我有一个看起来像这样的数据框:
public interface BluetoothMessageListener {
void onMessageReceived(String message);
}
我想找出编号。用户提供的天数之间存在差距,因此我希望每个用户的每一行都有一列,并且我的数据框应如下所示:
UserId Date_watched Days_not_watch
1 2010-09-11 5
1 2010-10-01 8
1 2010-10-28 1
2 2010-05-06 12
2 2010-05-18 5
3 2010-08-09 10
3 2010-09-25 5
我已经在数据框的列名称旁边提到了用于计算Gap的公式。
答案 0 :(得分:2)
这是使用groupby
+ shift
的一种方法:
# sort by date first
df['Date_watched'] = pd.to_datetime(df['Date_watched'])
df = df.sort_values(['UserId', 'Date_watched'])
# calculate groupwise start dates, shifted
grp = df.groupby('UserId')
starts = grp['Date_watched'].shift() + \
pd.to_timedelta(grp['Days_not_watch'].shift(), unit='d')
# calculate timedelta gaps
df['Gap'] = (df['Date_watched'] - starts).fillna(pd.Timedelta(0))
# convert to days and then integers
df['Gap'] = (df['Gap'] / pd.Timedelta('1 day')).astype(int)
print(df)
UserId Date_watched Days_not_watch Gap
0 1 2010-09-11 5 0
1 1 2010-10-01 8 15
2 1 2010-10-28 1 19
3 2 2010-05-06 12 0
4 2 2010-05-18 5 0
5 3 2010-08-09 10 0
6 3 2010-09-25 5 37