Question

从num_dict返回所有具有（值大于或等于min_cutoff。

参数：

num_dict：字典。 num_dict中的所有值都是数字。
min_cutoff：浮动。与num_dict值进行比较。返回所有键，其中它们的值> = min_cutoff。
设置：num_dict中所有键的值均满足截止条件的键。

示例：

keys_geq_cutoff({'Alice': 21, 'Brett': 20, 'Carlos': 31}, 21)
{'Alice', 'Carlos'}

我的代码：

def keys_geq_cutoff(num_dict, min_cutoff):
    for k, v in num_dict.items():
        if (v >= min_cutoff):
            return(keys_geq_cutoff(num_dict, min_cutoff))

错误：

test_keys_geq_cutoff (test_methods.TestPython1)
Traceback (most recent call last):
  File "/usr/src/app/test_methods.py", line 13, in test_keys_geq_cutoff
    result1 = main.keys_geq_cutoff(test_d, 0)
  File "/usr/src/app/main.py", line 4, in keys_geq_cutoff
    return(keys_geq_cutoff(num_dict, min_cutoff))
  File "/usr/src/app/main.py", line 4, in keys_geq_cutoff
    return(keys_geq_cutoff(num_dict, min_cutoff))
  File "/usr/src/app/main.py", line 4, in keys_geq_cutoff
    return(keys_geq_cutoff(num_dict, min_cutoff))
  [Previous line repeated 956 more times]
  File "/usr/src/app/main.py", line 3, in keys_geq_cutoff
    if (v >= min_cutoff):
RecursionError: maximum recursion depth exceeded in comparison

Answer 1

现在，您将在return语句中再次使用完全相同的参数（因此是RecursionError）调用该函数。您需要将密钥收集到集合中，然后返回该集合：

def keys_geq_cutoff(num_dict, min_cutoff):
    res = set()
    for k, v in num_dict.items():
        if (v >= min_cutoff):
            res.add(k)
    return res

或者，可以使用一种不错的设置理解：

def keys_geq_cutoff(num_dict, min_cutoff):
    return {k for k, v in num_dict.items() if v >= min_cutoff}

Answer 2

基本上，您遇到了递归错误，这意味着您要在函数主体中调用函数本身而没有提供exit子句，因此该函数将永远继续下去。为了解决这个问题，您必须创建一个exit子句（例如if语句）

但是您可能不想使用该技术

我认为这是您想要做的：

def keys_geq_cutoff(num_dict, min_cutoff):
    l = []
    for k, v in num_dict.items():
        if v >= min_cutoff:
            l.append(k)
    return set(l)

Answer 3

您真的需要一个功能吗？这个问题看起来非常类似于： Return all the keys (as set) from num_dict that have value greater than or equal to min_cutoff

您可以为此使用一个编码器：

num_dict =  {'Alice': 21, 'Brett': 20, 'Carlos': 31}
min_cutoff = 21
num_dict_keys = [k for k, v in num_dict.items() if v >= min_cutoff ]
print(num_dict_keys)

我的解决方案和算法出了什么问题

3 个答案: