Question

我正在分析随时间推移的PGA巡回赛数据。出于机器学习的目的，我希望列数据可以代表几周的统计数据。以下是原始数据结构的示例。

import pandas as pd
import numpy as np

data = {'Player Name':['Tiger','Tiger','Tiger','Tiger','Tiger','Tiger','Jack',
                       'Jack','Jack','Jack','Jack','Jack','Jack'], 
        'Date':[1, 2, 4, 6, 7, 9, 1, 3, 4, 6, 9, 10, 11],
        'SG Total':[13, 2, 14, 6, 8, 1, 1, 3, 8, 4, 9, 2, 1]}

df_original = pd.DataFrame(data)

我想以以下格式获取数据。

data = {'Player Name':['Tiger','Tiger','Tiger','Jack','Jack',
                   'Jack','Jack'], 
    'Date':[6, 7, 9, 6, 9, 10, 11],
    'SG Total (Date t-3)':[13, 2, 14, 1, 3, 8, 4],
    'SG Total (Date t-2)':[2, 14, 6, 3, 8, 4, 9],
    'SG Total (Date t-1)':[14, 6, 8, 8, 4, 9, 2],
    'SG Total (Date y)':  [6, 8, 1, 4, 9, 2, 1]}
df_correct = pd.DataFrame(data)

在我使用的真实数据集中，我大约有1000列。因此，新的所需数据集将可能具有4000列。正如您在所需数据集中看到的那样，我删除了每个玩家的前三周。我使用个人的前3周来填写（t-3），（t-2）和（t-1）

，因此我从该数据的第4周开始输入日期

最初，我每周都创建一个数据集，无论玩家是否玩过，并使用此代码创建所需的DataFrame。

#%% Creates weekly dataframes & predictions dataframes

#Creates dataframes of each week
dict_of_weeks = {}

for i in range(1,df_numeric_combined['Date'].nunique()+1):
    dict_of_weeks['Week_{}_df'.format(i)] = df_numeric_combined[df_numeric_combined['Date'] == i]
    dict_of_weeks['Week_{}_df'.format(i)].columns += ' (Week ' + str(i) + ')'
    dict_of_weeks['Week_{}_df'.format(i)].rename(columns={'Player Name (Week ' + str(i) + ')' : 'Player Name' , 'Date (Week ' + str(i) + ')' : 'Date'},inplace=True)


#Creating dataframes for prediction of each week
import functools

dict_of_predictions = {}

df_weeks = []

for i in range(4,df_numeric_combined['Date'].nunique()+1):
    dfs = [dict_of_weeks['Week_'+str(i-3)+'_df'], dict_of_weeks['Week_'+str(i-2)+'_df'], dict_of_weeks['Week_'+str(i-1)+'_df'], dict_of_weeks['Week_'+str(i)+'_df']]

    dict_of_predictions['Week_{}_predictions'.format(i)] = functools.reduce(lambda left,right: pd.merge(left,right,on=['Player Name'], how='outer'), dfs)

    cols = []
    count = 1
    for column in dict_of_predictions['Week_{}_predictions'.format(i)].columns:
        if column == 'Date_y':
            cols.append('Date_y_'+ str(count))
            count+=1
            continue
        cols.append(column)

    dict_of_predictions['Week_{}_predictions'.format(i)].columns = cols

    dict_of_predictions['Week_{}_predictions'.format(i)].drop(columns = ['Date_x', 'Date_y_1'],inplace = True)

    dict_of_predictions['Week_{}_predictions'.format(i)].rename(columns={'Date_y_2':'Date'},inplace=True)

    dict_of_predictions['Week_{}_predictions'.format(i)].columns = dict_of_predictions['Week_{}_predictions'.format(i)].columns.str.replace('(Week ' + str(i-3)+ ')', 'Week t-3').str.replace('(Week ' + str(i-2)+ ')', 'Week t-2').str.replace('(Week ' + str(i-1)+ ')', 'Week t-1').str.replace('(Week ' + str(i)+ ')', 'Week y')

    df_weeks.append(dict_of_predictions['Week_{}_predictions'.format(i)])

#Combines predictions dataframes
df = pd.concat(dict_of_predictions.values(), axis=0, join='inner')

然而，我创建的此代码仅在玩家连续玩了几周的情况下才有效，因为它取决于周数并减去3、2和1。

最终目标是以df_correct格式获取数据。

谢谢！

Answer 1

如果我正确理解了您的要求，则可以在shift的排序数据框中使用groupby，以完成每个玩家的previous周成绩：


## Sort first by player and date
df_corrected = df_original.sort_values(['Player Name','Date'])

your_columns = ['SG Total'] ## name your 4000 columns here

for col in your_columns:
    for s in [3,2,1,0]: ### time lapses
        df_corrected[f'{col} (Date t-{s})'] = df_corrected.groupby('Player Name')[col].shift(s)

df_corrected.drop(your_columns, axis=1, inplace=True)

哪个输出

Out[12]: 
   Player Name  Date  SG Total (Date t-3)  SG Total (Date t-2)  \
6         Jack     1                  NaN                  NaN   
7         Jack     3                  NaN                  NaN   
8         Jack     4                  NaN                  1.0   
9         Jack     6                  1.0                  3.0   
10        Jack     9                  3.0                  8.0   
11        Jack    10                  8.0                  4.0   
12        Jack    11                  4.0                  9.0   
0        Tiger     1                  NaN                  NaN   
1        Tiger     2                  NaN                  NaN   
2        Tiger     4                  NaN                 13.0   
3        Tiger     6                 13.0                  2.0   
4        Tiger     7                  2.0                 14.0   
5        Tiger     9                 14.0                  6.0   

    SG Total (Date t-1)  SG Total (Date t-0)  
6                   NaN                    1  
7                   1.0                    3  
8                   3.0                    8  
9                   8.0                    4  
10                  4.0                    9  
11                  9.0                    2  
12                  2.0                    1  
0                   NaN                   13  
1                  13.0                    2  
2                   2.0                   14  
3                  14.0                    6  
4                   6.0                    8  
5                   8.0                    1

Python行到时间序列列

1 个答案: