在世界杯比賽組內生成配對

Question

我匯總了2015年FIFA女足世界杯的一些數據：

import pandas as pd

df = pd.DataFrame({
    'team':['Germany','USA','France','Japan','Sweden','England','Brazil','Canada','Australia','Norway','Netherlands','Spain',
       'China','New Zealand','South Korea','Switzerland','Mexico','Colombia','Thailand','Nigeria','Ecuador','Ivory Coast','Cameroon','Costa Rica'],
    'group':['B','D','F','C','D','F','E','A','D','B','A','E','A','A','E','C','F','F','B','D','C','B','C','E'],
    'fifascore':[2168,2158,2103,2066,2008,2001,1984,1969,1968,1933,1919,1867,1847,1832,1830,1813,1748,1692,1651,1633,1485,1373,1455,1589],
    'ftescore':[95.6,95.4,92.4,92.7,91.6,89.6,92.2,90.1,88.7,88.7,86.2,84.7,85.2,82.5,84.3,83.7,81.1,78.0,68.0,85.7,63.3,75.6,79.3,72.8]
    })

df.groupby(['group', 'team']).mean()

現在，我想生成一個新的數據框，其中包含來自df每個group中的6種可能的配對或匹配，格式如下：

group    team1        team2
A        Canada       China
A        Canada       Netherlands
A        Canada       New Zealand
A        China        Netherlands
A        China        New Zealand
A        Netherlands  New Zealand
B        Germany      Ivory Coast
B        Germany      Norway
...     

一種簡潔明了的方法是什么？ 我可以在每個group和team進行一堆循環，但是我覺得應該有一種更簡潔的矢量化方法來處理pandas和拆分應用組合模式。

編輯：我也歡迎任何R的答案，認為在這里比較R和Pandas的方式會很有趣。 添加了r標簽。

以下是“注釋”中要求的R形式的數據：

team <- c('Germany','USA','France','Japan','Sweden','England','Brazil','Canada','Australia','Norway','Netherlands','Spain',
      'China','New Zealand','South Korea','Switzerland','Mexico','Colombia','Thailand','Nigeria','Ecuador','Ivory Coast','Cameroon','Costa Rica')
group <- c('B','D','F','C','D','F','E','A','D','B','A','E','A','A','E','C','F','F','B','D','C','B','C','E')
fifascore <- c(2168,2158,2103,2066,2008,2001,1984,1969,1968,1933,1919,1867,1847,1832,1830,1813,1748,1692,1651,1633,1485,1373,1455,1589)
ftescore <- c(95.6,95.4,92.4,92.7,91.6,89.6,92.2,90.1,88.7,88.7,86.2,84.7,85.2,82.5,84.3,83.7,81.1,78.0,68.0,85.7,63.3,75.6,79.3,72.8)

df <- data.frame(team, group, fifascore, ftescore)

Answer 1

這是兩行解決方案：

import itertools

for grpname,grpteams in df.groupby('group')['team']:
    # No need to use grpteams.tolist() to convert from pandas Series to Python list
    print list(itertools.combinations(grpteams, 2))

[('Canada', 'Netherlands'), ('Canada', 'China'), ('Canada', 'New Zealand'), ('Netherlands', 'China'), ('Netherlands', 'New Zealand'), ('China', 'New Zealand')]
[('Germany', 'Norway'), ('Germany', 'Thailand'), ('Germany', 'Ivory Coast'), ('Norway', 'Thailand'), ('Norway', 'Ivory Coast'), ('Thailand', 'Ivory Coast')]
[('Japan', 'Switzerland'), ('Japan', 'Ecuador'), ('Japan', 'Cameroon'), ('Switzerland', 'Ecuador'), ('Switzerland', 'Cameroon'), ('Ecuador', 'Cameroon')]
[('USA', 'Sweden'), ('USA', 'Australia'), ('USA', 'Nigeria'), ('Sweden', 'Australia'), ('Sweden', 'Nigeria'), ('Australia', 'Nigeria')]
[('Brazil', 'Spain'), ('Brazil', 'South Korea'), ('Brazil', 'Costa Rica'), ('Spain', 'South Korea'), ('Spain', 'Costa Rica'), ('South Korea', 'Costa Rica')]
[('France', 'England'), ('France', 'Mexico'), ('France', 'Colombia'), ('England', 'Mexico'), ('England', 'Colombia'), ('Mexico', 'Colombia')]

---

說明：

首先，我們使用df.groupby('group')獲得每個組中團隊的團隊列表，對其進行迭代並訪問其“團隊”系列，以獲取每個組中4個團隊的列表：

for grpname,grpteams in df.groupby('group')['team']:
    teamlist = grpteams.tolist()
... 
['Canada', 'Netherlands', 'China', 'New Zealand']
['Germany', 'Norway', 'Thailand', 'Ivory Coast']
['Japan', 'Switzerland', 'Ecuador', 'Cameroon']
['USA', 'Sweden', 'Australia', 'Nigeria']
['Brazil', 'Spain', 'South Korea', 'Costa Rica']
['France', 'England', 'Mexico', 'Colombia']

然后，我們生成團隊元組的全部播放列表。 David Arenburg的帖子提醒我使用itertools.combinations(..., 2) 。 但是我們可以使用生成器或嵌套的for循環：

def all_play_all(teams):
  for team1 in teams:
    for team2 in teams:
      if team1 < team2: # [Note] We don't need to generate indices then index into teamlist, just use direct string comparison
        yield (team1,team2)

>>> [match for match in all_play_all(grpteams)]
[('France', 'Mexico'), ('England', 'France'), ('England', 'Mexico'), ('Colombia', 'France'), ('Colombia', 'England'), ('Colombia', 'Mexico')]

請注意，我們采用了一種捷徑，首先生成所有可能的索引元組，然后使用這些元組將其索引到團隊列表中：

>>> T = len(teamlist) + 1
>>> [(i,j) for i in range(T) for j in range(T) if i<j]
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]

（請注意：如果我們使用了直接比較球隊名稱的方法，那么會對（按字母順序）重新使用組名（它們最初是由種子而不是按字母順序排序）產生輕微的副作用，例如，“ China” <'荷蘭”，因此他們的配對將顯示為（'荷蘭'，'中國'）而不是（'中國'，荷蘭'）

Answer 2

使用R，這是在GitHub上使用開發版本的可能的data.table解決方案

#### To install development version
## library(devtools)
## install_github("Rdatatable/data.table", build_vignettes = FALSE)

library(data.table) ## v >= 1.9.5
setDT(df)[, transpose(combn(team, 2L, simplify = FALSE)), keyby = group]
#    group          V1          V2
# 1:     A      Canada Netherlands
# 2:     A      Canada       China
# 3:     A      Canada New Zealand
# 4:     A Netherlands       China
# 5:     A Netherlands New Zealand
# 6:     A       China New Zealand
# 7:     B     Germany      Norway
# 8:     B     Germany    Thailand
...