Drop column by index in polars

Python is zero-indexed so when you say “first” do you mean index == 0 or index == 1? In any case you can use a selector to drop a column:

import polars as pl
import polars.selectors as cs

df = pl.DataFrame(
    {
        "a": [1,2,3],
        "b": [4,5,6],
        "c": [7,8,9],
    }
)

df.drop(cs.by_index(0))  # drops 'a'; 1 would drop 'b'

shape: (3, 2)
┌─────┬─────┐
│ b   ┆ c   │
│ --- ┆ --- │
│ i64 ┆ i64 │
╞═════╪═════╡
│ 4   ┆ 7   │
│ 5   ┆ 8   │
│ 6   ┆ 9   │
└─────┴─────┘

It seems exclude() needs column's name and you can use df.columns[1] to get its name.

index = 1

result = df.select(pl.exclude( df.columns[index] ))

result = df.drop( df.columns[index] )

It seems works also:

result = df.drop( pl.nth(index) )

Minimal working code for tests:

It allows to run ie. main.py 2 to remove 3rd column.

import polars as pl
import sys

# Creating a new Polars DataFrame
technologies = {
    'Courses': ["Spark", "Pandas", "Hadoop", "Python", "Pandas", "Spark"],
    'Fees': [22000, 26000, 25000, 20000, 26000, 22000],
    'Duration': ['30days', '60days', '50days', '40days', '60days', '30days'],
    'Discount': [1000, 2000, 1500, 1200, 2000, 1000]
}

df = pl.DataFrame(technologies)
print(df)

if len(sys.argv) > 1:
    index = int(sys.argv[1])
else:
    index = 1

print('index:', index)

name = df.columns[index]
print('name:', name)

# --- version 1 ---

print('--- version 1 ---')
result = df.select([col for idx, col in enumerate(df.columns) if idx != index])
print(result)

#result = df.select(pl.exclude(pl.nth(1)))

# --- version 2 ---

print('--- version 2 ---')
#result = df.select(pl.exclude(df.columns[index]))
name = df.columns[index]
result = df.select(pl.exclude(name))
print(result)

# --- version 3 ---

print('--- version 3 ---')
#result = df.drop(df.columns[index])
name = df.columns[index]
result = df.drop(name)
print(result)

# --- version 4 ---

print('--- version 4 ---')
result = df.drop(pl.nth(index))
print(result)

# ---

#print(df)