使用apply with python每月或每小时使用RMSE

Question

我有这个数据框：

dates,AA,BB,CC
2018-01-01 00:00:00,45.73,47.63,3.45625
2018-01-01 01:00:00,44.16,44.42,3.45625
2018-01-01 02:00:00,42.24,42.34,3.45625
2018-01-01 03:00:00,39.29,38.36,3.45625
2018-01-01 04:00:00,36,36.87,3.45625
2018-01-01 05:00:00,41.99,39.79,3.45625
2018-01-01 06:00:00,42.25,42.08,3.45625
2018-01-01 07:00:00,44.97,51.19,3.45625
2018-01-01 08:00:00,45,59.69,3.45625
2018-01-01 09:00:00,44.94,56.67,3.45625
2018-01-01 10:00:00,45.04,53.54,3.45625
2018-01-01 11:00:00,46.67,52.6,3.45625
2018-01-01 12:00:00,46.99,50.77,3.45625
2018-01-01 13:00:00,44.16,50.27,3.45625
2018-01-01 14:00:00,45.26,50.64,3.45625
2018-01-01 15:00:00,47.84,54.79,3.45625
2018-01-01 16:00:00,50.1,60.17,3.45625
2018-01-01 17:00:00,54.3,59.47,3.45625
2018-01-01 18:00:00,51.91,60.16,3.45625
2018-01-01 19:00:00,51.38,70.81,3.45625
2018-01-01 20:00:00,49.2,62.65,3.45625
2018-01-01 21:00:00,45.73,59.71,3.45625
2018-01-01 22:00:00,44.84,50.96,3.45625
2018-01-01 23:00:00,38.11,46.52,3.45625
2018-01-02 00:00:00,19.19,49.62,3.405
2018-01-02 01:00:00,14.99,45.05,3.405
2018-01-02 02:00:00,11,45.18,3.405
2018-01-02 03:00:00,10,37.12,3.405
2018-01-02 04:00:00,11.83,38.03,3.405
2018-01-02 05:00:00,14.99,46.17,3.405
2018-01-02 06:00:00,40.6,51.71,3.405
2018-01-02 07:00:00,46.99,54.37,3.405
2018-01-02 08:00:00,47.95,75.3,3.405
2018-01-02 09:00:00,49.9,68.48,3.405
2018-01-02 10:00:00,50,61.94,3.405
2018-01-02 11:00:00,49.7,63.26,3.405
2018-01-02 12:00:00,48.16,59.41,3.405
2018-01-02 13:00:00,47.24,60,3.405
2018-01-02 14:00:00,46.1,67.44,3.405
2018-01-02 15:00:00,47.6,66.82,3.405
2018-01-02 16:00:00,50.45,72.17,3.405
2018-01-02 17:00:00,54.9,70.28,3.405
2018-01-02 18:00:00,57.18,62.63,3.405

基本上，从2018-01-01到2018-12-31的每小时日期

我想用apply方法或同等方法做不同的事情。 首先，我想用AA作为参考解，计算BB和CC之间的月尺度RMSE（均方根误差）。 我的做法如下：

dfr = dfr.assign(month=lambda x: x.index.month).groupby('month')
rmseBB = dfr.apply(rmse,   s1='AA',s2='BB')
rmseCC = dfr.apply(rmse,   s1='AA',s2='CC')

这里是rmse函数：

def rmse(group,s1,s2):
    if len(group) == 0:
        return np.nan
    s = (group[s1] - group[s2]).pow(2).sum()
    print(len(group))
    rmseO = np.sqrt(s / len(group)) 
    return rmseO 

根据给定的结果，前面的步骤似乎工作正常

除此之外，我想做一些更复杂的事情，至少根据我的实际知识

我想计算属于同一个月的每个小时的RMSE。我的意思是一月的第一个小时有一个RMSE，一月的第二个小时有一个RMSE，依此类推。这意味着每个月的RMSE值为24。之后，我可以计算每个月的平均每小时RMSE。更重要的是，我希望能够选择每小时平均RMSE中考虑的时间。

这将意味着一种双分组，每月和每小时。我错了吗

我希望我已经讲清楚了

谢谢你的帮助

迭戈

Answer 1

如果你想使用group by，你可以像下面这样做

import pandas as pd
def rmse(df_time):
    RMSE_BB=pow(pow(df_time['AA']-df_time['BB'],2).mean(),0.5)
    RMSE_CC=pow(pow(df_time['AA']-df_time['CC'],2).mean(),0.5)
    return RMSE_BB,RMSE_CC

df=pd.read_csv("Dates.csv")
df['month']=df['dates'].apply(lambda x:x.split()[0][5:7])
df['time']=df['dates'].apply(lambda x: x.split()[1][0:2])
df1=df.groupby(['month','time']).apply(rmse)
df1

Answer 2

你可以按照下面的方法做

import pandas as pd
df=pd.read_csv("Dates.csv")
year=['01','02','03','04','05','06','07','08','09','10','11','12']
time=list('0'+str(x) for x in range(10))+list(str(x) for x in range(11,24))
for i in year:
    df_mon=df[(df['dates'].apply(lambda x:x.split()[0][5:7])==i)]
    if len(df_mon)==0:
        continue
    for j in time:
        df_time=df_mon[(df_mon['dates'].apply(lambda x:x.split()[1][0:2])==j)]
        RMSE_BB=pow(pow(df_time['AA']-df_time['BB'],2).mean(),0.5)
        RMSE_CC=pow(pow(df_time['AA']-df_time['CC'],2).mean(),0.5)
        print(i,j,RMSE_BB, RMSE_CC)