python - 如何从 pandas 数据框调用值到函数？

Question

我有一个包含两列的大型数据框和一个从每一行获取值并遍历数据框的函数。下面是数据框的头部。

xG_Team1  xG_Team2
0  1.440539  1.380095
1  2.123673  0.946116
2  1.819697  0.921660
3  1.132676  1.375717
4  1.244837  1.269933

x1, x2, x3 are constants.
    x1 = [1,0,0] 
    x2 = [0,1,0] 
    x3 = [0,0,1] 

For index 0, 
y  = np.array([1-(xG_Team1[0] + xG_Team2[0])/k, xG_Team1[0]/k, xG_Team2[0]/k])
i.e.   y  = np.array([1-(1.440539 + 1.380095)/k, 1.440539/k, 1.380095/k])


For index 1, 
        y  = np.array([1-(xG_Team1[1] + xG_Team2[1])/k, xG_Team1[1]/k, xG_Team2[1]/k])

其中 k 是 total_timeslot 和一个常量。

total_timeslot = 180 
Home_Goal = [] # No Goal
Away_Goal = [] # No Goal
    def sum_squared_diff(x1, x2, x3, y):
        ssd=[]
        for k in range(total_timeslot):
            if k in Home_Goal:
                ssd.append( sum((x2 - y)**2))
            elif k in Away_Goal:
                ssd.append(sum((x3 - y)**2))
            else:
                ssd.append(sum((x1 - y)**2))
        return ssd

y_0 =  sum_squared_diff(x1, x2, x3, y)

计划是对所有 y 的 sum_squared_diff 的输出求和。比如，for all i sum(y_i).

So for i = 0,
    y_0 =  sum_squared_diff(x1, x2, x3, y_0)
    len(y_0) = 180
    sum(y_0) = 0.0663099498972334
Then I will have n numbers of sum(y_i) for n xGs.
using @Dillon code, for the above datframe, n=5
sum(results.sum()) = 0.31885730707076826

Answer 1

data = {'xG_Team1': {0: 1.440539, 1: 2.123673, 2: 1.819697, 3: 1.132676, 4: 1.244837},
 'xG_Team2': {0: 1.380095, 1: 0.946116, 2: 0.92166, 3: 1.375717, 4: 1.269933}}

df = pd.DataFrame(data)

x1 = [1,0,0] 
x2 = [0,1,0] 
x3 = [0,0,1]

# Constants
total_timeslot = 180
k = 180

# Measures
Home_Goal = [] # No Goal
Away_Goal = [] # No Goal

def sum_squared_diff(x1, x2, x3, y):
    ssd = []
    for k in range(total_timeslot):  # k will take multiple values
        if k in Home_Goal:
            ssd.append(sum((x2 - y) ** 2))
        elif k in Away_Goal:
            ssd.append(sum((x3 - y) ** 2))
        else:
            ssd.append(sum((x1 - y) ** 2))
    return ssd

def my_function(row):
    xG_Team1 = row.xG_Team1
    xG_Team2 = row.xG_Team2
    return np.array([1-(xG_Team1 + xG_Team2)/k, xG_Team1/k, xG_Team2/k])

# You can use the apply function
results = df.apply(lambda row: sum_squared_diff(x1, x2, x3, my_function(row)), axis=1)

# Each item in results is a 180 item list
results
Out[]: 
0    [0.0003683886105401867, 0.0003683886105401867,...
1    [0.0004576767592872215, 0.0004576767592872215,...
2    [0.00036036396694006056, 0.0003603639669400605...
3    [0.00029220949467635905, 0.0002922094946763590...
4    [0.00029279065228265494, 0.0002927906522826549...

# For each list, calculate the sum
results.map(lambda x: sum(x))
Out[]: 
0    0.066310
1    0.082382
2    0.064866
3    0.052598
4    0.052702

# Get the sum of all these values
results.map(lambda x: sum(x)).sum()
Out[]: 
0.3188573070707662