pandas replacing by function output (Applying a function to rows based on regex)

Question

Basically I have a dataframe that could look like this:

ID NAME                      PAINT
0  some_name:target          blue
1  some_other_name           pink
2  other_name: other_target  yellow
3  other_name                black

And only want to replace values that follow a certain regex by applying a function to them.

def f(x):
  name, target = x.split(":")
  return "[" + target + "]" + " " + name

ID NAME                        PAINT
0  [target] some_name          blue
1  some_other_name             pink
2  [other_target] other_name   yellow
3  other_name                  black

I imagine it would look something like this but whatever works

df.replace(to_replace=strings_found_by_regex, value=f(strings_found_by_regex))

This could probably be done by iterating over rows and seing if those cells match the regex and then appplying f(x) but that looks rather ugly and I wondered whether there is a better way.

sushanth · Accepted Answer · 2020-08-15 11:22:20Z

3

try this, using Series.str.replace

Find out regex explanation here, regex101.com

df.NAME.str.replace("(.+)\s*:\s*(.+)", r"[\2] \1")

0           [target] some_name
1              some_other_name
2    [other_target] other_name
3                   other_name
Name: NAME, dtype: object

edited Aug 15, 2020 at 11:22

answered Aug 15, 2020 at 10:32

sushanth

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4 Comments

DarrylG Over a year ago

Shouldn't \2 be stripped? Row index 2 NAME column desired answer is [other_target] other_name while this answer provides [ other_target] other_name (i.e. extra space in [ other_target].

Matej Novosad Over a year ago

care to explain the raw string part? Thanks!

sushanth Over a year ago

@MatejNovosad, update regex link with substitution & here is link to thread that discuss about backreference

sushanth Over a year ago

@MatejNovosad Here is another interesting link that explains what "\1" does.. Hope it clarifies your queries.

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