返回由定界符分隔的字符串列表

Question

Im having some problems trying to solve this question. Its from a practice exam and I just can't seem to get it right. Im supposed to write a python function that takes in a string and a delimiter, and return a list back with the string stripped of the delimiter. We are not allowed to use the split function or "any such function". The example we received in the question was this

StringToken("this is so fun! I love it!", "!")

Outputs

["this is so fun", "I love it"]

This is the code I made up, its super simple.

def tokenizer(string, tmp):
    newStr = []
    for i in range(len(string)):
        if string[i] != tmp:
            newStr.append(string[i])
    return newStr

and the output is this

['T', 'h', 'i', 's', ' ', 'i', 's', ' ', 's', 'o', ' ', 'f', 'u', 'n', ' ', 'I', ' ', 'l', 'o', 'v', 'e', ' ', 'i', 't']

How can I rejoin each word?

Answer 1

If you join all the elements in the list you will get a single string which may not be what you are looking for.

Create a string before append it to the list like;

>>> def StringToken(string, tmp):
    newStrlist = []
    newStr = ''
    for i in range(len(string)):
        if string[i] != tmp:
            newStr += string[i]
        elif newStr != '':
            newStrlist.append(newStr)
            newStr = ''
    return newStrlist
... ... ... ... ... ... ... ... ... ... 
>>> StringToken("this is so fun! I love it!", "!")
['this is so fun', ' I love it']

Answer 2

See comments in the code for a description.

def StringToken(string, tmp):
    newStr = ""   # A string to build upon
    lst = []      # The list to return
    for c in string: # Iterate over the characters
        if tmp == c: # Check for the character to strip
            if newStr != "":   # Prevent empty strings in output
                lst.append(newStr.strip())   # add to the output list
                newStr = ""                  # restart the string
                continue                     # move to the next character
        newStr += c  # Build the string
    return lst   # Return the list

Output

StringToken("this is so fun! I love it!", "!")
# ['this is so fun', 'I love it']

Answer 3

Instead of looping over the all the letters in the string, you can use find to get the index of the next occurrence of the delimiter and then build your list accordingly:

def tokenizer(string, delim):
    new_list = []
    while True:
        index = string.find(delim)  # use find to next occurrence of delimiter
        if index > -1:
            new_list.append(string[:index])
            string = string[index + len(delim):]
        else:
            new_list.append(string)
            break              # break because there is no delimiter present anymore

    # remove whitespaces and trim the existing strings 
    return [item.strip() for item in new_list if item.strip()]

Usage:

>>> tokenizer("this is so fun! I love it!", "!")
["this is so fun", "I love it"]

Answer 4

Here's an alternative that's a little shorter than the current answers:

def StringToken(string, tmp):
    newStr = []
    start = 0
    for ind, char in enumerate(string):
        if char == tmp:
            newStr.append(string[start:ind])
            start = ind + 1
    return newStr

Output

>>> StringToken("this is so fun! I love it!", "!")
['this is so fun', ' I love it']

---

Edit: If you would like to remove leading or trailing spaces like in your example, that can be done using strip():

def StringToken(string, tmp):
    newStr = []
    start = 0
    for ind, char in enumerate(string):
        if char == tmp:
            newStr.append(string[start:ind].strip())
            start = ind + 1
    return newStr

Output

>>> StringToken("this is so fun! I love it!", "!")
['this is so fun', 'I love it']

Answer 5

simply use join operator this will join entire list with a given delimiter. Here in this you can use empty delimiter ''. try:

a=['T', 'h', 'i', 's', ' ', 'i', 's', ' ', 's', 'o', ' ', 'f', 'u', 'n', ' ', 'I', ' ', 'l', 'o', 'v', 'e', ' ', 'i', 't']
''.join(a)

output will be

'This is so fun I love it'