Python无尽循环

Question

我有以下代码：

for i in range(N):

    # Dispersed Source
    theta = np.random.uniform(0, np.pi, 1)
    phi = np.random.uniform(0, 2 * np.pi, 1)
    R = np.random.uniform(0, Ro, 1)

    x = R * np.sin(theta) * np.cos(phi)
    y = R * np.sin(theta) * np.sin(phi)  # ?
    z = R * np.cos(theta)

    x_vec, y_vec, z_vec, nu = particle_func(x, y, z)

    particle_trace = go.Scatter3d(
                        x=x_vec,
                        y=y_vec,
                        z=z_vec,
                        mode='lines'
                    )

    data.append(particle_trace)

    for j in range(int(nu)):
        x = x_vec[-1]
        y = y_vec[-1]
        z = z_vec[-1]

        x_vec_1, y_vec_1, z_vec_1, nu1 = particle_func(x, y, z)

        particle_trace_fiss = go.Scatter3d(
            x=x_vec_1,
            y=y_vec_1,
            z=z_vec_1,
            mode='lines'
        )

        data.append(particle_trace_fiss)

问题是for循环需要继续这样：

for j in range(int(nu)):
        x = x_vec[-1]
        y = y_vec[-1]
        z = z_vec[-1]

        x_vec_1, y_vec_1, z_vec_1, nu1 = particle_func(x, y, z)

        particle_trace_fiss = go.Scatter3d(
            x=x_vec_1,
            y=y_vec_1,
            z=z_vec_1,
            mode='lines'
        )

        data.append(particle_trace_fiss)

 for k in range(int(nu1)):
        x = x_vec_1[-1]
        y = y_vec_1[-1]
        z = z_vec_1[-1]

        x_vec_2, y_vec_2, z_vec_2, nu2 = particle_func(x, y, z)

        particle_trace_fiss = go.Scatter3d(
            x=x_vec_2,
            y=y_vec_2,
            z=z_vec_2,
            mode='lines'
        )

        data.append(particle_trace_fiss)

 for l in range(int(nu2)):
        x = x_vec_2[-1]
        y = y_vec_2[-1]
        z = z_vec_2[-1]

        x_vec_3, y_vec_3, z_vec_3, nu3 = particle_func(x, y, z)

        particle_trace_fiss = go.Scatter3d(
            x=x_vec_3,
            y=y_vec_3,
            z=z_vec_3,
            mode='lines'
        )

        data.append(particle_trace_fiss)

...

因为nu的值是一个随机数或零（从particle_func函数给出），我需要的循环数可能会永远持续下去。也就是说，我可以复制/粘贴

for k in range(int(nu#)):...
     data.append(particle_trace_fiss)

无数次并获得我想要的东西（其中nu＃表示前一循环提供的nu的值。

请参阅下面的算法以获得澄清：

道歉，如果我在问题中遗漏了清晰度，我将很乐意回复任何要求澄清的评论。所有帮助都提前感谢，谢谢!!

Answer 1

您似乎只需要将for包裹在while循环中，也可以使用任意截止值。

此代码：

for i in range(N):

    # Dispersed Source
    theta = np.random.uniform(0, np.pi, 1)
    phi = np.random.uniform(0, 2 * np.pi, 1)
    R = np.random.uniform(0, Ro, 1)

    x = R * np.sin(theta) * np.cos(phi)
    y = R * np.sin(theta) * np.sin(phi)  # ?
    z = R * np.cos(theta)

    x_vec, y_vec, z_vec, nu = particle_func(x, y, z)

    particle_trace = go.Scatter3d(
                        x=x_vec,
                        y=y_vec,
                        z=z_vec,
                        mode='lines'
                    )

    data.append(particle_trace)

    for j in range(int(nu)):
        x = x_vec[-1]
        y = y_vec[-1]
        z = z_vec[-1]

        x_vec_1, y_vec_1, z_vec_1, nu1 = particle_func(x, y, z)

        particle_trace_fiss = go.Scatter3d(
            x=x_vec_1,
            y=y_vec_1,
            z=z_vec_1,
            mode='lines'
        )

        data.append(particle_trace_fiss)

成为这段代码：

for i in range(N):

    # Dispersed Source
    theta = np.random.uniform(0, np.pi, 1)
    phi = np.random.uniform(0, 2 * np.pi, 1)
    R = np.random.uniform(0, Ro, 1)

    x = R * np.sin(theta) * np.cos(phi)
    y = R * np.sin(theta) * np.sin(phi)  # ?
    z = R * np.cos(theta)


    arbitrary_limit = 10

    x_vec = [x]
    y_vec = [y]
    z_vec = [z]
    nu_vec = [1]

    nu = nu_vec[-1]    # Updated - keep a list of values of nu

    while int(nu) != 0:

        for j in range(int(nu)):
            x = x_vec[-1]
            y = y_vec[-1]
            z = z_vec[-1]

            x_vec, y_vec, z_vec, nu = particle_func(x, y, z)

            nu_vec.append(nu)

            particle_trace = go.Scatter3d(
                            x=x_vec,
                            y=y_vec,
                            z=z_vec,
                            mode='lines'
                        )

            data.append(particle_trace)

我想问题是，您将如何决定何时停止？您可以采取哪些措施来限制nu的价值吗？

更新：根据您的评论，我删除了arbitrary_limit代码。我还添加了nu_vec来跟踪nu的所有值，根据您在另一个尝试的答案中的解释。

Answer 2

我以为我可以使用以下内容：

for i in range(N):

    # Central Point Source
    # z = 0
    # x = 0
    # y = 0

    # Dispersed Source
    theta = np.random.uniform(0, np.pi, 1)
    phi = np.random.uniform(0, 2 * np.pi, 1)
    R = np.random.uniform(0, Ro, 1)

    x = R * np.sin(theta) * np.cos(phi)
    y = R * np.sin(theta) * np.sin(phi)  # ?
    z = R * np.cos(theta)

    x_vec, y_vec, z_vec, nu = particle_func(x, y, z)

    particle_trace = go.Scatter3d(
                        x=x_vec,
                        y=y_vec,
                        z=z_vec,
                        mode='lines'
                    )

    data.append(particle_trace)

    while nu != 0:

        x = x_vec[-1]
        y = y_vec[-1]
        z = z_vec[-1]

        for j in range(int(nu)):

            x_vec, y_vec, z_vec, nu = particle_func(x, y, z)

            particle_trace_fiss = go.Scatter3d(
                x=x_vec,
                y=y_vec,
                z=z_vec,
                mode='lines'
            )

            data.append(particle_trace_fiss)

因为它表现得像我想的那样。但是，我很困惑：

for j in range(int(nu)):

            x_vec, y_vec, z_vec, nu = particle_func(x, y, z)

因为nu的新值将与for循环范围内的nu值一样多。每个nu值都很重要。也就是说，我打算生成一定数量的nu值。上述问题是每个nu值被for循环中的下一个nu值覆盖。如果生成5个nu值，我在技术上需要独立地对每个nu值运行另一个for循环，并且循环继续。我希望这有道理，有什么想法吗？我想我可以将所有的nu值附加到一个向量中并得到它们吗？