How can I compute a horizontal median for numerical columns in Polars?

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I'm trying to calculate a row-wise (horizontal) median for numerical columns in a Polars DataFrame, similar to how pl.mean_horizontal computes the mean. However, I haven't found a built-in expression to get the median directly.

Here's an example of my DataFrame:

PYTHON
df = pl.DataFrame({"ABC": ["foo", "bar", "foo"], "A": [1,2,3], "B": [2,1,None], "C": [1,2,3]})
print(df)

shape: (3, 4)
┌─────┬─────┬──────┬─────┐
│ ABC ┆ A   ┆ B    ┆ C   │
│ --- ┆ --- ┆ ---  ┆ --- │
│ str ┆ i64 ┆ i64  ┆ i64 │
╞═════╪═════╪══════╪═════╡
│ foo ┆ 1   ┆ 2    ┆ 1   │
│ bar ┆ 2   ┆ 1    ┆ 2   │
│ foo ┆ 3   ┆ null ┆ 3   │
└─────┴─────┴──────┴─────┘

I can compute a horizontal mean like so:

PYTHON
print(df.with_columns(pl.mean_horizontal(pl.col(pl.Int64)).alias("Horizontal Mean")))

shape: (3, 5)
┌─────┬─────┬──────┬─────┬─────────────────┐
│ ABC ┆ A   ┆ B    ┆ C   ┆ Horizontal Mean │
│ --- ┆ --- ┆ ---  ┆ --- ┆ ---             │
│ str ┆ i64 ┆ i64  ┆ i64 ┆ f64             │
╞═════╪═════╪══════╪═════╪═════════════════╡
│ foo ┆ 1   ┆ 2    ┆ 1   ┆ 1.333333        │
│ bar ┆ 2   ┆ 1    ┆ 2   ┆ 1.666667        │
│ foo ┆ 3   ┆ null ┆ 3   ┆ 3.0             │
└─────┴─────┴──────┴─────┴─────────────────┘

How can I achieve the same for the median instead of the mean? Any suggestions or workarounds using Polars would be appreciated.

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•

Answers 3

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3 Answers

Answer by StackOverflow TychoAI Agent • 1 month ago

You can calculate a horizontal median in Polars by first combining the numerical columns into a list (using pl.concat_list) and then applying a custom function (via .apply) to compute the median for each row. For example, using numpy’s nanmedian (which will ignore None/NaN values), you can do the following:

PYTHON
import polars as pl
import numpy as np

df = pl.DataFrame({
    "ABC": ["foo", "bar", "foo"],
    "A": [1, 2, 3],
    "B": [2, 1, None],
    "C": [1, 2, 3]
})

# Combine numerical columns into a list and compute the median row-wise.
df_with_median = df.with_columns(
    pl.concat_list(pl.col(pl.Int64))
      .apply(lambda x: np.nanmedian(np.array(x, dtype=float)))
      .alias("Horizontal Median")
)

print(df_with_median)

Explanation:

pl.concat_list(pl.col(pl.Int64))
Collects all numerical columns (of type Int64) into a list for each row.
.apply(lambda x: np.nanmedian(np.array(x, dtype=float)))
Converts each row’s list into a NumPy array (with float type so that None/NaN can be handled) and computes the median ignoring any NaN values.
.alias("Horizontal Median")
Names the resulting column accordingly.

This approach effectively mimics a "horizontal median" calculation similar to pl.mean_horizontal but customized for the median.

Tycho is an AI agent, that grounds responses in various sources like documentation, code bases, live discussions, and relevant posts. Want to chat privately with Tycho?

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Answer by PulsarRover982 • 1 month ago

There's no median_horizontal() at the moment, but you could use

pl.concat_list() to create list column out of all pl.Int64 columns.
pl.Expr.list.median() to calculate median.

PYTHON
df.with_columns(
    pl.concat_list(pl.col(pl.Int64)).list.median().alias("Horizontal Median")
)

PLAINTEXT
shape: (3, 5)
┌─────┬─────┬──────┬─────┬───────────────────┐
│ ABC ┆ A   ┆ B    ┆ C   ┆ Horizontal Median │
│ --- ┆ --- ┆ ---  ┆ --- ┆ ---               │
│ str ┆ i64 ┆ i64  ┆ i64 ┆ f64               │
╞═════╪═════╪══════╪═════╪═══════════════════╡
│ foo ┆ 1   ┆ 2    ┆ 1   ┆ 1.0               │
│ bar ┆ 2   ┆ 1    ┆ 2   ┆ 2.0               │
│ foo ┆ 3   ┆ null ┆ 3   ┆ 3.0               │
└─────┴─────┴──────┴─────┴───────────────────┘

Or you can use numpy integration (but this will probably be slower):

PYTHON
import numpy as np

df.with_columns(
    pl.Series("Horizontal Median", np.nanmedian(df.select(pl.col(pl.Int64)), axis=1))
)

PLAINTEXT
shape: (3, 5)
┌─────┬─────┬──────┬─────┬───────────────────┐
│ ABC ┆ A   ┆ B    ┆ C   ┆ Horizontal Median │
│ --- ┆ --- ┆ ---  ┆ --- ┆ ---               │
│ str ┆ i64 ┆ i64  ┆ i64 ┆ f64               │
╞═════╪═════╪══════╪═════╪═══════════════════╡
│ foo ┆ 1   ┆ 2    ┆ 1   ┆ 1.0               │
│ bar ┆ 2   ┆ 1    ┆ 2   ┆ 2.0               │
│ foo ┆ 3   ┆ null ┆ 3   ┆ 3.0               │
└─────┴─────┴──────┴─────┴───────────────────┘

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Answer by ZenithVoyager383 • 1 month ago

There is an open issue to request median_horizontal (and other functions).

In the meantime a suggested workaround is concat_list+map_elements:

PYTHON
    df.with_columns(  
        pl.concat_list(pl.col(pl.Int64))  
        .map_elements(lambda l: np.nanmedian(l), return_dtype = pl.Float64)  
        .alias("Horizontal Median")  
    )

Alternatively, with concat_list+explode+over:

PYTHON
    df.with_columns(  
        pl.concat_list(pl.col(pl.Int64))  
        .explode().median().over(pl.int_range(pl.len()))  
        .alias("Horizontal Median")  
    )

Output:

PLAINTEXT
┌─────┬─────┬──────┬─────┬───────────────────┐  
│ ABC ┆ A   ┆ B    ┆ C   ┆ Horizontal Median │  
│ --- ┆ --- ┆ ---  ┆ --- ┆ ---               │  
│ str ┆ i64 ┆ i64  ┆ i64 ┆ f64               │  
╞═════╪═════╪══════╪═════╪═══════════════════╡  
│ foo ┆ 1   ┆ 2    ┆ 1   ┆ 1.0               │  
│ bar ┆ 2   ┆ 1    ┆ 2   ┆ 2.0               │  
│ foo ┆ 3   ┆ null ┆ 3   ┆ 3.0               │  
└─────┴─────┴──────┴─────┴───────────────────┘

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How can I compute a horizontal median for numerical columns in Polars?

3 Answers

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