使用Python使用函数返回多个值

Question

我有一个代码列表，用于更新每5个元素的时间值。这是代码：

msg = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
theta=1
time=4.5

def get_new_theta(msg, theta,time):
    new_theta = [theta]
    new_time = [time]
    for a, b in zip(msg[3::5], msg[4::5]):
        new_theta.append(new_theta[-1] + a + b)
        new_time.append(new_time[-1]+time)
    return new_theta
    return new_time

for i, theta in enumerate(get_new_theta(msg, theta,time)[:-1]):
    for j in range(5):
        print(theta, msg[i*5+j],time)

但是，我没有得到想要的输出。所需的输出应为：

1 1 4.5
1 2 4.5
1 3 4.5
1 4 4.5
1 5 4.5
10 6 9
10 7 9
10 8 9
10 9 9
10 10 9
29 11 13.5
29 12 13.5
29 13 13.5
29 14 13.5
29 15 13.5

需要此论坛的帮助。谢谢。

Answer 1

def get_new_theta(msg, theta, time):
    new_theta = [theta]
    new_time = [time]
    for a, b in zip(msg[3::5], msg[4::5]):
        new_theta.append(new_theta[-1] + a + b)
        new_time.append(new_time[-1]+time)
    return new_theta[:-1], new_time[:-1]

msg = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
theta = 1
time = 4.5

for i, (theta, time) in enumerate( zip(*get_new_theta(msg, theta, time)) ):
    for j in range(5):
        print(theta, msg[i*5+j], time)

给予：

1 1 4.5
1 2 4.5
1 3 4.5
1 4 4.5
1 5 4.5
10 6 9.0
10 7 9.0
10 8 9.0
10 9 9.0
10 10 9.0
29 11 13.5
29 12 13.5
29 13 13.5
29 14 13.5
29 15 13.5

get_new_theta(msg, theta, time)返回列表(new_theta, new_time)的元组。 zip(*get_new_theta(msg, theta, time))给出了元组(theta, time)上的迭代器