在世界杯组内生成配对
在世界杯组内生成配对
提问于 2015-06-02 21:04:47
我收集了一些2015年国际足联女足世界杯的数据：
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()
现在，我想生成一个新的数据格式，其中包含来自group的每个df中的6个可能的配对或匹配，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
... 做这件事的简明扼要的方法是什么？我可以在每个group和team中执行一串循环，但是我觉得应该有一种更清晰的矢量化方法来使用pandas和分割-应用-联合范例来实现这一点。
编辑：我也欢迎任何R的答案，认为在这里比较R和Pandas的方式会很有趣。添加了r标记。
以下是R格式的数据，如注释中所要求的那样：
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)回答 2
Stack Overflow用户
回答已采纳
发布于 2015-06-02 21:23:23
以下是两行解决方案：
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')]解释：
首先，我们使用df.groupby('group')获得每个组中的团队列表，迭代并访问其'team‘系列，以获得每个组中的4个团队的列表：
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，我们生成了全场所有的元组的列表。大卫·阿伦堡的帖子提醒我要使用itertools.combinations(..., 2)。但我们可以使用生成器或嵌套的for-循环：
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')]注意，我们正在采取一种捷径，首先生成所有可能的索引元组，然后使用这些元组索引到工作组列表中：
>>> 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)](注:如果我们使用直接比较团队名称的方法，就会产生(按字母顺序)组名的轻微副作用(它们最初是按种子排序，而不是按字母顺序排序)。‘中国’<‘荷兰’，所以他们的配对将出现在(‘荷兰’，‘中国’)而不是(‘中国’，荷兰)
Stack Overflow用户
发布于 2015-06-02 21:34:35
使用R，这里有一个可能的data.table解决方案，使用它在GitHub上的开发版本
#### 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
...页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/30606566
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