将自然范围之外的角度（对于 theta：0 到 180 度）转换为该范围内的等效角度

Question

如何缩短以下代码？这是一个查看角度值并确保它们不会超出范围的简单代码。问题是，虽然这样可以完成工作，但我真的想要一些更 Python 的、更容易使用且编写起来不那么麻烦的东西。

theta_min_lim = 0.0
theta_max_lim = 180.0
if theta_min<theta_min_lim:
    new_theta_min = theta_max_lim-abs(theta_min)
    theta_row_index_along_new = np.array(np.where(sph_pos_count[:,1]>=new_theta_min)).flatten()
    theta_row_index_along = np.concatenate((theta_row_index_along_new, theta_row_index_along))

if theta_max>theta_max_lim:
    new_theta_max = theta_max-theta_max_lim
    theta_row_index_along_new = np.array(np.where(sph_pos_count[:,1]<=new_theta_max)).flatten()
    theta_row_index_along = np.concatenate((theta_row_index_along_new, theta_row_index_along))

if phi_min<phi_min_lim:
    new_phi_min = phi_max_lim-abs(phi_min)
    phi_row_index_along_new = np.array(np.where(sph_pos_count[:,2]>=new_phi_min)).flatten()
    phi_row_index_along = np.concatenate((phi_row_index_along_new, phi_row_index_along))

if phi_max>phi_max_lim:
    new_phi_max = phi_max-phi_max_lim
    phi_row_index_along_new = np.array(np.where(sph_pos_count[:,2]<=new_phi_max)).flatten()
    phi_row_index_along = np.concatenate((phi_row_index_along_new, phi_row_index_along))

#theta_row_index_along = np.concatenate(theta_row_index_along_new, theta_row_index_along)
row_index_along = np.intersect1d(theta_row_index_along,phi_row_index_along)

sph_pos_count_along = sph_pos_count[row_index_along]

#GIving range for theta and phi in the direction along the velocity
 theta_min_opp = vtheta_opp - 2.0*max(sph_cord[:,1])/dtheta
 theta_max_opp = vtheta_opp + 2.0*max(sph_cord[:,1])/dtheta
 phi_min_opp = vphi_opp - 2.0*max(sph_cord[:,2])/dphi
 phi_max_opp = vphi_opp + 2.0*max(sph_cord[:,2])/dphi

 #Finding index of spheres opposite to the direction of motion

 theta_row_index_opp = np.array(np.where(np.logical_and(sph_pos_count[:,1]>=theta_min_opp, sph_pos_count[:,1]<=theta_max_opp))).flatten()
phi_row_index_opp = np.array(np.where(np.logical_and(sph_pos_count[:,2]>=phi_min_opp,  sph_pos_count[:,2]<=phi_max_opp))).flatten()

if theta_min_opp<theta_min_lim:
    new_theta_min_opp = theta_max_lim-abs(theta_min_opp)
    theta_row_index_opp_new = np.array(np.where(sph_pos_count[:,1]>=new_theta_min_opp)).flatten()
    theta_row_index_opp = np.concatenate((theta_row_index_opp_new, theta_row_index_opp))

if theta_max_opp>theta_max_lim:
    new_theta_max_opp = theta_max_opp-theta_max_lim
    theta_row_index_opp_new = np.array(np.where(sph_pos_count[:,1]<=new_theta_max_opp)).flatten()
    theta_row_index_opp = np.concatenate((theta_row_index_opp_new, theta_row_index_opp))

if phi_min_opp<phi_min_lim:
    new_phi_min_opp = phi_max_lim-abs(phi_min_opp)
    phi_row_index_opp_new = np.array(np.where(sph_pos_count[:,2]>=new_phi_min_opp)).flatten()
    phi_row_index_opp = np.concatenate((phi_row_index_opp_new, phi_row_index_opp))

if phi_max_opp>phi_max_lim:
    new_phi_max_opp = phi_max_opp-phi_max_lim
    phi_row_index_opp_new = np.transpose(np.where(sph_pos_count[:,2]<=new_phi_max_opp)).flatten()
    phi_row_index_opp = np.concatenate((phi_row_index_opp_new, phi_row_index_opp))

Answer 1

只需使用模运算（Python 中的模% 运算符）：

    theta = theta % 180

对于-30，这将为您提供150。对于190，这将为您提供10。