python - 如何在函数中循环遍历 pandas 数据框中的列表

Question

这是我的数据框，

 df = pd.DataFrame({'Id': [102,103,104,303,305],'ExpG_Home':[1.8,1.5,1.6,1.8,2.9],
                  'ExpG_Away':[2.2,1.3,1.2,2.8,0.8],
                  'HomeG_Time':[[93, 109, 187],[169], [31, 159],[176],[16, 48, 66, 128]], 
                  'AwayG_Time':[[90, 177],[],[],[123,136],[40]]})

首先，我需要创建一个数组 y，对于给定的 ID 号，它从同一行(ExpG_Home & ExpG_Away)获取值。

y = [1 - (ExpG_Home + ExpG_Away), ExpG_Home, ExpG_Away]

其次，我发现这更难，对于用于创建 y 的 Id，下面的函数从 HomeG_Time & AwayG_Time 中获取相应的列表> 并创建一个数组。不幸的是，我的函数一次只占用一行。我需要为大型数据集执行此操作。

x1 = [1,0,0]    
x2 = [0,1,0]    
x3 = [0,0,1]    
total_timeslot = 200      # number of timeslot per game.
k = 1    # constant



#For Id=102 with ExpG_Home=2.2 and ExpG_Away=1.8
HomeG_Time = [93, 109, 187]  
AwayG_Time = [90, 177]
y = np.array([1-(2.2 + 1.8)/k, 2.2/k, 1.8/k])
  # output of y = [0.98 , 0.011, 0.009]

def squared_diff(x1, x2, x3, y):
    ssd = []
    for k in range(total_timeslot):  
        if k in HomeG_Time:
            ssd.append(sum((x2 - y) ** 2))
        elif k in AwayG_Time:
            ssd.append(sum((x3 - y) ** 2))
        else:
            ssd.append(sum((x1 - y) ** 2))
    return ssd

sum(squared_diff(x1, x2, x3, y))
Out[37]: 7.880400000000012

此输出仅适用于第一行。

Answer 1

这是给出的完整片段，

>>> import numpy as np
>>> x1 = np.array( [1,0,0] )
>>> x2 = np.array( [0,1,0] )
>>> x3 = np.array( [0,0,1] )
>>> total_timeslot = 200
>>> HomeG_Time = [93, 109, 187]
>>> AwayG_Time = [90, 177]
>>> ExpG_Home=2.2 
>>> ExpG_Away=1.8
>>> y = np.array( [1 - (ExpG_Home + ExpG_Away), ExpG_Home, ExpG_Away] )
>>> def squared_diff(x1, x2, x3, y):
...     ssd = []
...     for k in range(total_timeslot):  
...         if k in HomeG_Time:
...             ssd.append(sum((x2 - y) ** 2))
...         elif k in AwayG_Time:
...             ssd.append(sum((x3 - y) ** 2))
...         else:
...             ssd.append(sum((x1 - y) ** 2))
...     return ssd
... 
>>> sum(squared_diff(x1, x2, x3, y))
4765.599999999989

假设这个。使用 pandas.DataFrame.apply 将 y 计算为 (N,3)

>>> y = np.array( df.apply(lambda row: [1 - (row.ExpG_Home + row.ExpG_Away),
...                row.ExpG_Home, row.ExpG_Away ], 
...              axis=1).tolist() )
>>> y.shape
(5, 3)

现在计算给定 x 的平方误差

>>> def squared_diff(x, y):
...     return np.sum( np.square(x - y), axis=1)

在你的例子中，如果 error2 是 squared_diff(x2,y) 你添加的是 HomeG_Time 的出现次数

>>> n3 = df.AwayG_Time.apply(len)
>>> n2 = df.HomeG_Time.apply(len)
>>> n1 = 200 - (n2 + n3)

最终的误差平方和是(根据你的计算)

>>> squared_diff(x1, y) * n1 + squared_diff(x2, y) * n2 + squared_diff(x3, y) * n3
0    4766.4
1    2349.4
2    2354.4
3    6411.6
4    4496.2
dtype: float64
>>>