我有这个人。 dataframe(每小时时间戳索引):
relative_humidity condition fid
2017-08-02 10:00:00 0.49 Chance of a Thunderstorm 1
2017-08-02 11:00:00 0.50 Chance of a Thunderstorm 1
2017-08-02 12:00:00 0.54 Partly Cloudy 1
2017-08-02 13:00:00 0.58 Partly Cloudy 2
2017-08-02 14:00:00 0.68 Partly Cloudy 2
如何计算每天最常出现的情况,并将其放在以日期为索引的数据框中。还需要按fid
分隔?
我试过了:
df.groupby(['fid', pd.Grouper(freq='D')])['condition']
答案 0 :(得分:2)
index[0]
需要value_counts
,因为数据已排序且第一个值为top:
d = {'level_1':'date'}
df1 = df.groupby(['fid', pd.Grouper(freq='D')])['condition'] \
.apply(lambda x: x.value_counts().index[0]).reset_index().rename(columns=d)
print (df1)
fid date condition
0 1 2017-08-02 Chance of a Thunderstorm
1 2 2017-08-02 Partly Cloudy
答案 1 :(得分:1)
df.groupby(['fid',pd.Grouper(freq='D'),'condition']).size().groupby(level=[0,1]).head(1)
输出:
fid condition
1 2017-08-02 Chance of a Thunderstorm 2
2 2017-08-02 Partly Cloudy 2
dtype: int64