4

I have a polars DataFrame with a number of Series that look like:

pl.Series(['cow', 'cat', '', 'lobster', ''])

and I'd like them to be

pl.Series(['cow', 'cat', pl.Null, 'lobster', pl.Null])

A simple string replacement won't work since pl.Null is not of type PyString:

pl.Series(['cow', 'cat', '', 'lobster', '']).str.replace('', pl.Null)

What's the idiomatic way of doing this for a Series/DataFrame in polars?

  • Note that `pl.Null` is a `DataType`. If you want to set a null value, you should use pythons builtin `None`. – ritchie46 May 23 '22 at 06:46

1 Answers1

10

Series

For a single Series, you can use the set method.

import polars as pl
my_series = pl.Series(['cow', 'cat', '', 'lobster', ''])
my_series.set(my_series.str.lengths() == 0, None)

shape: (5,)
Series: '' [str]
[
        "cow"
        "cat"
        null
        "lobster"
        null
]

DataFrame

For DataFrames, I would suggest using when/then/otherwise. For example, with this data:

df = pl.DataFrame({
    'str1': ['cow', 'dog', "", 'lobster', ''],
    'str2': ['', 'apple', "orange", '', 'kiwi'],
    'str3': ['house', '', "apartment", 'condo', ''],
})
df

shape: (5, 3)
┌─────────┬────────┬───────────┐
│ str1    ┆ str2   ┆ str3      │
│ ---     ┆ ---    ┆ ---       │
│ str     ┆ str    ┆ str       │
╞═════════╪════════╪═══════════╡
│ cow     ┆        ┆ house     │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ dog     ┆ apple  ┆           │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│         ┆ orange ┆ apartment │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ lobster ┆        ┆ condo     │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│         ┆ kiwi   ┆           │
└─────────┴────────┴───────────┘

We can run a replacement on all string columns as follows:

df.with_columns([
    pl.when(pl.col(pl.Utf8).str.lengths() ==0)
    .then(None)
    .otherwise(pl.col(pl.Utf8))
    .keep_name()
])

shape: (5, 3)
┌─────────┬────────┬───────────┐
│ str1    ┆ str2   ┆ str3      │
│ ---     ┆ ---    ┆ ---       │
│ str     ┆ str    ┆ str       │
╞═════════╪════════╪═══════════╡
│ cow     ┆ null   ┆ house     │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ dog     ┆ apple  ┆ null      │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ null    ┆ orange ┆ apartment │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ lobster ┆ null   ┆ condo     │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ null    ┆ kiwi   ┆ null      │
└─────────┴────────┴───────────┘

The above should be fairly performant.

If you only want to replace empty strings with null on certain columns, you can provide a list:

only_these = ['str1', 'str2']
df.with_columns([
    pl.when(pl.col(only_these).str.lengths() == 0)
    .then(None)
    .otherwise(pl.col(only_these))
    .keep_name()
])

shape: (5, 3)
┌─────────┬────────┬───────────┐
│ str1    ┆ str2   ┆ str3      │
│ ---     ┆ ---    ┆ ---       │
│ str     ┆ str    ┆ str       │
╞═════════╪════════╪═══════════╡
│ cow     ┆ null   ┆ house     │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ dog     ┆ apple  ┆           │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ null    ┆ orange ┆ apartment │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ lobster ┆ null   ┆ condo     │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ null    ┆ kiwi   ┆           │
└─────────┴────────┴───────────┘