Question

I am trying to find things in a string – all of them are before a number, for example:

"Diablo Lord Of Destruction 9.2"

This is an index from a file such that file[2] = "Diablo Lord Of Destruction 9.2"

how can I write code that will select only the text and leave out the numbers and any white space before those numbers (as below)?

"Diablo Lord Of Destruction"

I understand you can easily do this by doing something like this:

contents = file[2]
print contents[0:-2]

Since the values will be changing, I need a more robust solution that can handle different sized numbers and different lengths of white space.

Answer 1

You can utilize regular expressions and the sub() method:

Return the string obtained by replacing the leftmost non-overlapping
occurrences of pattern in string by the replacement repl. If the
pattern isn’t found, string is returned unchanged. repl can be a
string or a function; if it is a string, any backslash escapes in it
are processed.

>>> import re
>>> re.sub('[0-9.]*', '', 'Diablo Lord of Destruction 9.2')[:-1]
'Diablo Lord of Destruction'
>>> re.sub('[\d.]*', '', 'Diablo Lord of Destruction 9.2')[:-1]
'Diablo Lord of Destruction'

The code above will find all number occurrences, [0-9.] or [\d.], and replace them with ”. In addition, it trims the last character, which was a space.