熊貓：休息日期時間間隔由

Question

天

我有日期時間間隔數據框，像這樣的：

 
    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 
.... 

我想覆蓋超過一天的「分裂」的時間間隔，以確保每個行落在同一天：

 
    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

可以使用.dt accessor創建的地方進行更新布爾索引，然後做出相應的調整：

# 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) 

上面的過程假設在start_date和end_date跨度之間的日期差異之間僅有一天。如果有可能，有多天的跨度，你可以用上面的過程中while循環：

# 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) 

從您的樣本數據的輸出產生：

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

這工作：

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 

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請注意，這是一般性的方法，並且可以處理兩者之間的任何天數。只有您的索引位置會有所不同。