mikeL mikeL - 29 days ago 13
Python Question

Get average counts per minute by hour

I have a dataframe with a time stamp as the index and a column of labels

df=DataFrame({'time':[ datetime(2015,11,2,4,41,10), datetime(2015,11,2,4,41,39), datetime(2015,11,2,4,41,47),
datetime(2015,11,2,4,41,59), datetime(2015,11,2,4,42,4), datetime(2015,11,2,4,42,11),
datetime(2015,11,2,4,42,15), datetime(2015,11,2,4,42,30), datetime(2015,11,2,4,42,39),
datetime(2015,11,2,4,42,41),datetime(2015,11,2,5,2,9),datetime(2015,11,2, 5,2,10),
datetime(2015,11,2,5,2,16),datetime(2015,11,2,5,2,29),datetime(2015,11,2, 5,2,51),
datetime(2015,11,2,5,9,1),datetime(2015,11,2,5,9,21),datetime(2015,11,2,5,9,31),
datetime(2015,11,2,5,9,40),datetime(2015,11,2,5,9,55)],
'Label':[2,0,0,0,1,0,0,1,1,1,1,3,0,0,3,0,1,0,1,1]}).set_index(['time'])


I want to get the avergae number of times that a label appears in a distinct minute
in a distnct hour.

For example, Label 0 appears 3 times in hour 4 in minute 41, 2 times in hour 4
in minute 42,

2 times in hour 5 in in minute 2, and 2 times in hour 5 in minute 9 so its average count per
minute in hour 4 is

(2+3)/2=2.5


and its count per minute in hour 5 is

(2+2)/2=2


The output I am looking for is

Hour 1
Label avg
0 2.5
1 2
2 .5
3 0


Hour 2
Label avg
0 2
1 1.5
2 0
3 1


What I have so far is

df['hour']=df.index.hour

hour_grp=df.groupby(['hour'], as_index=False)


then I can deo something like

res=[]
for key, value in hour_grp:
res.append(value)


then group by minute

res[0].groupby(pd.TimeGrouper('1Min'))['Label'].value_counts()


but this is where I'm stuck, not to mention it is not very efficient

Answer

Accessing minute of DateTimeIndex:

mn = df.index.minute

Accessing hour of DateTimeIndex:

hr = df.index.hour

Perform Groupby w.r.t to them and compute value_counts. unstack by filling missing values with 0 and groupby w.r.t minute and compute it's mean thereafter.

df.groupby([mn,hr])['Label'].value_counts().unstack(fill_value=0).groupby(level=1).mean()

Image

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