison, on my machine this loads the words from the Gutenberg version of Moby Dick (220231 words) in 0.11 seconds.

edited Feb 12, 2016 at 4:51

answered Feb 12, 2016 at 4:40

Hugh Bothwell

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