两段代码只有几行不一样，且分散，如何优化？

Question

比如:

if d_m > d_o:
    d1 = d_m
    sf_0 = scale_factor * 0.5
    sf_1 = scale_factor
    for i in range(20):
        # ... 省略
        if d < d1:
            # ... 省略
            sf_0 = sf
        else:
            sf_1 = sf
            break
else:
    d1 = d_o
    sf_0 = 0
    sf_1 = scale_factor * 0.5
    for i in range(20):
        # ... 省略
        if d < d1:
            # ... 省略
            sf_1 = sf
        else:
            sf_0 = sf
            break

相似度很高，但逻辑顺序不能变，这种能优化吗？

Answer 1

d1, sf_0, sf_1 = (d_m, scale_factor * 0.5, scale_factor) if d_m > d_o else (d_o, 0, scale_factor * 0.5)
for i in range(20):
    # ... 省略
    if d < d1:
        # ... 省略
        sf_0 = sf
    else:
        sf_1 = sf
        break

Answer 2

def common(d1_v, sf_0_v, sf_1_v, d, sf, dm_gt_do):
    d1 = d1_v
    sf_0 = sf_0_v
    sf_1 = sf_1_v

    for i in range(20):
        # ... 省略
        if d < d1:
            # ... 省略
            sf_0 = sf if dm_gt_do else sf_1 = sf
        else:
            sf_1 = sf if dm_gt_do else sf_0 = sf
            break


if d_m > d_o:
    common(d_m, scale_factor * 0.5, scale_factor, d, sf, True)
else:
    common(d_o, 0, scale_factor * 0.5, d, sf, False)

Answer 3

优化如下:

# 初始化变量  
d1 = d_m if d_m > d_o else d_o  
sf_0 = scale_factor * 0.5 if d_m > d_o else 0  
sf_1 = scale_factor if d_m > d_o else scale_factor * 0.5  
  
# 创建一个通用的循环  
for i in range(20):  
    if d < d1:  
        # 更新sf_0或sf_1，取决于之前的条件  
        if d_m > d_o:  
            sf_0 = sf  
        else:  
            sf_1 = sf  
    else:  
        # 更新sf_1或sf_0，并跳出循环  
        if d_m > d_o:  
            sf_1 = sf  
        else:  
            sf_0 = sf  
        break

Answer 4

Python 不熟，代码仅示意

def ffff(d1, sf_0, sf_1, posi):
    for i in range(20):
            if (d < d1 == posi)：
                ef_0 = sf
            else:
                ef_1 = sf
                break

if d_m > d_o:
    ffff(d_m, scale_factor * 0.5, scale_factor, True)
else:
    ffff(d_o, 0, scale_factor, False)