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)]
我希望以下列格式转换为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]
}
}
因此每个播放器词典都有本地总值，并且根据它的全局总值，它是百分比。
目前，我这样做：
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]
问题在于这不是很灵活，因为我必须做类似的事情， 但有近12个类别（A，B，......）
有更好的方法吗？也许这对大熊猫来说是微不足道的？
编辑以澄清：
每个玩家没有重复的类别，每个玩家都有相同的序列（有些可能有0但类别是唯一的）
答案 0 :(得分：2)
每个人似乎都被dict解决方案所吸引，但为什么不尝试转换为pandas？
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，你需要的只是'df'数据框，其余的你可以 稍后计算并添加，无需将其保存到字典中。
以下是OP请求的更新：
# 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)
答案 1 :(得分：1)
一种解决方案是使用groupby对来自同一玩家的连续玩家得分进行分组
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]))
然后去更新所有总计
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！需要！所有玩家1的得分都在您的元组中组合在一起，并且所有玩家2的得分都被分组等
答案 2 :(得分：0)
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]
现在我们有一个关于玩家数量的词典，但是唯一有效的百分比是总数而且我们没有全局计数。
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)现在为每个事件产生有效的全局计数。
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)通过并构建百分比。
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]}})
如果您需要完全指定，则可以删除全局总计中的None条目和dict(result)，以将defaultdict转换为香草dict 1}}。
答案 3 :(得分：0)
在Python 3.6 +中使用more_itertools中的重新映射工具：
<强>鉴于
for (int i = 0; i < newDt.Rows.Count;i++ )
{
SqlCommand command = new SqlCommand("update batch set sold_qty=@soldqty2 where id=@id2",con);
command.Parameters.AddWithValue("@soldqty2", Convert.ToInt32(newDt.Rows[i]["QTY"]));
command.Parameters.AddWithValue("@id2", Convert.ToInt32(newDt.Rows[i]["QTY"]) + Convert.ToInt32(newDt.Rows[i]["BATCH NUM"]));
rexe=command.ExecuteNonQuery();
}
<强>代码
import copy as cp
import collections as ct
import more_itertools as mit
data = [
("Player1", "A", 1, 100),
("Player1", "B", 15, 100),
("Player2", "A", 7, 100),
("Player2", "B", 65, 100),
('Global Total', None, 88, 100)
]
# Discard the last entry
data = data[:-1]
# Key functions
kfunc = lambda tup: tup[0]
vfunc = lambda tup: tup[1:]
rfunc = lambda x: {item[0]: [item[1]] for item in x}
<强>详情
第1步 - 构建# Step 1
remapped = mit.map_reduce(data, kfunc, vfunc, rfunc)
# Step 2
intermediate = ct.defaultdict(list)
for d in remapped.values():
for k, v in d.items():
intermediate[k].extend(v)
# Step 3
remapped["Global Total"] = {k: [sum(v)] for k, v in intermediate.items()}
final = cp.deepcopy(remapped)
for name, d in remapped.items():
for lbl, v in d.items():
stat = (v[0]/remapped["Global Total"][lbl][0]) * 100
final[name][lbl].append(stat)
组的新词典。
这是通过定义指示如何处理键和值的关键函数来完成的。 reduce函数将值处理为子字典。另请参阅docs for more details on more_itertools.map_reduce。
remapped第2步 - 为查找构建>>> remapped
defaultdict(None,
{'Player1': {'A': [1], 'B': [15]},
'Player2': {'A': [7], 'B': [65]}})
dict
intermediate第3步 - 从后面的词典中构建>>> intermediate
defaultdict(list, {'A': [1, 7], 'B': [15, 65]})
词典
final