删除多个列表的最后一个值的更优雅的方法？

Question

我有多个列表，即X、Y、VX、VY、R、T、KE、PE、TE，对于这些列表，我需要在迭代中断时删除它们的第（I+1）个值。我现在能做的就是：

del X[i+1:]
del Y[i+1:]
del VX[i+1:]
del VY[i+1:]
del R[i+1:]
del T[i+1:]
del KE[i+1:]
del PE[i+1:]
del TE[i+1:]

我觉得它看起来很难看。在这种情况下，有没有办法尽量减少线路的使用。可能是一个循环，但只有1或2行？你知道吗

编辑： 我不想包含我的代码主体，因为我知道我的很多同事都在做同样的问题，我不希望任何人抄袭。不管怎样，我就是这么做的。你知道吗

def F_1(X):
    return VX

def F_2(X):
    return VY

def F_3(X, Y):
    return -G*EM*X/((X**2 + Y**2)**(3/2))

def F_4(X, Y):
    return -G*EM*Y/((X**2 + Y**2)**(3/2))

def RKutta(F_1, F_2, F_3, F_4, XSTART, YSTART, VXSTART, VYSTART):
    N = 100000
    X, Y = np.zeros([N+1]), np.zeros([N+1])
    VX, VY = np.zeros([N+1]), np.zeros([N+1])
    R, T =  np.zeros([N+1]), np.zeros([N+1])
    KE, PE, TE = np.zeros([N+1]), np.zeros(N+1), np.zeros([N+1])

    X[0], Y[0] = XSTART, YSTART
    VX[0], VY[0] = VXSTART, VYSTART
    T[0] = 0
#    KE[0], PE[0], TE[0] = KE, PE, TE

    for i in range (1,N):

        K_1X = F_1(VX[i-1])
        K_1Y = F_2(VY[i-1])
        K_1VX = F_3(X[i-1], Y[i-1])
        K_1VY = F_4(X[i-1], Y[i-1])

        K_2X = F_1(VX[i-1] + (H*K_1VX/2))
        K_2Y = F_2(VY[i-1] + (H*K_1VY/2))
        K_2VX = F_3(X[i-1] + (H*K_1X/2), Y[i-1] + (H*K_1Y/2))
        K_2VY = F_4(X[i-1] + (H*K_1X/2), Y[i-1] + (H*K_1Y/2))

        K_3X = F_1(VX[i-1] + (H*K_2VX/2))
        K_3Y = F_2(VY[i-1] + (H*K_2VY/2))
        K_3VX = F_3(X[i-1] + (H*K_2X/2), Y[i-1] + (H*K_2Y/2))
        K_3VY = F_4(X[i-1] + (H*K_2X/2), Y[i-1] + (H*K_2Y/2))

        K_4X = F_1(VX[i-1] + H*K_3VX)
        K_4Y = F_2(VY[i-1] + H*K_3VY)
        K_4VX = F_3(X[i-1] + H*K_3X, Y[i-1] + H*K_3Y)
        K_4VY = F_4(X[i-1] + H*K_3X, Y[i-1] + H*K_3Y)

        X[i] = X[i-1] + (H/6)*(K_1X + 2*K_2X + 2*K_3X + K_4X)
        Y[i] = Y[i-1] + (H/6)*(K_2Y + 2*K_2Y + 2*K_3Y + K_4Y)
        VX[i] = VX[i-1] + (H/6)*(K_1VX + 2*K_2VX + 2*K_3VX + 2*K_4VX)
        VY[i] = VY[i-1] + (H/6)*(K_1VY + 2*K_2VY + 2*K_3VY + 2*K_4VY)
        R[i] = ((X[i])**2 + (Y[i])**2)**(0.5)
        T[i] = T[i-1] + H

        KE[i] = ((VX[i]**2 + VY[i]**2)/2)
        PE[i] = (-1*G*EM/R[i])
        TE[i] = KE[i] + PE[i]

        if R[i] < 6.371E6:  #if orbit radius is less than Earth radius
            break

        for sublist in [X, Y, VX, VY, R, T, KE, PE, TE]:
            del sublist[i+1:]

    return X, Y, VX, VY, R, T, KE, PE, TE

X = 3.84E8
Y = 0
X, Y, VX, VY, R, T, KE, PE = RKutta(F_1, F_2, F_3, F_4, X, Y, VX, VY)

plt.plot(X, Y, label = "")
plt.xlabel("X distance (m)")
plt.ylabel("Y distance (m)")
plt.title("X against Y")
plt.legend()
plt.show()

我正在尝试用四阶龙格库塔方法来模拟一个绕地球运行的火箭。目标是最终打印出火箭的轨道。我还在想我的代码，所以他们还没开始工作。使用类将如何改善这一点？你知道吗

Answer 1

以下是我在评论中试图解释的内容。虽然我不知道你在做什么的细节，但希望这足以让你大致了解：

# Define a class to hold each group of values.
class Example:
    def __init__(x, y, vx, vy, r, t, ke, pe, te):
        self.x  = x
        self.y  = y
        self.vx = vx
        self.vy = vy
        self.r  = r
        self.t  = t
        self.ke = ke
        self.pe = pe
        self.te = te

# Somehow create a list of Example instances.
my_data = [Example(x1, y1, vx1, vy1, r1, t1, ke1, pe1, te1),
           Example(x2, y2, vx2, vy2, r2, t2, ke2, pe2, te2),
             ...
           Example(xN, yN, vxN, vyN, rN, tN, keN, peN, teN)]

# Then you could do what you asked about like this:
for i, elem in enumerate(my_data):
    if elem.ke > elem.pe:  # Done?
        break

del my_data[i+1:]  # Delete remaining elements in my_data.