缩短长条件测试`if find\ num\ U 3！=无并找到\u num \u 3！=我找到了第三个！=j:

Question

我有这样一个链条件测试：

if find_num_3 != None and find_num_3 != i and find_num_3 != j: #don't reproduce itself

在代码中：

for i in range(len(nums)):
    num_1= nums[i]
    sub_target = target - num_1
    logging.debug(f"level_1_lookup: {lookup}")

    for j in range(i+1, len(nums)):
        num_2 = nums[j] #
        num_3 = sub_target - num_2              
        find_num_3 = lookup.get(num_3) #             

        if find_num_3 != None and find_num_3 != i and find_num_3 != j: #don't reproduce itself    

            result = [num_1, num_2, num_3]
            triplets.append(result)

            logging.debug(f"lookup: {lookup} sub_nums: {nums[j:]} \nresult: {result}")
            logging.info(f"\ttriplets: {triplets}\n\n\n\n")                    
return triplets 

如何将长链转变为紧凑的短链结构。你知道吗

Answer 1

如果只想减小行的大小，可以将条件括在括号中：

if (find_num_3 is not None
        and find_num_3 != i
        and find_num_3 != j):
    # ...

---

您可以将每个条件值存储在变量中，这很方便，因为它命名了条件并使最终条件易于理解：

defined = find_num_3 is not None
not_i = find_num_3 != i
not_j = find_num_3 != j
if defined and not_i and not_j:
    #...

或：

if all(defined, not_i, not_j):
    # ...

---

您还可以检查find_num_3是否在值列表中：

if find_num_3 not in (None, i, j):
    # ...