Question

我会尽力解释我的情况。

我有一个包含名称的列表

 name_list = ["Bob" , "Henry" , "Jeff" , "Harry"]

和另一个包含数字的列表

number_list = ["1,4,2,3" , "4,3,1,2" , "3,2,4,1" , "1,2,3,4" , "4,3,2,1" , "1,2,4,3"]

我正在努力想出一个代码，它将创建一个将每个名称与name_list中相应位置相关联的记录

即

 "result" = Bob  - 1:3 , 2:0 , 3:1 , 4:2
            Henry- 1:0 , 2:3 , 3:2 , 4:1
            Jeff - 1:1 , 2:2 , 3:1 , 4:2
            Harry- 1:2 , 2:1 , 3:2 , 4:1

这实际上是否可以实现？因为我觉得我现在已经死了一段时间，我拼命地需要帮助。我对此代码的思考过程是计算出现的每个数字的出现次数，然后（因为此代码用于即时径流投票系统）确定谁具有最高的1个计数，依此类推。

这是即时决选投票代码的一部分，关于我是否在正确的轨道上的任何信息都是非常可取的

Answer 1

你可以试试这个：

name_list = ["Bob" , "Henry" , "Jeff" , "Harry"]
number_list = ["1,4,2,3" , "4,3,1,2" , "3,2,4,1" , "1,2,3,4" , "4,3,2,1" , "1,2,4,3"] 
new_number_list = [list(map(int, i.split(","))) for i in number_list]
final_data = {a:{c:sum(d[i] == c for d in [e for e in new_number_list if e != a]) for c in new_number_list[i]} for i, a in enumerate(name_list)}

输出：

{'Bob': {1: 3, 2: 0, 3: 1, 4: 2}, 'Jeff': {1: 1, 2: 2, 3: 1, 4: 2}, 'Harry': {1: 2, 2: 1, 3: 2, 4: 1}, 'Henry': {1: 0, 2: 3, 3: 2, 4: 1}}

编辑：要保持所有值的排序，您可以使用：

new_final_data = {a:sorted(b.items(), key=lambda x:x[0]) for a, b in final_data.items()}

输出：

{'Henry': [(1, 0), (2, 3), (3, 2), (4, 1)], 'Bob': [(1, 3), (2, 0), (3, 1), (4, 2)], 'Harry': [(1, 2), (2, 1), (3, 2), (4, 1)], 'Jeff': [(1, 1), (2, 2), (3, 1), (4, 2)]}

Answer 2

我是怎么做的，我不确定你是否需要一个功能，因为你没有指定

from collections import Counter
from itertools import chain

# first lets create a list of dictionaries mapping each name to a number
maps = [dict(zip(name_list, chunk.split(','))) for chunk in number_list]

[{'Bob': '1', 'Harry': '3', 'Henry': '4', 'Jeff': '2'},
 {'Bob': '4', 'Harry': '2', 'Henry': '3', 'Jeff': '1'},
 {'Bob': '3', 'Harry': '1', 'Henry': '2', 'Jeff': '4'},
 {'Bob': '1', 'Harry': '4', 'Henry': '2', 'Jeff': '3'},
 {'Bob': '4', 'Harry': '1', 'Henry': '3', 'Jeff': '2'},
 {'Bob': '1', 'Harry': '3', 'Henry': '2', 'Jeff': '4'}]

for name in name_list:
    temp = Counter()
    # counter object where name is the key, value is another counter object
    temp.update({name: Counter(map(lambda x: x.get(name), maps))})
    num = len(temp.values()[0].keys())
    # create string to format
    fmt = '{} has {} occurences, ' * num
    print '{} - '.format(name) + fmt.format(*list(chain.from_iterable(temp.values()[0].items())))

Bob - 1 has 3 occurences, 3 has 1 occurences, 4 has 2 occurences, 
Henry - 3 has 2 occurences, 2 has 3 occurences, 4 has 1 occurences, 
Jeff - 1 has 1 occurences, 3 has 1 occurences, 2 has 2 occurences, 4 has 2 occurences, 
Harry - 1 has 2 occurences, 3 has 2 occurences, 2 has 1 occurences, 4 has 1 occurences,

计算列表列表并将其与另一个列表相关联

2 个答案: