python-polars string concatenation of two existing columns - python-polars

Question

I want to concatenate the last letter from two existing columns and create a new column from this using polars.LazyFrame
for example in pandas can achieve this with the following code
import pandas as pd
df = pd.DataFrame({"col1":["abc","def"], "col2":["ghi","jkl"]})
df["last_letters_concat"]=df["col1"].str.strip().str[-1]+df["col2"].str.strip().str[-1]
print(df)
My attempt in polars
import polars as pl
from polars import col
#using same df
df.lazy().with_column(
(pl.col("col1")[-1] + pl.col('col2'))[-1].alias("last_letters_concat")).collect()
How can i do this?

Answer 1

You can use the str.slice expression for that. Below I show to examples that produce the same result.
pl.DataFrame({
"col1": ["abc","def"],
"col2":["ghi","jkl"]
})
# concat all last letters
out1 = df.select(
pl.concat_str([pl.col("col1").str.slice(-1), pl.col("col2").str.slice(-1)])
)
# concat only two specific columns
out2 = df.select(
pl.col("col1").str.slice(-1) + pl.col("col2").str.slice(-1)
)
assert out1.frame_equal(out2)
print(out1)
shape: (2, 1)
┌──────┐
│ col1 │
│ --- │
│ str │
╞══════╡
│ ci │
├╌╌╌╌╌╌┤
│ fl │
└──────┘
I recommend using the concat_str expression as this has O(n) complexity where n is the number of columns you add, whereas the addition operator has O(n^2) complexity.
EDIT: as of polars >= 0.14.12 the optimizer will ensure it always is linear complexity