How to convert string with a delimiter to list of strings in Python?

Question

I'm querying a database and get this text:

"('username', '192.168.1.1', datetime.datetime(2013, 1, 3, 7, 18))"

I want to split into list like this:

['username', '192.168.1.1', datatime.datetime(2013, 1, 3, 7, 18)]

So, which way is the best way? I'm using Python 2.4.

do you want datetime.datetime(..) to become a datetime object, or should it stay a string? — wastl, Nov 10 '14 at 16:15
datetime is between quotes? Because this way a simple `split` operation won't work. — Alberto Coletta, Nov 10 '14 at 16:17
If it weren't for the `datetime` bit, I'd suggest `thetext[1:-1].split(",")` for this. Perhaps you could use a [regex](https://docs.python.org/2/library/re.html) to replace the commas inside the brackets with something else (such as `;`) first? — GoBusto, Nov 10 '14 at 16:19
What database are you using, and why do you get a string back? — jonrsharpe, Nov 10 '14 at 16:20
possible duplicate of [Python - lexical analysis and tokenization](http://stackoverflow.com/questions/2358890/python-lexical-analysis-and-tokenization) — ivan_pozdeev, Nov 10 '14 at 16:22
@SoundOfSilent What format is the data officially is? And why is it in such a format? The best solution is to get the data that's already split into its meaningful parts in the first place. — ivan_pozdeev, Nov 10 '14 at 16:27

Alex Riley · Answer 1 · 2014-11-10T16:33:11.223

One way would be to use eval and list:

>>> import datetime
>>> string = "('username', '192.168.1.1', datetime.datetime(2013, 1, 3, 7, 18))"
>>> list(eval(string))
['username', '192.168.1.1', datetime.datetime(2013, 1, 3, 7, 18)]

eval evaluates the string as a Python expression, here producing a tuple containing the two strings and a datetime instance. list converts this tuple to a list.

Caution: eval will potentially execute any valid Python code, even malicious code. There are safer alternatives such as ast.literal_eval which could be used if you're not sure whether you trust what's in your database.

As always with `eval`, insecure unless the input is trusted. — ivan_pozdeev, Nov 10 '14 at 16:23

score 0 · Answer 2 · answered Nov 10 '14 at 17:31

You can use maxsplit parameter of the split function:

your_string = "('username', '192.168.1.1', datetime.datetime(2013, 1, 3, 7, 18))"
your_string[1:-1].split(",", 2)

# Returns:
# ["'username'", " '192.168.1.1'", ' datetime.datetime(2013, 1, 3, 7, 18)']

score 0 · Answer 3 · answered Nov 10 '14 at 17:31

As noted in jonrsharpe comment, it looks like you are falling in a trap you have just digged yourself.

Any correct database can give you the individual columns, and if you correctly do the query, you get them in proper format.

Assuming you can use a DB-API 2.0 described in PEP 249, and assuming conn is an open connection to your database, you should do something like :

curs = conn.cursor()
curs.execute('SELECT name, ip, dat FROM tab')
t = curs.fetchone()

and t should directly be the tuple ('username', '192.168.1.1', datatime.datetime(2013, 1, 3, 7, 18)) (use list(t) if you need a list ...)

How to convert string with a delimiter to list of strings in Python?

3 Answers3