如何解决计算同一熊猫列中两个不同值之间的时间
我有如下数据
Device Time Condition
D1 01/11/2019 00:00 issue
D1 01/11/2019 00:15 issue
D1 01/11/2019 00:30 issue
D1 01/11/2019 00:45 issue
D1 01/11/2019 01:00 issue
D1 01/11/2019 01:15 Resolved
D1 01/11/2019 01:30 Resolved
D2 01/11/2019 01:45 issue
D2 01/11/2019 02:00 Resolved
D1 01/11/2019 01:45 issue
D1 01/11/2019 02:00 Resolved
我需要创建一个新列，以查找第一个问题和第一个解决问题之间的时间。 我需要一个groupby声明，该声明将保留所有问题的第一个问题，并保留所有问题的第一个问题。然后找到时间-当我按设备分组并设置条件时，每个设备只保留了一个问题。
所需的输出如下所示
Device Time Condition durationTofix
D1 01/11/2019 00:00 issue
D1 01/11/2019 00:15 issue
D1 01/11/2019 00:30 issue
D1 01/11/2019 00:45 issue
D1 01/11/2019 01:00 issue
D1 01/11/2019 01:15 Resolved 01:15
D1 01/11/2019 01:30 Resolved
D2 01/11/2019 01:45 issue
D2 01/11/2019 02:00 Resolved 00:15
D1 01/11/2019 01:45 issue
D1 01/11/2019 02:00 Resolved 00:15
由于groupby Device and Condition不够，我认为无法创建索引列
data["index"] = data.groupby(['Device','condition']).??? #Something like cumcount() but it is not cumcount in this case
然后使用数据透视表进行时间计算
H = data.pivot_table(index=['index','Device'],columns=['condition'],values='Timestamp',aggfunc=lambda x: x)
H['durationTofix'] = H['Resolved']- H['issue']
解决方法
在每个Resolved之前的每个连续组中，Device之前始终至少存在一个问题：
#converting to datetimes
df['Time'] = pd.to_datetime(df['Time'])
#consetutive groups
g = df['Device'].ne(df['Device'].shift()).cumsum()
#test issue values
m = df['Condition'].eq('issue')
#replace not issue to missing values
i = df['Time'].where(m)
#get first duplicated rows by consecutive groups and condition column
mask = ~df.assign(g=g,i=i).duplicated(['g','Condition'])
#forward filling Time by first issue per groups
s = df['Time'].where(mask & m).groupby(g).ffill()
#subtract and filter only first Resolved per groups
df['durationTofix'] = df['Time'].sub(s).where(mask & df['Condition'].eq('Resolved'))
print (df)
Device Time Condition durationTofix
0 D1 2019-01-11 00:00:00 issue NaT
1 D1 2019-01-11 00:15:00 issue NaT
2 D1 2019-01-11 00:30:00 issue NaT
3 D1 2019-01-11 00:45:00 issue NaT
4 D1 2019-01-11 01:00:00 issue NaT
5 D1 2019-01-11 01:15:00 Resolved 01:15:00
6 D1 2019-01-11 01:30:00 Resolved NaT
7 D2 2019-01-11 01:45:00 issue NaT
8 D2 2019-01-11 02:00:00 Resolved 00:15:00
9 D1 2019-01-11 01:45:00 issue NaT
10 D1 2019-01-11 02:00:00 Resolved 00:15:00
最大的问题是如何正确分组/解决问题，这可以通过颠倒的cumsum完成：
df["Time"] = pd.to_datetime(df["Time"])
df["group"] = (df["Condition"].eq("Resolved")&df["Condition"].shift(-1).eq("issue"))[::-1].cumsum()[::-1]
df["diff"] = (df[~df.duplicated(["Condition","group"])].groupby("group")["Time"].transform(lambda d: d.diff()))
print (df)
Device Time Condition group diff
0 D1 2019-01-11 00:00:00 issue 2 NaT
1 D1 2019-01-11 00:15:00 issue 2 NaT
2 D1 2019-01-11 00:30:00 issue 2 NaT
3 D1 2019-01-11 00:45:00 issue 2 NaT
4 D1 2019-01-11 01:00:00 issue 2 NaT
5 D1 2019-01-11 01:15:00 Resolved 2 01:15:00
6 D1 2019-01-11 01:30:00 Resolved 2 NaT
7 D2 2019-01-11 01:45:00 issue 1 NaT
8 D2 2019-01-11 02:00:00 Resolved 1 00:15:00
9 D1 2019-01-11 01:45:00 issue 0 NaT
10 D1 2019-01-11 02:00:00 Resolved 0 00:15:00
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