Python while循环未遍历delta_omega数组，无限循环问题求助

问题根源 #

你的while循环里初始化了i=0，但全程没更新i的值——i永远是0，一直满足i < len(delta_omega)的条件，所以会无限重复处理数组第一个元素。加break的话会直接终止循环，自然只算一次。

两种解决方法 #

方法1：修复while循环，手动递增计数器 #

在while循环的末尾加上i += 1，让计数器每次循环后加1，遍历数组所有元素：

delta_omega = np.array([-5*10**6, -10*10**6, -30*10**6]) #range of frequencies
i = 0

while i<len(delta_omega):
    # 保留原有计算逻辑
    B_p = (h/mu_eff) * (delta_omega[i] + (1/lmbda)*(vf_oven**2 - (2*a*length_slow))**0.5)
    B_n = (h/mu_eff) * (delta_omega[i] - (1/lmbda)*(vf_oven**2 - (2*a*length_slow))**0.5)
    delta_n = delta_omega[i] + (k*vf_oven) - (mu_eff*B_n)/hbar
    delta_p = delta_omega[i] - (k*vf_oven) + (mu_eff*B_p)/hbar
    F = (hbar*k*L)/2 * ((II_sat/(1+II_sat+(2*delta_n/L)**2))
                        - (II_sat/(1+II_sat+(2*delta_p/L)**2)))
    accn = abs(F/m_Rb)
    vf_slower = (vf_oven**2 - (2*accn*length_slow))**0.5
    t_d = 1/accn * (vf_oven - vf_slower) #time taken during slower
    
    # After slower
    da = 0.1 #distance from end of slower to the middle of the MOT
    vf_MOT = (vf_slower**2 - (2*accn*da))**0.5
    t_a = da/vf_MOT #time taken after slower

    r0 = 0.01 #MOT capture radius
    vr_max = r0/(t_b+t_d+t_a)
    
    # Flux of atoms captured
    f_oven = ((n*A)/4) * (2/(np.pi)**0.5) * ((2*k_b*T)/m_Rb)**0.5
    f = f_oven * (1 - np.exp(-vr_max**2/vp**2))*(1 - np.exp(-vz_max**2/vp**2))
    print('The rate of capture at a detuning of', delta_omega[i], 'is', format(f, '.1E'))
    
    # 关键：让计数器递增
    i += 1

方法2：改用for循环（更推荐） #

直接遍历delta_omega的每个元素，不用手动管理计数器，代码更简洁，也不容易出错：

delta_omega = np.array([-5*10**6, -10*10**6, -30*10**6]) #range of frequencies

# 直接遍历数组中的每个detuning值
for detuning in delta_omega:
    # 把原来的delta_omega[i]换成detuning即可
    B_p = (h/mu_eff) * (detuning + (1/lmbda)*(vf_oven**2 - (2*a*length_slow))**0.5)
    B_n = (h/mu_eff) * (detuning - (1/lmbda)*(vf_oven**2 - (2*a*length_slow))**0.5)
    delta_n = detuning + (k*vf_oven) - (mu_eff*B_n)/hbar
    delta_p = detuning - (k*vf_oven) + (mu_eff*B_p)/hbar
    F = (hbar*k*L)/2 * ((II_sat/(1+II_sat+(2*delta_n/L)**2))
                        - (II_sat/(1+II_sat+(2*delta_p/L)**2)))
    accn = abs(F/m_Rb)
    vf_slower = (vf_oven**2 - (2*accn*length_slow))**0.5
    t_d = 1/accn * (vf_oven - vf_slower) #time taken during slower
    
    # After slower
    da = 0.1 #distance from end of slower to the middle of the MOT
    vf_MOT = (vf_slower**2 - (2*accn*da))**0.5
    t_a = da/vf_MOT #time taken after slower

    r0 = 0.01 #MOT capture radius
    vr_max = r0/(t_b+t_d+t_a)
    
    # Flux of atoms captured
    f_oven = ((n*A)/4) * (2/(np.pi)**0.5) * ((2*k_b*T)/m_Rb)**0.5
    f = f_oven * (1 - np.exp(-vr_max**2/vp**2))*(1 - np.exp(-vz_max**2/vp**2))
    print('The rate of capture at a detuning of', detuning, 'is', format(f, '.1E'))

内容的提问来源于stack exchange，提问作者Rainydays123