Python 熊猫分块分组
我有一个数据集:Python 熊猫分块分组,python,pandas,dataset,Python,Pandas,Dataset,我有一个数据集: df = pd.DataFrame({ 'service': ['a', 'a', 'a', 'b', 'c', 'a', 'a'], 'status': ['problem', 'problem', 'ok', 'problem', 'ok', 'problem', 'ok'], 'created': [ datetime(2019, 1, 1, 1, 1, 0), datetime(2019, 1, 1, 1, 1,
df = pd.DataFrame({
'service': ['a', 'a', 'a', 'b', 'c', 'a', 'a'],
'status': ['problem', 'problem', 'ok', 'problem', 'ok', 'problem', 'ok'],
'created': [
datetime(2019, 1, 1, 1, 1, 0),
datetime(2019, 1, 1, 1, 1, 10),
datetime(2019, 1, 1, 1, 2, 0),
datetime(2019, 1, 1, 1, 3, 0),
datetime(2019, 1, 1, 1, 5, 0),
datetime(2019, 1, 1, 1, 10, 0),
datetime(2019, 1, 1, 1, 20, 0),
],
})
print(df.head(10))
service status created
0 a problem 2019-01-01 01:01:00 # -\
1 a problem 2019-01-01 01:01:10 # --> one group
2 a ok 2019-01-01 01:02:00 # -/
3 b problem 2019-01-01 01:03:00
4 c ok 2019-01-01 01:05:00
5 a problem 2019-01-01 01:10:00 # -\
6 a ok 2019-01-01 01:20:00 # - --> one group
service downtime_seconds
0 a 60 # `created` difference between 2 and 0
1 a 600 # `created` difference between 6 and 5
您可以看到a
服务更改状态问题->正常(0,2项;5,6项)。您还可以看到3
,4
项没有变化(只有1条记录-没有组/块)。我需要创建下一个数据集:
df = pd.DataFrame({
'service': ['a', 'a', 'a', 'b', 'c', 'a', 'a'],
'status': ['problem', 'problem', 'ok', 'problem', 'ok', 'problem', 'ok'],
'created': [
datetime(2019, 1, 1, 1, 1, 0),
datetime(2019, 1, 1, 1, 1, 10),
datetime(2019, 1, 1, 1, 2, 0),
datetime(2019, 1, 1, 1, 3, 0),
datetime(2019, 1, 1, 1, 5, 0),
datetime(2019, 1, 1, 1, 10, 0),
datetime(2019, 1, 1, 1, 20, 0),
],
})
print(df.head(10))
service status created
0 a problem 2019-01-01 01:01:00 # -\
1 a problem 2019-01-01 01:01:10 # --> one group
2 a ok 2019-01-01 01:02:00 # -/
3 b problem 2019-01-01 01:03:00
4 c ok 2019-01-01 01:05:00
5 a problem 2019-01-01 01:10:00 # -\
6 a ok 2019-01-01 01:20:00 # - --> one group
service downtime_seconds
0 a 60 # `created` difference between 2 and 0
1 a 600 # `created` difference between 6 and 5
我可以通过迭代
:
for i in range(len(df.index)):
# if df.loc[i]['status'] blablabla...
是否可以使用pandas
而无需迭代
?也许有更优雅的方法
谢谢。在您的情况下,我们需要通过颠倒顺序和cumsum
来创建groupby
键,然后我们只需要在分组之前过滤df,使用nunique
和transform
s=df.status.eq('ok').iloc[::-1].cumsum()
con=df.service.groupby(s).transform('nunique')==1
df_g=df[con].groupby(s).agg({'service':'first','created':lambda x : (x.iloc[-1]-x.iloc[0]).seconds})
Out[124]:
service created
status
1 a 600
3 a 60