有效分组到di中

Question

我有一个元组列表：

[('Player1', 'A', 1, 100),
('Player1', 'B', 15, 100),
('Player2', 'A', 7, 100),
('Player2', 'B', 65, 100),
('Global Total', None, 88, 100)]

我希望将其转换为以下格式的dict：

{
  'Player1': {
              'A': [1, 12.5],
              'B': [15, 18.75],
              'Total': [16, 18.18]
           },
  'Player2': {
              'A': [7, 87.5],
              'B': [65, 81.25],
              'Total': [72, 81.81]
           },
  'Global Total': {
            'A': [8, 100],
            'B': [80, 100]
         }
}

所以每个玩家都有它的本地总价值和它的百分比，根据它的全局总价值。你知道吗

目前，我是这样做的：

fixed_vals = {}
for name, status, qtd, prct in data_set: # This is the list of tuples var
  if name in fixed_vals:
    fixed_vals[name].update({status: [qtd, prct]})
  else:
    fixed_vals[name] = {status: [qtd, prct]}

fixed_vals['Global Total']['Total'] = fixed_vals['Global Total'].pop(None)
total_a = 0
for k, v in fixed_vals.items():
  if k != 'Global Total':
    total_a += v['A'][0]

fixed_vals['Global Total']['A'] = [
  total_a, total_a * 100 / fixed_vals['Global Total']['Total'][0]
]

fixed_vals['Global Total']['B'] = [
  fixed_vals['Global Total']['Total'][0] - total_a,
  fixed_vals['Global Total']['Total'][0] - fixed_vals['Global Total']['A'][1]
]

for player, vals in fixed_vals.items():
  if player != 'Global Total':
    vals['A'][1] = vals['A'][0] * 100 / fixed_vals['Global Total']['A'][0]
    vals['B'][1] = fixed_vals['Global Total']['A'][1] - vals['B'][1]

问题是这不是很灵活，因为我必须做类似的事情， 但几乎有12个类别（A，B，…）

有没有更好的办法？也许这对熊猫来说是微不足道的？你知道吗

编辑澄清：

每个玩家没有重复的类别，每个人都有相同的序列（有些可能有0，但类别是唯一的）

Answer 1

每个人似乎都被dict唯一的解决方案所吸引，但是为什么不尝试转换成pandas？你知道吗

import pandas as pd

# given
tuple_list = [('Player1', 'A', 1, 100),
('Player1', 'B', 15, 100),
('Player2', 'A', 7, 100),
('Player2', 'B', 65, 100),
('Global Total', None, 88, 100)]

# make a dataframe
df = pd.DataFrame(tuple_list , columns = ['player', 'game','score', 'pct'])
del df['pct']
df = df[df.player!='Global Total']
df = df.pivot(index='player', columns='game', values='score')
df.columns.name='' 
df.index.name='' 

# just a check 
assert df.to_dict() == {'A': {'Player1': 1, 'Player2': 7}, 
                        'B': {'Player1': 15, 'Player2': 65}}

#         A   B
#player        
#Player1  1  15
#Player2  7  65
print('Obtained dataset:\n', df)

基本上，您所需要的只是“df”数据帧，其余的都可以 计算和添加以后，不需要保存到字典。你知道吗

以下内容应OP请求更新：

# the sum across columns is this - this was the 'Grand Total' in the dicts
#  A     8
#  B    80
sum_col = df.sum(axis=0)

# lets calculate the share of each player score:
shares = df / df.sum(axis=0) * 100
assert shares.transpose().to_dict() == {'Player1': {'A': 12.5, 'B': 18.75}, 
                                        'Player2': {'A': 87.5, 'B': 81.25}}
# in 'shares' the columns add to 100%:
#         A     B
#player             
#Player1 12.50 18.75
#Player2 87.50 81.25

# lets mix up a dataframe close to original dictionary structure 
mixed_df = pd.concat([df.A, shares.A, df.B, shares.B], axis=1)
totals = mixed_df.sum(axis=0)
totals.name = 'Total'
mixed_df = mixed_df.append(totals.transpose())
mixed_df.columns = ['A', 'A_pct', 'B', 'B_pct']    
print('\nProducing some statistics\n', mixed_df)

Answer 2

一种解决方案是使用groupby对来自同一玩家的连续玩家分数进行分组

tup = [('Player1', 'A', 1, 100),('Player1', 'B', 15, 100),('Player2', 'A', 7, 100),    ('Player2', 'B', 65, 100),    ('Global Total', None, 88, 100)]`

然后导入我们的groupby

from itertools import groupby

result = dict((name,dict((x[1],x[2:]) for x in values)) for name,values in groupby(tup,lambda x:x[0]))

那就去更新所有的总数

for key in result:
    if key == "Global Total": continue # skip this one ...
    # sum up our player scores
    result[key]['total'] = [sum(col) for col in zip(*result[key].values())]

# you can print the results too
print result

# {'Player2': {'A': (7, 100), 'total': [72, 200], 'B': (65, 100)}, 'Player1': {'A': (1, 100), 'total': [16, 200], 'B': (15, 100)}, 'Global Total': {'total': [88, 100], None: (88, 100)}}

注意此解决方案！要求！所有player1的分数在元组中分组，所有player2的分数在元组中分组等

Answer 3

A）将代码分解为可管理的块：

from collections import defaultdict
result = defaultdict(dict)
for (cat, sub, num, percent) in input_list:
    result[cat][sub] = [num, percent]

现在我们有了一个玩家计数的dict，但是唯一有效的百分比是total，我们没有全局计数。你知道吗

from collections import Counter
def build_global(dct):
    keys = Counter()
    for key in dct:
        if key == "Global Total":
            continue
        for sub_key in dct[key]:
            keys[sub_key] += dct[key][sub_key][0]
    for key in keys:
        dct["Global Total"][key] = [keys[key], 100]

build_global(result)现在为每个事件生成有效的全局计数。你知道吗

最后：

def calc_percent(dct):
    totals = dct["Global Total"]
    for key in dct:
        local_total = 0
        if key == "Global Total":
            continue
        for sub_key in dct[key]:
            local_total += dct[key][sub_key][0]
            dct[key][sub_key][1] = (dct[key][sub_key][0]/float(totals[sub_key][0])) * 100
        dct[key]['Total'] = [local_total, (local_total/float(dct['Global Total'][None][0])) * 100]

calc_percent(result)遍历并构建百分比。你知道吗

结果是：

defaultdict(<type 'dict'>, 
    {'Player2': {'A': [7, 87.5], 'B': [65, 81.25], 'Total': [72, 81.81818181818183]}, 
     'Player1': {'A': [1, 12.5], 'B': [15, 18.75], 'Total': [16, 18.181818181818183]}, 
     'Global Total': {'A': [8, 100], None: [88, 100], 'B': [80, 100]}})

如果您确实需要它，您可以删除global total中的None条目，并dict(result)将defaultdict转换为香草dict。你知道吗