calculating rowwise minimum of two series? - python-polars

Question

Hi;
Is there any function that can generate a serie by calculating rowwise minimum of two series.? Functionality will be similar to np.minimum
a = [1,4,2,5,2]
b= [5,1,4,2,5]
np.minimum(a,b) -> [1,1,2,2,2]
Thanks.

Answer 1

q =df.lazy().with_column(pl.when(pl.col("a")>pl.col("b")).then(pl.col("b")).otherwise(pl.col("a")).alias("minimum"))
df = q.collect()
i didn't tested it but this should work i think

Answer 2

As the accepted answer states, you can use pl.when -> then -> otherwise expression.
If you have a wider DataFrame, you can use the DataFrame.min() method, pl.min expression, or a pl.fold for more control.
import polars as pl
df = pl.DataFrame({
"a": [1,4,2,5,2],
"b": [5,1,4,2,5],
"c": [3,2,5,7,2]
})
df.min(axis=1)
This outputs:
shape: (5,)
Series: 'a' [i64]
[
1
1
2
2
2
]
Min expression
When given multiple expression inputs to a pl.min the minimum is determined row-wise instead of column-wise.
df.select(pl.min(["a", "b", "c"]))
Outputs:
shape: (5, 1)
┌─────┐
│ min │
│ --- │
│ i64 │
╞═════╡
│ 1 │
├╌╌╌╌╌┤
│ 1 │
├╌╌╌╌╌┤
│ 2 │
├╌╌╌╌╌┤
│ 2 │
├╌╌╌╌╌┤
│ 2 │
└─────┘
Fold expression
Or with a fold expression:
df.select(
pl.fold(int(1e9), lambda acc, a: pl.when(acc > a).then(a).otherwise(acc), ["a", "b", "c"])
)
shape: (5, 1)
┌─────────┐
│ literal │
│ --- │
│ i64 │
╞═════════╡
│ 1 │
├╌╌╌╌╌╌╌╌╌┤
│ 1 │
├╌╌╌╌╌╌╌╌╌┤
│ 2 │
├╌╌╌╌╌╌╌╌╌┤
│ 2 │
├╌╌╌╌╌╌╌╌╌┤
│ 2 │
└─────────┘
The fold allows for more cool things, because you operate over expressions.
So we could for instance compute the min of the squared columns:
pl.fold(int(1e9), lambda acc, a: pl.when(acc > a).then(a).otherwise(acc), [pl.all()**2])
Or we could compute the min of square root of column "a" and the rest of the columns is squared.
pl.fold(int(1e9), lambda acc, a: pl.when(acc > a).then(a).otherwise(acc), [pl.col("a").sqrt(), pl.all().exclude("a")**2])
You get the idea.