R：计数的一个因素的累积长度data.frame

-1

Time = c("2016-03-01","2016-03-02","2016-03-03","2016-03-02","2016-03-03","2016-03-02") 
match = c("a","b","c","a","b","c") 
names = c("julien","julien","julien", "mathieu","mathieu","simon") 
df = data.frame(Time, names, match) 
df = df[order(Time),] 
df 
     Time names match 
1 2016-03-01 julien  a 
2 2016-03-02 julien  b 
4 2016-03-02 mathieu  a 
6 2016-03-02 simon  c 
3 2016-03-03 julien  c 
5 2016-03-03 mathieu  b

而且我想通过时间由每个玩家扮演一个新列匹配的累计数。我想知道，在任何时候，每个球员打了多少比赛。像这样：

 Time names match nb.of.match.played 
1 2016-03-01 julien  a     1 
2 2016-03-02 julien  b     2 
4 2016-03-02 mathieu  a     1 
6 2016-03-02 simon  c     1 
3 2016-03-03 julien  c     3 
5 2016-03-03 mathieu  b     2

这似乎很容易做到，但我尝试了一次失败的结果的事情每次。感谢您的帮助！

来源

2016-03-29 Julien Céré

显示一些尽可能地和它 – IceFire

对不起，我没有任何有形的铅（代码段）值得炫耀的，我们可以看看你的工作代码段的....我想这解决方案：http://stackoverflow.com/questions/22843286/r-cumulative-count-over-multiple-columns-by-factor但我不认为目标是相似的。 –

我解决我的问题与趋势cumsum using ddply

但我认为cumsum不符合的因素长度工作，所以我有“1”的列上cumsum可以工作。

Time = c("2016-03-01","2016-03-02","2016-03-03","2016-03-02","2016-03-03","2016-03-02") 
match = c("a","b","c","a","b","c") 
names = c("julien","julien","julien", "mathieu","mathieu","simon") 
df = data.frame(Time, names, match) 
df = df[order(Time),] 
df$nb = 1 
df 
     Time names match nb 
1 2016-03-01 julien  a 1 
2 2016-03-02 julien  b 1 
4 2016-03-02 mathieu  a 1 
6 2016-03-02 simon  c 1 
3 2016-03-03 julien  c 1 
5 2016-03-03 mathieu  b 1 

within(df, { 
    nb.match <- ave(nb, names, FUN = cumsum) 
}) 
df 
     Time names match nb nb.match 
1 2016-03-01 julien  a 1  1 
2 2016-03-02 julien  b 1  2 
4 2016-03-02 mathieu  a 1  1 
6 2016-03-02 simon  c 1  1 
3 2016-03-03 julien  c 1  3 
5 2016-03-03 mathieu  b 1  2

来源

2016-03-31 14:15:32

R：计数的一个因素的累积长度data.frame

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