Python:函数按十位对数字进行分组。

Question

获取一个数字列表，并将数字按其数十个位置分组，给出它自己的子列表中的每一个十位。

例：

$ group_by_10s([1, 10, 15, 20])
[[1], [10, 15], [20]]
$ group_by_10s([8, 12, 3, 17, 19, 24, 35, 50])
[[3, 8], [12, 17, 19], [24], [35], [], [50]]

我的方法：

limiting = 10
ex_limiting = 0
result = []
for num in lst:
    row = []
    for num in lst:
        if num >= ex_limiting and num <= limiting:
            row.append(num)
            lst.remove(num)
    result.append(row)
    ex_limiting = limiting
    limiting += 10

但它返回[1，10，20]。我的方法有什么问题，我该如何解决？

Answer 1

您可能已经有了正确的答案，但这里有一个替代解决方案：

def group_by_10s(mylist):
    result = []
    decade = -1
    for i in sorted(mylist):
        while i // 10 != decade:
            result.append([])
            decade += 1
        result[-1].append(i)
    return result

group_by_10s([8, 12, 3, 17, 19, 24, 35, 50])
#[[3, 8], [12, 17, 19], [24], [35], [], [50]]

它只使用普通的Python，不使用额外的模块。

Answer 2

您可以使用列表理解：

def group_by_10s(_d):
  d = sorted(_d)
  return [[c for c in d if c//10 == i] for i in range(min(_d)//10, (max(_d)//10)+1)]

print(group_by_10s([1, 10, 15, 20]))
print(group_by_10s([8, 12, 3, 17, 19, 24, 35, 50]))
print(group_by_10s(list(range(20))))

输出：

[[1], [10, 15], [20]]
[[3, 8], [12, 17, 19], [24], [35], [], [50]]
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]]

Answer 3

请注意循环期间不迭代列表的建议。我最终得到了这个答案，对于任何想要答案的人来说。谢谢你的支持！

def group_by_10s(numbers):
  external_loop = int(max(numbers)/10)
  limiting = 10
  ex_limiting = 0
  result = []
  for external_loop_count in range(external_loop+1):
    row = []
    for num in numbers:
        if num >= ex_limiting and num < limiting:
            row.append(num)
            row.sort()
    result.append(row)
    ex_limiting = limiting
    limiting += 10
  return(result)