Question

如果我输入n，将使用密钥count_dict创建大小为n+1的字典0...n。

我想解析一个名为binary_list的列表，其中包含大小为n的字符串，每个字符串由0或1组成。每个解析将计算字符串中1的数量，并将每个字符串的append计数为1到适当的键

例如：

{0:['000'], 1:['001','010','100'] , 2:['011', '101', '110'] , 3:['111']}

执行此操作的代码：

count_dict = dict.fromkeys(range(0,n+1))

for element in binary_list:
    count_dict[element.count('1')].append(element)

错误“

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "facebook_prob.py", line 23, in sortbit
    count_dict[element.count('1')].append(element)
AttributeError: 'NoneType' object has no attribute 'append'

Answer 1

如果不指定第二个参数，默认值为None。（见dict.fromkeys）

>>> n = 3
>>> count_dict = dict.fromkeys(range(0,n+1))
>>> count_dict
{0: None, 1: None, 2: None, 3: None}

除此之外，dict.fromkeys创建的值由所有条目共享。

>>> count_dict = dict.fromkeys(range(0,n+1), [])
>>> count_dict
{0: [], 1: [], 2: [], 3: []}
>>> count_dict[0] is count_dict[1]
True

collections.defaultdict更合适。

>>> from collections import defaultdict
>>> count_dict = defaultdict(list)
>>> count_dict[0].append('000')
>>> count_dict[1].append('001')
>>> count_dict[1].append('010')
>>> count_dict
defaultdict(<type 'list'>, {0: ['000'], 1: ['001', '010']})

Nonetype对象没有附加属性追加列表到dict

1 个答案: