如何按条件删除Pandas值并相应左移行内排名值

Hey there! Let's work through this problem together. First, I'll start with a sample DataFrame that aligns with your description—this will make it easier to see the solution in action.

Sample Input DataFrame #

Let's create a representative dataset:

import pandas as pd

data = {
    'status': ['in', 'out', 'in', 'out'],
    'A': [10, 20, 30, 40],
    'rank1': [1, 2, 3, 4],
    'rank2': [5, 6, 7, 8],
    'rank3': [9, 10, 11, 12]
}
df = pd.DataFrame(data)

This gives us:

Method 1: Efficient Vectorized Operation (Best for Large Data) #

If you're working with a large DataFrame, vectorized operations are way faster than row-wise loops. Here's how to directly update the relevant rows:

# Identify rows where status is 'out'
out_mask = df['status'] == 'out'

# Shift values left: A gets rank1, rank1 gets rank2, rank2 gets rank3, rank3 becomes NA
df.loc[out_mask, 'A'] = df.loc[out_mask, 'rank1']
df.loc[out_mask, 'rank1'] = df.loc[out_mask, 'rank2']
df.loc[out_mask, 'rank2'] = df.loc[out_mask, 'rank3']
df.loc[out_mask, 'rank3'] = pd.NA

Method 2: Flexible Row-wise Apply (Good for Dynamic Column Counts) #

If you have a variable number of rank columns and don't want to hardcode each shift, use apply to handle rows dynamically:

def process_row(row):
    if row['status'] == 'out':
        # Extract non-status columns, drop 'A', then shift left and pad with NA
        non_status_cols = row.drop('status')
        shifted_vals = non_status_cols.drop('A').tolist() + [pd.NA]
        # Reconstruct the row with status first, then shifted values
        return pd.Series([row['status']] + shifted_vals, index=row.index)
    return row

result_df = df.apply(process_row, axis=1)

Expected Output #

Either method will give you this result:

Notes #

• If your DataFrame has more rank columns, Method 2 will automatically handle the shift without needing to update code.
• Use Method 1 for performance if your dataset is large, as vectorized operations are optimized in pandas.

内容的提问来源于stack exchange，提问作者wflwo