Question

我有一个带有日期时间间隔的DataFrame，如下所示：

   id            start_date              end_date
1   1   2016-10-01 00:00:00   2016-10-01 03:00:00
2   1   2016-10-03 05:30:00   2016-10-03 06:30:00
3   2   2016-10-03 23:30:00   2016-10-04 01:00:00  # This line should be splitted
4   1   2016-10-04 05:00:00   2016-10-04 06:00:00
5   2   2016-10-04 05:50:00   2016-10-04 06:00:00
6   1   2016-10-05 18:50:00   2016-10-06 02:00:00  # This one too
....

我想“分割”超过一天的间隔，以确保每一行都在同一天：

     id            start_date              end_date
1     1   2016-10-01 00:00:00   2016-10-01 03:00:00
2     1   2016-10-03 05:30:00   2016-10-03 06:30:00
3     2   2016-10-03 23:30:00   2016-10-03 23:59:59 # Splitted
4     2   2016-10-04 00:00:00   2016-10-04 01:00:00 # Splitted
5     1   2016-10-04 05:00:00   2016-10-04 06:00:00
6     2   2016-10-04 05:50:00   2016-10-04 06:00:00
7     1   2016-10-05 18:50:00   2016-10-05 23:59:59 # Splitted
8     1   2016-10-06 00:00:00   2016-10-06 02:00:00 # Splitted
....

Answer 1

您可以使用.dt accessor创建执行更新的布尔索引，然后相应地进行调整：

# Get the rows to split.
split_rows = (df['start_date'].dt.date != df['end_date'].dt.date)

# Get the new rows to append, adjusting the start_date to the next day.
new_rows = df[split_rows].copy()
new_rows['start_date'] = new_rows['start_date'].dt.date + pd.DateOffset(days=1)

# Adjust the end_date of the existing rows.
df.loc[split_rows, 'end_date'] = df.loc[split_rows, 'start_date'].dt.date + pd.DateOffset(days=1, seconds=-1)

# Append the new rows to the existing dataframe.
df = df.append(new_rows).sort_index().reset_index(drop=True)

上述过程假设start_date和end_date之间的日期差异只有一天。如果有可能存在多天跨度，您可以将上述过程包装在while循环中：

# Get the rows to split.
split_rows = (df['start_date'].dt.date != df['end_date'].dt.date)

while split_rows.any():
    # Get the new rows, adjusting the start_date to the next day.
    new_rows = df[split_rows].copy()
    new_rows['start_date'] = new_rows['start_date'].dt.date + pd.DateOffset(days=1)

    # Adjust the end_date of the existing rows.
    df.loc[split_rows, 'end_date'] = df.loc[split_rows, 'start_date'].dt.date + pd.DateOffset(days=1, seconds=-1)

    # Append the new rows to the existing dataframe.
    df = df.append(new_rows).sort_index().reset_index(drop=True)

    # Get new rows to split (if the start_date to end_date span is more than 1 day).
    split_rows = (df['start_date'].dt.date != df['end_date'].dt.date)

样本数据的结果输出：

   id          start_date            end_date
0   1 2016-10-01 00:00:00 2016-10-01 03:00:00
1   1 2016-10-03 05:30:00 2016-10-03 06:30:00
2   2 2016-10-03 23:30:00 2016-10-03 23:59:59
3   2 2016-10-04 00:00:00 2016-10-04 01:00:00
4   1 2016-10-04 05:00:00 2016-10-04 06:00:00
5   2 2016-10-04 05:50:00 2016-10-04 06:00:00
6   1 2016-10-05 18:50:00 2016-10-05 23:59:59
7   1 2016-10-06 00:00:00 2016-10-06 02:00:00

Answer 2

这有效：

def date_split(row):
    starts = pd.Series(pd.date_range(row['start_date'].date(),
                                     periods=row['diff']+1, freq='D'))
    starts[0] = row['start_date']
    ends = starts[1:] - pd.to_timedelta(1, unit='s')
    ends.loc[len(ends)+1] = row['end_date']
    ends.reset_index(drop=True, inplace=True)

    ret = pd.concat([starts, ends], axis=1, keys=['start_date', 'end_date'])
    ret['id'] = row['id']
    return ret

df['diff'] = df['end_date'].dt.day - df['start_date'].dt.day

req = pd.concat([df[df['diff'] == 0]] +\
                [date_split(row) for _, row in df[df['diff'] > 0].iterrows()])
req = req.drop('diff', axis=1).reset_index(drop=True)
req

请注意，这是一般方法，可以处理两者之间的任意天数。只有你的指数位置会有所不同。

熊猫：按天打破日期时间间隔

2 个答案: