使用日期列 pandas 数据框创建额外的行

Question

假设我有以下数据：

ID  Leave Type  Start Date    End Date
1   Sick        2022-01-01    2022-01-01
1   Holiday     2023-03-28  
2   Holiday     2023-01-01    2023-01-02
3   Work        2023-01-01    2023-01-01

我需要找到一种方法来确认 Start Date 和 End Date 具有相同的值。如果不是，它需要计算结束日期提前的天数，并为每一天创建一个行，添加 1 天并始终匹配开始日期和结束日期。如果结束日期为空，它应该创建行直到到达 2023-03-30 为止。这种方式产生了这个数据：

ID  Leave Type  Start Date    End Date
1   Sick        2022-01-01    2022-01-01
1   Holiday     2023-03-28    2023-03-28
1   Holiday     2023-03-29    2023-03-29
1   Holiday     2023-03-30    2023-03-30
2   Holiday     2023-01-01    2023-01-01
2   Holiday     2023-01-02    2023-01-02
3   Work        2023-01-01    2023-01-01

谢谢！

Answer 1

假设您在输出中错误地粘贴了额外的一行（第 5 行）。你也可以试试这个：

import pandas as pd
from datetime import timedelta, datetime

# create the dataframe
df = pd.DataFrame({'ID': [1, 1, 2, 3], 
                   'Leave Type': ['Sick', 'Holiday', 'Holiday', 'Work'], 
                   'Start Date': ['2022-01-01', '2023-03-28', '2023-01-01', '2023-01-01'], 
                   'End Date': ['2022-01-01', '', '2023-01-02', '2023-01-01']})

# convert date columns to datetime format
df['Start Date'] = pd.to_datetime(df['Start Date'])
df['End Date'] = pd.to_datetime(df['End Date'])

# fill in blank end dates with a maximum date value
df['End Date'] = df['End Date'].fillna(datetime(2023, 3, 30))

# create an empty list to store new rows
create_rows = []

# loop through each row in the dataframe
for index, row in df.iterrows():
    
    # if the start and end dates are not the same, add rows for each day in between
    if row['Start Date'] != row['End Date']:
        delta = row['End Date'] - row['Start Date']
        for i in range(delta.days + 1):
            date = row['Start Date'] + timedelta(days=i)
            create_rows.append({'ID': row['ID'], 'Leave Type': row['Leave Type'], 'Start Date': date, 'End Date': date})
    
    # if the start and end dates are the same, append the original row
    else:
        create_rows.append({'ID': row['ID'], 'Leave Type': row['Leave Type'], 'Start Date': row['Start Date'], 'End Date': row['End Date']})
    
# create a new dataframe with the original rows and the new rows
output_df = pd.DataFrame(create_rows)

# sort the dataframe by ID and Start Date
output_df = output_df.sort_values(['ID', 'Start Date'])

# reset the index
output_df = output_df.reset_index(drop=True)

print(output_df)

Answer 2

您可以使用：

# ensure datetime and fill NA with default date
df[['Start Date', 'End Date']] = df[['Start Date', 'End Date']].apply(pd.to_datetime)
df['End Date'] = df['End Date'].fillna('2023-03-30')

# repeat index and create output
idx = df.index.repeat(df['End Date'].sub(df['Start Date']).dt.days.add(1))
out = df.loc[idx]

# increment days
out['Start Date'] += pd.TimedeltaIndex(out.groupby(level=0).cumcount(), unit='D')
out['End Date'] = out['Start Date']

输出：

   ID Leave Type Start Date   End Date
0   1       Sick 2022-01-01 2022-01-01
1   1    Holiday 2023-03-28 2023-03-28
1   1    Holiday 2023-03-29 2023-03-29
1   1    Holiday 2023-03-30 2023-03-30
2   2    Holiday 2023-01-01 2023-01-01
2   2    Holiday 2023-01-02 2023-01-02
3   3       Work 2023-01-01 2023-01-01

可重复输入：

df = pd.DataFrame({'ID': [1, 1, 2, 3],
                   'Leave Type': ['Sick', 'Holiday', 'Holiday', 'Work'],
                   'Start Date': ['2022-01-01', '2023-03-28', '2023-01-01', '2023-01-01'],
                   'End Date': ['2022-01-01', None, '2023-01-02', '2023-01-01']})

Answer 3

import pandas as pd
from pandas.tseries.offsets import MonthEnd

df = pd.DataFrame({'ID': ['Sick', 'Holiday', 'Holiday', 'Work'],
                   'Leave Type': ['2022-01-01', '2023-03-28', '2023-01-01', '2023-01-01'],
                   'Start Date': ['2022-01-01', '', '2023-01-02', '2023-01-01'],
                   })
# Converts columns 'Leave Type' and 'Start Date' to datetime
df[['Leave Type', 'Start Date']] = \
    df[['Leave Type', 'Start Date']].apply(pd.to_datetime, errors='coerce')
# Fill NaT values with the last day of the month
df['Start Date'] = df['Start Date'].fillna(df['Leave Type'] + MonthEnd(0))
# Replace 'Start Date' values with list of date ranges
df['Start Date'] = \
    [pd.date_range(s, e, freq='D').tolist() for s,e in zip(df['Leave Type'], df['Start Date'])]
# Explode the list
df = df.explode('Start Date')

print(df)

结果

        ID Leave Type Start Date
0     Sick 2022-01-01 2022-01-01
1  Holiday 2023-03-28 2023-03-28
1  Holiday 2023-03-28 2023-03-29
1  Holiday 2023-03-28 2023-03-30
1  Holiday 2023-03-28 2023-03-31
2  Holiday 2023-01-01 2023-01-01
2  Holiday 2023-01-01 2023-01-02
3     Work 2023-01-01 2023-01-01