基于出现频率的概率预测

Probabilistic prediction based on occurrence frequency

我有一个 2011-2013 年降雨量的时间序列,其中降雨量数据为 1(无雨)和 0(有雨)格式。原始数据间隔为 1 小时,每天上午 10 点至下午 3 点。我不想预测 2014 年的降雨量,但我想根据降雨量列中出现 1 或 0 来预测同一时间间隔全年降雨的可能性。目前,我使用以下代码通过计算出现 1 次或 0 次来预测下雨的可能性:

import pandas as pd
 
b = {'year':[2011,2011,2011,2011,2011,2011,2011,2011,2011,2011,2011,2011,
             2012,2012,2012,2012,2012,2012,2012,2012,2012,2012,2012,2012,
             2013,2013,2013,2013,2013,2013,2013,2013,2013,2013,2013,2013],
     'month': [1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12],
     'rain':[1,0,0,0,1,1,0,1,1,0,0,1,0,0,1,0,0,0,1,1,1,1,1,0,0,1,1,0,1,0,1,0,1,0,1,0]}

b = pd.DataFrame(b,columns = ['year','month','rain'])

def X(b):
    if (b['month'] == 1):
        return 'Jan'
    elif (b['month']==2):
        return 'Feb'
    elif (b['month']==3):
        return 'Mar'
    elif (b['month']==4):
        return 'Apr'
    elif (b['month']==5):
        return 'May'
    elif (b['month']==6):
        return 'Jun'
    elif (b['month']==7):
        return 'Jul'
    elif (b['month']==8):
        return 'Aug'
    elif (b['month']==9):
        return 'Sep'
    elif (b['month']==10):
        return 'Oct'
    elif (b['month']==11):
        return 'Nov'
    elif (b['month']==12):
        return 'Dec' 

b['X'] = b.apply(X,axis =1)

mask_x = (b['X']=='Jul')

mask_y = b['rain'].loc[mask_x]

mask_y.value_counts()

我认为这种方法不适用于大型数据集,有人可以建议我一种有效且稳健的方法来预测此类数据集的降雨量。

数据是通过每小时随机选择 [0,1] 创建的。我们通过在日期列中按时间分组来计算总数和案例数。现在您可以通过 total/number 个事件来计算降雨率。我正在按照您的代码创建年、月和月的缩写名称,但这并不是真正必要的。

import pandas as pd
import numpy as np
import random

random.seed(20200817)

date_rng = pd.date_range('2013-01-01', '2016-01-01', freq='1H')
rain = random.choices([0,1], k=len(date_rng))
b = pd.DataFrame({'date':pd.to_datetime(date_rng), 'rain':rain})

hour_rain = b.groupby([b.date.dt.month, b.date.dt.day, b.date.dt.hour])['rain'].agg([sum,np.size])
hour_rain.index.names = ['month','day','hour']

hour_rain.reset_index()

month   day hour    sum size
0   1   1   0   0   4
1   1   1   1   2   3
2   1   1   2   3   3
3   1   1   3   1   3
4   1   1   4   1   3
... ... ... ... ... ...
8755    12  31  19  2   3
8756    12  31  20  2   3
8757    12  31  21  2   3
8758    12  31  22  0   3
8759    12  31  23  0   3

我正在尝试执行的操作如下所示:

import pandas as pd
import numpy as np
import random

random.seed(20200817)
date_rng = pd.date_range('2013-01-01', '2015-12-31', freq='1H')
rain = random.choices([0,1], k=len(date_rng))
b = pd.DataFrame({'date':pd.to_datetime(date_rng), 'rain':rain})
b['year'] = b['date'].dt.year
b['month'] = b['date'].dt.month
b['day'] = b['date'].dt.day
b['hour'] = b['date'].dt.hour
b['X'] = b['date'].dt.strftime('%b')

b['hour']= b['hour'].astype(str).str.zfill(2)
b['day']= b['day'].astype(str).str.zfill(2)


# Joint the Month, Date, Hour and Minute together
b['var'] = b['X']+b['day'].astype(str)+b['hour'].astype(str)


cols = b.columns.tolist()
cols = cols[-1:] + cols[:-1]
b = b[cols]


# drop the unwanted columns
b = b.drop(["date","month","X","hour","day","year"], axis=1)


# now lets say I wanna predict 20 January 15.00 chance of rain

mask_x = (b['var']=='Jan2015')

mask_y = b['rain'].loc[mask_x]

mask_y.value_counts()

output:
0    2
1    1

# means the chance of rain is 33.33% and no chance of rain is 66.67% 

当我对大型数据集(超过 20 年)执行此操作时,我觉得效果不是很好。