Subseting python dataframe using position values from lists

Question

I have a dataframe with raw data and I would like to select different range of rows for each column, using two different lists: one containing the first row position to select and the other the last.

INPUT

| Index    | Column A | Column B |
|:--------:|:--------:|:--------:|
|      1   |    2     |    8     |
|      2   |    4     |    9     |
|      3   |    1     |    7     |

first_position=[1,2]
last_position=[2,3]

EXPECTED OUTPUT

| Index    | Column A | Column B |
|:--------:|:--------:|:--------:|
|      1   |    2     |    9     |
|      2   |    4     |    7     |

Which function can I use? Thanks!

I tried df.filter but I think it does not accept list as input.

I´ll have a dataframe with multiple columns, and I have calculated somewhere else the start and end positions of the rows I want to subset per each column. This data has been store in two different lists (start_list, end_list). Probably I can solve this using a for loop... — Claudia Salat
– Claudia Salat, Commented Dec 1, 2022 at 22:09

alphamu · Accepted Answer · 2022-12-01 21:22:12Z

Basically, as far as I can see, you have two meaningful columns in your DataFrame.

Thus, I would suggest using "Index" column as the index indeed:

df.set_index(df.columns[0], inplace=True)

That way you might use .loc:

df_out = pd.concat(
    [
        df.loc[first_position, "Column A"].reset_index(drop=True),
        df.loc[last_position, "Column B"].reset_index(drop=True)
    ],
    axis=1
)

However, having indexes stored in separate lists you would need to watch them yourselves, which may be not too convenient.

Instead, I would re-organize it with slicing:

df_out = pd.concat(
    [
        df[["Column A"]][:-1].reset_index(drop=True),
        df[["Column B"]][1:].reset_index(drop=True)
    ],
    axis=1
)

In either cases, index is being destroyed. If that matters, then the scenario without .reset_index(drop=True) would be required.

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