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Python 情节:由多只colmuns熊猫分组_Python_Pandas_Matplotlib_Group By - Fatal编程技术网

Python 情节:由多只colmuns熊猫分组

Python 情节:由多只colmuns熊猫分组,python,pandas,matplotlib,group-by,Python,Pandas,Matplotlib,Group By,嗨,我有一个功能rmse: id date load Instant DayType Month Day sort 0 2013-02-02 4667.341595 46 6 2 2 4667.341595 1 2013-02-02 4620.702889 47 6 2 2 4620.702889 2 2013-05-12 -4439.33362

嗨,我有一个功能rmse:

id  date        load          Instant  DayType  Month  Day  sort
0   2013-02-02  4667.341595   46       6        2      2    4667.341595
1   2013-02-02  4620.702889   47       6        2      2    4620.702889
2   2013-05-12  -4439.333624  3        0        5      12   4439.333624
3   2013-05-12  -4409.947044  4        0        5      12   4409.947044
4   2013-05-12  -4369.322473  5        0        5      12   4369.322473

Load是我们的x,我想绘制每天每个瞬间的rmse返回值类型是指我想在x瞬间和y中绘制我的函数返回值和7个不同的绘图(每天类型的绘图)。

这就是你想要的吗

def RMSE2(x):
    return(np.sqrt(np.mean(x**2)))
输出:

df.groupby('DayType')['load'].apply(lambda x:np.sqrt(np.mean(x**2)))

请修改问题,最后一句话模棱两可。谢谢,我找到了解决方案。如果你找到了解决方案,请本着stackoverflow的精神在这里发布答案,这可能会帮助其他人。@suvy完成了我发布的答案
DayType
0    4406.294544
6    4644.080789
Name: load, dtype: float64
columns = np.unique(final.DayType)
b = []
for i in columns:
    daytype = final[final['DayType'] == i]
    a = daytype[['load']].groupby(daytype['Instant']).apply(RMSE2)
    b.append(a)


import matplotlib.pyplot as plt
plt.xticks(rotation=70)
fig_size = plt.rcParams["figure.figsize"]
fig_size[0] = 8
fig_size[1] = 8
for i in range(9):
    x = range(48)  


    y = b[i]['load']
    plt.plot(x,y)

    plt.ylabel("Rmse")
    plt.xlabel("Day")

    plt.legend(range(9),loc="upper left", bbox_to_anchor=[0, 1],
           ncol=2, shadow=True, title="Legend", fancybox=True)




plt.show()