python-polars string concatenation of two existing columns

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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?

python-polars
1 Answers

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

ritchie460
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