Split a string by a delimiter in python

How to split this string where __ is the delimiter

MATCHES__STRING

To get an output of ['MATCHES', 'STRING']?

You can use the str.split method: string.split('__')

>>> "MATCHES__STRING".split("__")
['MATCHES', 'STRING']

You may be interested in the csv module, which is designed for comma-separated files but can be easily modified to use a custom delimiter.

import csv
csv.register_dialect( "myDialect", delimiter = "__", <other-options> )
lines = [ "MATCHES__STRING" ]

for row in csv.reader( lines ):
    ...

When you have two or more elements in the string (in the example below there are three), then you can use a comma to separate these items:

date, time, event_name = ev.get_text(separator='@').split("@")

After this line of code, the three variables will have values from three parts of the variable ev.

So, if the variable ev contains this string and we apply separator @:

Sa., 23. März@19:00@Klavier + Orchester: SPEZIAL

Then, after the split operation the variable

• date will have value Sa., 23. März
• time will have value 19:00
• event_name will have value Klavier + Orchester: SPEZIAL

For Python 3.8, you actually don't need the get_text method, you can just go with ev.split("@"), as a matter of fact the get_text method is throwing an att. error. So if you have a string variable, for example:

filename = 'file/foo/bar/fox'

You can just split that into different variables with comas as suggested in the above comment but with a correction:

W, X, Y, Z = filename.split('_') 
W = 'file' 
X = 'foo'
Y = 'bar'
Z = 'fox'

Besides split and rsplit, there is partition/rpartition. It separates string once, but the way question was asked, it may apply as well.

Example:

>>> "MATCHES__STRING".partition("__")
('MATCHES', '__', 'STRING')

>>> "MATCHES__STRING".partition("__")[::2]
('MATCHES', 'STRING')

And a bit faster then split("_",1):

$ python -m timeit "'validate_field_name'.split('_', 1)[-1]"
2000000 loops, best of 5: 136 nsec per loop

$ python -m timeit "'validate_field_name'.partition('_')[-1]"
2000000 loops, best of 5: 108 nsec per loop

Timeit lines are based on this answer