我雖然這很簡單，但不幸的是，事實(shí)并非如此。我正在嘗試構(gòu)建一個(gè)函數(shù)來獲取可迭代的字典（即唯一字典列表）并返回字典唯一分組列表的列表。如果我有x球員，我想組建規(guī)模k較大的團(tuán)隊(duì)n。這個(gè)問題和來自 CMSDK 的一組答案是我能找到的最接近解決方案的東西。在從處理字母字符串到字典的過程中，我發(fā)現(xiàn)我的 Python 技能不足。我正在改編的原始函數(shù)來自第二個(gè)答案：import itertools as itdef unique_group(iterable, k, n): """Return an iterator, comprising groups of size `k` with combinations of size `n`.""" # Build separate combinations of `n` characters groups = ("".join(i) for i in it.combinations(iterable, n)) # 'AB', 'AC', 'AD', ... # Build unique groups of `k` by keeping the longest sets of characters return (i for i in it.product(groups, repeat=k) if len(set("".join(i))) == sum((map(len, i)))) # ('AB', 'CD'), ('AB', 'CE'), ... 我目前的改編（TypeError: object of type 'generator' has no len()由于調(diào)用，完全失敗并出現(xiàn)錯(cuò)誤map(len, i)）：def unique_group(iterable, k, n): groups = [] groups.append((i for i in it.combinations(iterable, n))) return ( i for i in it.product(groups, repeat=k) if len(set(i)) == sum((map(len, i))) )對(duì)于一些上下文：我試圖根據(jù)他們的技能以編程方式將一組玩家劃分為圣誕節(jié)瑣事的團(tuán)隊(duì)。字典列表由一個(gè) yaml 文件組成，看起來像- name: Patricia skill: 4- name: Christopher skill: 6- name: Nicholas skill: 7- name: Bianca skill: 4其中，在yaml.load生成字典列表之后：players = [{'name':'Patricia', 'skill':4},{'name':'Christopher','skill':6}, {'name':'Nicholas','skill':7},{'name':'Bianca','skill':4}]所以我希望輸出看起來像這些（ wherek = 2和n = 2）的列表：( # Team assignment grouping 1 ( # Team 1 ( {'name': 'Patricia', 'skill': 4}, {'name': 'Christopher', 'skill': 6} ), # Team 2 ( {'name': 'Nicholas', 'skill': 7}, {'name': 'Bianca', 'skill': 4} ) ),)每個(gè)團(tuán)隊(duì)分配分組需要在團(tuán)隊(duì)中具有唯一的球員（即，團(tuán)隊(duì)分配分組中的多個(gè)團(tuán)隊(duì)中不能有相同的球員），并且每個(gè)團(tuán)隊(duì)分配分組都需要是唯一的。一旦我有了團(tuán)隊(duì)分配組合列表，我將總結(jié)每個(gè)組的技能，取最高技能和最低技能之間的差異，并選擇最高技能和最低技能之間差異最小的分組（有方差）。我承認(rèn)我不完全理解這段代碼。我理解第一個(gè)任務(wù)是創(chuàng)建一個(gè)字符串中所有字母組合的列表，以及在產(chǎn)品不包含不同組中相同字母的情況下查找產(chǎn)品的返回語句。我最初的嘗試是簡單地采用 ，it.product(it.combinations(iterable, n), repeat=k)但這并不能實(shí)現(xiàn)跨組的唯一性（即，我在一個(gè)分組的不同團(tuán)隊(duì)中獲得相同的球員）。
6 回答
慕桂英4014372
TA貢獻(xiàn)1871條經(jīng)驗(yàn) 獲得超13個(gè)贊
這是我利用集合的新數(shù)據(jù)類的地方。您可以通過frozen=True在裝飾器中設(shè)置來使數(shù)據(jù)類可散列。首先，您要將玩家添加到一個(gè)集合中以獲得獨(dú)特的玩家。然后你會(huì)得到 n 大小球隊(duì)的所有球員組合。然后你可以創(chuàng)建一組獨(dú)特的團(tuán)隊(duì)。然后創(chuàng)建有效的分組，而沒有球員在團(tuán)隊(duì)中出現(xiàn)多次。最后，您可以計(jì)算整個(gè)分組的總團(tuán)隊(duì)技能水平的最大差異（再次利用組合），并使用它來對(duì)您的有效分組進(jìn)行排序。所以像這樣的事情。
from dataclasses import dataclass
from itertools import combinations
from typing import FrozenSet
import yaml
@dataclass(order=True, frozen=True)
class Player:
name: str
skill: int
@dataclass(order=True, frozen=True)
class Team:
members: FrozenSet[Player]
def total_skill(self):
return sum(p.skill for p in self.members)
def is_valid(grouping):
players = set()
for team in grouping:
for player in team.members:
if player in players:
return False
players.add(player)
return True
def max_team_disparity(grouping):
return max(
abs(t1.total_skill() - t2.total_skill())
for t1, t2 in combinations(grouping, 2)
)
def best_team_matchups(player_file, k, n):
with open(player_file) as f:
players = set(Player(p['name'], p['skill']) for p in yaml.load(f))
player_combs = combinations(players, n)
unique_teams = set(Team(frozenset(team)) for team in player_combs)
valid_groupings = set(g for g in combinations(unique_teams, k) if is_valid(g))
for g in sorted(valid_groupings, key=max_team_disparity):
print(g)
best_team_matchups('test.yaml', k=2, n=4)
示例輸出：
(
Team(members=frozenset({
Player(name='Chr', skill=6),
Player(name='Christopher', skill=6),
Player(name='Nicholas', skill=7),
Player(name='Patricia', skill=4)
})),
Team(members=frozenset({
Player(name='Bia', skill=4),
Player(name='Bianca', skill=4),
Player(name='Danny', skill=8),
Player(name='Nicho', skill=7)
}))
)
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