Question

我有这四个问题。列表：

x1 = [1,0,0] ; prob. of No goal scored per timeslot.
x2 = [0,1,0] ; prob. of Home Team scoring per timeslot.
x3 = [0,0,1] ; prob. of Away Team scoring per timeslot.
y = [0.97, 0.02, 0.01]; constant prob. per timeslot.

我需要x1和y的平方差（sse）之和，直到得分为目标。然后，sse由x2或x3和y计算，具体取决于首先得分的人。然后它会循环回x1和y，直到任何一支球队再次对目标进行评分等等......

Here is what I have attempted so far.

Total timeslot = 15.
Order of timeslot
Home_Goal_timeslot = 8th, 14th. i.e. at two different timeslots.
Away_Goal_timeslot = 11th
No_Goal = 12. i.e. number of timeslots 12


x1 = np.array([1,0,0])
x2 = np.array([0,1,0])
x3 = np.array([0,0,1])
y = np.array([0.97, 0.02, 0.01])

def sum_squared_diff(x1, x2, x3, y, k):
    for k in range(15):
        if k == 'No_Goal':
           return sum((x1 - y)**2)
        elif k == 'Home_Goal':    
            return  sum((x2 - y)**2)
        else:
            return  sum((x3 - y)**2)





sum_squared_diff(x1, x2, x3, y, k=15)

Out[1912]: 1.9213999999999998     

print(sum((x1 - y)**2)**12, sum((x2 - y)**2)**2, sum((x3 - y)**2)**1)

    5.669391237529683e-35 3.6153219599999997 1.9213999999999998
add up to 5.5367219599999995

为什么上述解决方案有所不同？我更喜欢在每个时段之后添加sse，并在进球时记下。所以我可以返回并添加/减去一些infinitesimal epsilon到y值并测量测试......

This is the out come i want

 0.0014000000000000017,
 0.0014000000000000017,
 0.0014000000000000017,
 0.0014000000000000017,
 0.0014000000000000017,
 0.0014000000000000017,
 0.0014000000000000017,
 0.0014000000000000017,
 1.9014,
 0.0014000000000000017,
 0.0014000000000000017,
 1.9213999999999998,
 0.0014000000000000017,
 0.0014000000000000017,
 1.9014

Answer 1

解决方案是不同的，因为最后一个else语句永远是真的，因为正如@slybloty所说，你将int k与str进行比较。所以你的

def sum_squared_diff(x1, x2, x3, y, k):
        for k in range(15):
            if k == 'No_Goal':
               return sum((x1 - y)**2)
            elif k == 'Home_Goal':    
                return  sum((x2 - y)**2)
            else:
                return  sum((x3 - y)**2)

和

def sum_squared_diff(x1, x2, x3, y, timeslot):
    return sum((x3 - y)**2)

做同样的工作

Answer 2

我想要做的就是遍历您的total timeslot，如果它与您的Home_Goal或Away_Goal匹配，那么就这样计算。
你正在做的是：

将int与str（k == 'No_Goal'）进行比较，您将获得False几乎所有时间。
在计算后使用return在第一次迭代时结束循环。

根据我的理解，这就是你要找的东西：

total_timeslot = 15
Home_Goal = [8, 14]
Away_Goal = [11]

def sum_squared_diff(x1, x2, x3, y):
    r1, r2, r3 = 0, 0, 0
    for k in range(total_timeslot):
        if k in Home_Goal:
            r2 += sum((x2 - y)**2)
        elif k in Away_Goal:
            r3 += sum((x3 - y)**2)
        else:
            r1 += sum((x1 - y)**2)
    return r1, r2, r3

UPDATE：

在您更新了问题后，似乎您希望在每次迭代后print获得所有结果，而不仅仅是最终结果。在这种情况下，您可以在每个循环结束时添加print语句并删除增量。

更新2：

以下是反映变化的代码：

def sum_squared_diff(x1, x2, x3, y):
    r1, r2, r3 = 0, 0, 0
    for k in range(total_timeslot):
        if k in Home_Goal:
            r2 = sum((x2 - y)**2)
        elif k in Away_Goal:
            r3 = sum((x3 - y)**2)
        else:
            r1 = sum((x1 - y)**2)
    print(r1, r2, r3)

在不同的循环python中迭代四个列表

2 个答案:

UPDATE：

更新2：