有效迭代3.311031748 E+；Python中的12个组合
我收集了一个庞大的口袋妖怪数据集，我的目标是根据我构建的比率来确定“十大团队”——口袋妖怪BST（基本统计总数）：平均弱点。对于那些关心的人，我将平均弱点计算为每个类型的口袋妖怪弱点之和（0.25到飞行+1到水+2到钢铁+4到火，等等），然后除以18（游戏中可用的类型总数） 举一个简单的例子——由以下三个口袋妖怪组成的团队：金勒、米米丘、马格尼宗，团队比例为1604.1365384615383 由于数据将用于竞争性游戏，我删除了所有未完全进化的口袋妖怪以及传奇/神话中的口袋妖怪。以下是我目前的流程：有效迭代3.311031748 E+；Python中的12个组合,python,loops,bigdata,combinations,Python,Loops,Bigdata,Combinations,我收集了一个庞大的口袋妖怪数据集，我的目标是根据我构建的比率来确定“十大团队”——口袋妖怪BST（基本统计总数）：平均弱点。对于那些关心的人，我将平均弱点计算为每个类型的口袋妖怪弱点之和（0.25到飞行+1到水+2到钢铁+4到火，等等），然后除以18（游戏中可用的类型总数） 举一个简单的例子——由以下三个口袋妖怪组成的团队：金勒、米米丘、马格尼宗，团队比例为1604.1365384615383 由于数据将用于竞争性游戏，我删除了所有未完全进化的口袋妖怪以及传奇/神话中的口袋妖怪。以下是我目前的流
import itertools
import pandas as pd
df = pd.read_csv("Downloads/pokemon.csv") # read in csv of fully-evolved Pokemon data
# list(df) # list of df column names - useful to see what data has been collected
df = df[df["is_legendary"] == 0] # remove legendary pokemon - many legendaries are allowed in competitive play
df = df[['abilities', # trim df to contain only the columns we care about
'against_bug',
'against_dark',
'against_dragon',
'against_electric',
'against_fairy',
'against_fight',
'against_fire',
'against_flying',
'against_ghost',
'against_grass',
'against_ground',
'against_ice',
'against_normal',
'against_poison',
'against_psychic',
'against_rock',
'against_steel',
'against_water',
'attack',
'defense',
'hp',
'name',
'sp_attack',
'sp_defense',
'speed',
'type1',
'type2']]
df["bst"] = df["hp"] + df["attack"] + df["defense"] + df["sp_attack"] + df["sp_defense"] + df["speed"] # calculate BSTs
df['average_weakness'] = (df['against_bug'] # calculates a Pokemon's 'average weakness' to other types
+ df['against_dark']
+ df['against_dragon']
+ df['against_electric']
+ df['against_fairy']
+ df['against_fight']
+ df['against_fire']
+ df['against_flying']
+ df['against_ghost']
+ df['against_grass']
+ df['against_ground']
+ df['against_ice']
+ df['against_normal']
+ df['against_poison']
+ df['against_psychic']
+ df['against_rock']
+ df['against_steel']
+ df['against_water']) / 18
df['bst-weakness-ratio'] = df['bst'] / df['average_weakness'] # ratio of BST:avg weakness - the higher the better
names = df["name"] # pull out list of all names for creating combinations
combinations = itertools.combinations(names, 6) # create all possible combinations of 6 pokemon teams
top_10_teams = [] # list for storing top 10 teams
for x in combinations:
ratio = sum(df.loc[df['name'].isin(x)]['bst-weakness-ratio']) # pull out sum of team's ratio
if(len(top_10_teams) != 10):
top_10_teams.append((x, ratio)) # first 10 teams will automatically populate list
else:
top_10_teams.append((x, ratio)) # add team to list
top_10_teams.sort(key=lambda x:x[1], reverse=True) # sort list by descending ratios
del top_10_teams[-1] # drop team with the lowest ratio - only top 10 remain in list
top_10_teams
在您的示例中，每个口袋妖怪都有一个bst_弱点比率，在计算团队价值时，您没有考虑成员之间相互抵消弱点，而是简单地将6个成员的比率相加？如果是这样的话，最好的团队不应该是拥有6个最好的口袋妖怪的团队吗？我不明白你为什么需要这些组合 尽管如此，我想在进入组合数学之前，你可以从你的列表中删除很多口袋妖怪。 如果您有一个布尔数组（n_pokemons，n_types），用True表示每个Pokemon的弱点，那么您可以检查是否有一个Pokemon具有相同的弱点，但具有更好的bst值
# Loop over all pokemon and check if there are other pokemon
# ... with the exact same weaknesses but better stats
# -name -weaknesses -bst
# pokemon A [0, 0, 1, 1, 0, ...], bst=34.85 -> delete A
# pokemon B [0, 0, 1, 1, 0, ...], bst=43.58
# ... with a subset of the weaknesses and better stats
# pokemon A [0, 0, 1, 1, 0, ...], bst=34.85 -> delete A
# pokemon B [0, 0, 1, 0, 0, ...], bst=43.58
n_pokemons = 1000
n_types = 18
n_min_weaknesses = 1 # number of minimal and maximal weaknesses for each Pokemon
n_max_weaknesses = 4
import numpy as np
# Generate pokemons
name_arr = np.array(['pikabra_{}'.format(i) for i in range(n_pokemons)])
# Random stats
bst_arr = np.random.random(n_pokemons) * 100
# Random weaknesses
weakness_array = np.zeros((n_pokemons, n_types), dtype=bool) # bool array indicating the weak types of each pokemon
for i in range(n_pokemons):
rnd_weaknesses = np.random.choice(np.arange(n_types), np.random.randint(n_min_weaknesses, n_max_weaknesses+1))
weakness_array[i, rnd_weaknesses] = True
# Remove unnecessary pokemons
i = 0
while i < n_pokemons:
j = i + 1
while j < n_pokemons:
del_idx = None
combined_weaknesses = np.logical_or(weakness_array[i], weakness_array[j])
if np.all(weakness_array[i] == weakness_array[j]):
if bst_arr[j] < bst_arr[i]:
del_idx = i
else:
del_idx = j
elif np.all(combined_weaknesses == weakness_array[i]) and bst_arr[j] < bst_arr[i]:
del_idx = i
elif np.all(combined_weaknesses == weakness_array[j]) and bst_arr[i] < bst_arr[j]:
del_idx = j
if del_idx is not None:
name_arr = np.delete(name_arr, del_idx, axis=0)
bst_arr = np.delete(bst_arr, del_idx, axis=0)
weakness_array = np.delete(weakness_array, del_idx, axis=0)
n_pokemons -= 1
if del_idx == i:
i -= 1
break
else:
j -= 1
j += 1
i += 1
print(n_pokemons)
将numpy导入为np
#生成口袋妖怪
name_arr=np.array（['pikabra_{}'。范围内i的格式（i）（n_pokemons）]）
#随机统计
bst_arr=np.random.random（n_口袋妖怪）*100
#随机弱点
弱点数组=np.zero（（n个口袋妖怪，n个类型），dtype=bool）#bool数组指示每个口袋妖怪的弱点类型
对于范围内的i（n_口袋妖怪）：
rnd_弱点=np.random.choice（np.arange（n_类型），np.random.randint（n_最小弱点，n_最大弱点+1））
弱点数组[i，rnd\u弱点]=真
#移除不必要的口袋妖怪
i=0
而我NumPypandas