如果性能很重要，则对新行使用Index.repeat和DataFrame.loc，创建date列，计数器为GroupBy.cumcount，最后聚合sum：

df['VALID_FROM'] = pd.to_datetime(df['VALID_FROM'])
df['VALID_TO'] = pd.to_datetime(df['VALID_TO'])
df1 = df.loc[df.index.repeat(df['VALID_TO'].sub(df['VALID_FROM']).dt.days + 1)]
df1['dates'] = df1['VALID_FROM'] + pd.to_timedelta(df1.groupby(level=0).cumcount(),unit='d')
df1 = df1.groupby('dates', as_index=False)['VALUE'].sum()
print (df1)
       dates  VALUE
0 2022-01-01      7
1 2022-01-02     14
2 2022-01-03     12
3 2022-01-04     10
4 2022-01-05      3
5 2022-01-06      3

一种选择是构建一个日期列表，从原始 Dataframe 的最小值到最大值，使用带有conditional_join的非相等连接来获得匹配，最后使用groupby和sum：

# pip install pyjanitor
import pandas as pd
import janitor
# build the date pandas object:
dates = df.filter(like='VALID').to_numpy()
dates = pd.date_range(dates.min(), dates.max(), freq='1D')
dates = pd.Series(dates, name='dates')
# compute the inequality join between valid_from and valid_to, 
# followed by the aggregation on a groupby:
(df
.conditional_join(
    dates, 
    ('VALID_FROM', 'dates', '<='),
    ('VALID_TO','dates', '>='), 
    # if you have numba installed, 
    # it can improve performance
    use_numba=False, 
    df_columns='VALUE')
.groupby('dates')
.VALUE
.sum()
) 
dates
2022-01-01     7
2022-01-02    14
2022-01-03    12
2022-01-04    10
2022-01-05     3
2022-01-06     3
Name: VALUE, dtype: int64