直到满足条件的两个日期之间的天数差 [英] Difference of days between two dates until a condition is met
问题描述
在找到条件之前，我很难计算连续的天数.
下表给出了Gap done是我用解决方案表
I have got trouble to make to compute the number of days in a row until a condition is found.
It is given in the following table were Gap done is the messy table I obtained with the solution form there , and Expected gap the output I want to obtain.
+--------+------------+---------------------+----------+----------------------------------------------------------------------------------------------+
| Player | Result | Date | Gap done | Expected Gap |
+--------+------------+---------------------+----------+----------------------------------------------------------------------------------------------+
| K2000 | Lose | 2015-11-13 13:42:00 | Nan | Nan/0 |
| K2000 | Lose | 2016-03-23 16:40:00 | 131.0 | 131.0 |
| K2000 | Lose | 2016-05-16 19:17:00 | 54.0 | 185.0 |
| K2000 | Win | 2016-06-09 19:36:00 | 54.0 | 239.0 #he always lose before |
| K2000 | Win | 2016-06-30 14:05:00 | 54.0 | 54.0 #because he won last time, it's 54 days btw this current date and the last time he won. |
| K2000 | Lose | 2016-07-29 16:20:00 | 29.0 | 29.0 |
| K2000 | Win | 2016-10-08 17:48:00 | 29.0 | 58.0 |
| Kssis | Lose | 2007-02-25 15:05:00 | Nan | Nan/0 |
| Kssis | Lose | 2007-04-25 16:07:00 | 59.0 | 59.0 |
| Kssis | Not ranked | 2007-06-01 16:54:00 | 37.0 | 96.0 |
| Kssis | Lose | 2007-09-09 14:33:00 | 99.0 | 195.0 |
| Kssis | Lose | 2008-04-06 16:27:00 | 210.0 | 405.0 |
+--------+------------+---------------------+----------+----------------------------------------------------------------------------------------------+
解决方案的问题有并不能真正计算日期.在此示例中，日期始终可能相隔1天.
The issue of the solution there is it does not really compute date. It has the chance that date in this example are always separate by 1 day.
确定我适应了
def sum_days_in_row_with_condition(g):
sorted_g = g.sort_values(by='date', ascending=True)
condition = sorted_g['Result'] == 'Win'
sorted_g['days-in-a-row'] = g.date.diff().dt.days.where(~condition).ffill()
return sorted_g
但是，正如我向您展示的那样，这很混乱.
But as I showed you, this is messy.
所以我考虑了一个解决方案，但是它需要全局变量(功能失效)，这有点讲究.
So I thought about a solution, but it needs global variables (out of function), and that's a little fastidious.
任何人都可以以更简单的方式帮助解决此问题吗?
Can anyone help to solve this problematic in a simpler way ?
Pandas版本:0.23.4 Python版本:3.7.4
Pandas version: 0.23.4 Python version: 3.7.4
推荐答案
IIUC，您需要找到布尔掩码m1，其中win的前一行也是win.在m1中创建一个groupID s以分隔组win.将它们分成组并cumsum
IIUC, you need to find the boolean mask m1 where win has previous row also win. From m1 create a groupID s to separate group win. Split them into group and cumsum
m = df.Result.eq('Win')
m1 = m & m.shift()
s = m1.ne(m1.shift()).cumsum()
df['Expected Gap'] = df.groupby(['Player', s])['Gap done'].cumsum()
Out[808]:
Player Result Date Gap done Expected Gap
0 K2000 Lose 2015-11-13 13:42:00 NaN NaN
1 K2000 Lose 2016-03-23 16:40:00 131.0 131.0
2 K2000 Lose 2016-05-16 19:17:00 54.0 185.0
3 K2000 Win 2016-06-09 19:36:00 54.0 239.0
4 K2000 Win 2016-06-30 14:05:00 54.0 54.0
5 K2000 Lose 2016-07-29 16:20:00 29.0 29.0
6 K2000 Win 2016-10-08 17:48:00 29.0 58.0
7 Kssis Lose 2007-02-25 15:05:00 NaN NaN
8 Kssis Lose 2007-04-25 6:07:00 59.0 59.0
9 Kssis Not-ranked 2007-06-01 16:54:00 37.0 96.0
10 Kssis Lose 2007-09-09 14:33:00 99.0 195.0
11 Kssis Lose 2008-04-06 16:27:00 210.0 405.0
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