otp.Operation.str.regex_replace#

regex_replace(pat, repl, *, replace_every=False, caseless=False)#

Search for occurrences (case dependent) of pat and replace with repl.

Parameters

pat (str or Column or Operation) – Pattern to replace specified via the POSIX extended regular expression syntax.
repl (str or Column or Operation) – Replacement string. \0 refers to the entire matched text. \1 to \9 refer to the text matched by the corresponding parenthesized group in pat.
replace_every (bool) – If replace_every flag is set to True, all matches will be replaced, if False only the first one.
caseless (bool) – If the caseless flag is set to True, matching is case-insensitive.

Returns

String with pattern pat replaced by repl.

Return type

Operation

Examples

>>> data = otp.Ticks(X=['A Table', 'A Chair', 'An Apple'])
>>> data['Y'] = data['X'].str.regex_replace('An? ', 'The ')
>>> otp.run(data)
                     Time         X          Y
0 2003-12-01 00:00:00.000   A Table  The Table
1 2003-12-01 00:00:00.001   A Chair  The Chair
2 2003-12-01 00:00:00.002  An Apple  The Apple

Parameter replace_every will replace all occurrences of pat in the string:

>>> data = otp.Ticks(X=['A Table, A Chair, An Apple'])
>>> data['Y'] = data['X'].str.regex_replace('An? ', 'The ', replace_every=True)
>>> otp.run(data)
        Time                           X                                Y
0 2003-12-01  A Table, A Chair, An Apple  The Table, The Chair, The Apple

Capturing groups in regular expressions is supported:

>>> data = otp.Ticks(X=['11/12/1992', '9/22/1993', '3/30/1991'])
>>> data['Y'] = data['X'].str.regex_replace(r'(\d{1,2})/(\d{1,2})/', r'\2.\1.')
>>> otp.run(data)
                     Time           X           Y
0 2003-12-01 00:00:00.000  11/12/1992  12.11.1992
1 2003-12-01 00:00:00.001   9/22/1993   22.9.1993
2 2003-12-01 00:00:00.002   3/30/1991   30.3.1991