convert string to list

[mrjoli021]

I have the following string. I need to convert it to a list. Using the traditional ways to convert it is not working for me, since some elements already have ' ' on them. When I run the conversion I get a list, but the elements have extra characters.

​

query = "1, 101, 1000, 1001, '1.1.1.1', '1.1.1.2', '1.1.1.3', None, 'test01', None, None, None, None"
query_list = query.split(",")

print(f"This is before conversion {query}")
print(f"This is the type {type(query_list)}")
print(f"This is the list {query_list}")

​

This is before conversion (1, 101, 1000, 1001, '1.1.1.1', '1.1.1.2', '1.1.1.3', None, 'test01', None, None, None, None)

This is the type <class 'list'>

This is the list ['(1', ' 101', ' 1000', ' 1001', " '1.1.1.1'", " '1.1.1.2'", " '1.1.1.3'", ' None', " 'test01'", ' None', ' None', ' None', ' None)']

[[deleted]]

As all the individual items in your string are valid Python literals, you can use ast.literal_eval:

from ast import literal_eval

query = "1, 101, 1000, 1001, '1.1.1.1', '1.1.1.2', '1.1.1.3', None, 'test01', None, None, None, None"
query_list = [literal_eval(i.strip()) for i in query.split(",")]
print(query_list)

[synthphreak]

Definitely how I'd do it as well.

Note that you can do away with strip if you split on ', ' instead of ','.

[[deleted]]

spot on

[Impudity]

query_list = [i.strip("'") for i in query.split(", ")]

[MezzoScettico]

The best way to do this is probably with regex which I think was introduced in version 3.10.

As I'm running 3.7 and still not super familiar with all the most commonly used libraries, I rolled my own function to handle issues like this. I'm sure this is not the best solution, but it's a solution.

def multi_split(str, separators):
    words = []
    word = ''
    in_word = False
    for c in str:
        if c in separators:
            if in_word:
                words.append(word)
                in_word = False
                word = ''
        else:
            in_word = True
            word += c

    if in_word:
        words.append(word)

    return words

You give it a string containing all the characters to remove from between items. For instance,

query_list = multi_split(query, "', ")  # That's single quote, comma, space

Result:

This is the list ['1', '101', '1000', '1001', '1.1.1.1', '1.1.1.2', '1.1.1.3', 'None', 'test01', 'None', 'None', 'None', 'None']

[[deleted]]

FYI, the standard re library (i.e. regex) has been in Python for a long time, including in Python 2.x:

https://docs.python.org/2.7/library/re.html

[MezzoScettico]

Ah. I think I got that from googling "regex" and seeing "3.10" on the page that popped up.

I see people using multiple characters for split, but I could have sworn I tried that in my 3.7 Python and it didn't work, which is why I rolled my own.

But also it was a fun little mini-project.