Allow dictionary in .str.replace() and .str.replace_all()

[lmocsi]

Description

It would be nice to have instead of chained replace_all's:

df = (df.with_columns(pl.col('field1').str.replace_all('aa','1').str.replace_all('bb','2'),
                      pl.col('field2').str.replace_all('aa','1').str.replace_all('bb','2')
      )

using a dict, something like:

conv = {'aa': '1','bb': '2'}
df = (df.with_columns(pl.col('field1').str.replace_all(conv),
                      pl.col('field2').str.replace_all(conv)
      )

[Julian-J-S]

hmm interesting idea but I can also see how this could be confusing.
For example: should the replacements build on top of each other?
Like:

• text = "ABC"
• replacements = {'AB': '1', 'BC': '2'}
• then the result would be 1C because after replacing AB then BC is not available anymore
• and if the order of the dict changes to {'BC': '2', 'AB': '1'} the result would be A2
• having chained replace_all is very clear to the reader what should happen and in what order

If you try to replace whole values (in your example 'aa' and 'bb') you can look into map_dict https://pola-rs.github.io/polars/py-polars/html/reference/expressions/api/polars.Expr.map_dict.html#polars-expr-map-dict

[cmdlineluser]

As @JulianCologne points out, they are slightly different operations.

Passing multiple patterns/replacements usually implies the replacements happen in isolation / independently of each other.

Whereas with chaining, each call is processing the output from the previous call.

As for chaining, I was trying to figure out if it was possible to build a helper function.

i.e. turn

replacements = {'one': 'two', 'three': 'four'}

expr = pl.col('field1', 'field2').str.replace_all

into:

pl.col('field1', 'field2').str.replace_all('one', 'two').str.replace_all('three', 'four')

But I'm not sure if that is possible?

With expr and string, we could split on . and loop through getattr()

expr, function = pl.col('field1', 'field2'), 'str.replace_all'

The actual str.replace_all function seems to live in pl.expr.string.ExprNameSpace which we could functools.reduce

expr, function = pl.col('field1', 'field2'), pl.expr.string.ExprStringNameSpace.replace_all

But that's somewhat awkward.

[lmocsi]

It seems that standard replace allows dicts:

import polars as pl
dff = pl.DataFrame({'a':['a','b','c']})
mapping = {'a':'apple', 'b':'banana'}

dff.with_columns(pl.col('a').replace(mapping, default=pl.lit('***')).alias('aa'))