按组选择具有最小值的行 [英] Select rows with min value by group

问题描述

我遇到了一个困扰我一段时间的问题……希望这里的任何人都能帮助我。

I got a problems that bugs me for some time… hopefully anybody here can help me.

我得到了以下数据框

f <- c('a','a','b','b','b','c','d','d','d','d')
v1 <- c(1.3,10,2,10,10,1.1,10,3.1,10,10)
v2 <- c(1:10)
df <- data.frame(f,v1,v2)

f是一个因素； v1和v2是值。
对于f的每个级别，我只想保留一行：在该因子级别中具有v1最低值的那一行。

f is a factor; v1 and v2 are values. For each level of f, I want only want to keep one row: the one that has the lowest value of v1 in this factor level.

f   v1  v2
a   1.3 1
b   2   3
c   1.1 6
d   3.1 8

我尝试了各种不同的操作，包括合计，ddply，by，tapply……但似乎没有任何效果。如有任何建议，我将非常感谢。

I tried various things with aggregate, ddply, by, tapply… but nothing seems to work. For any suggestions, I would be very thankful.

推荐答案

使用DWin解决方案，使用<$ c可以避免 tapply $ c> ave 。

Using DWin's solution, tapply can be avoided using ave.

df[ df$v1 == ave(df$v1, df$f, FUN=min), ]

这可以进一步提高速度，如下所示。请注意，这也取决于级别数。我给出这个信息是因为我注意到 ave 经常被遗忘，尽管它是R中功能更强大的功能之一。

This gives another speed-up, as shown below. Mind you, this is also dependent on the number of levels. I give this as I notice that ave is far too often forgotten about, although it is one of the more powerful functions in R.

f <- rep(letters[1:20],10000)
v1 <- rnorm(20*10000)
v2 <- 1:(20*10000)
df <- data.frame(f,v1,v2)

> system.time(df[ df$v1 == ave(df$v1, df$f, FUN=min), ])
   user  system elapsed 
   0.05    0.00    0.05 

> system.time(df[ df$v1 %in% tapply(df$v1, df$f, min), ])
   user  system elapsed 
   0.25    0.03    0.29 

> system.time(lapply(split(df, df$f), FUN = function(x) {
+             vec <- which(x[3] == min(x[3]))
+             return(x[vec, ])
+         })
+  .... [TRUNCATED] 
   user  system elapsed 
   0.56    0.00    0.58 

> system.time(df[tapply(1:nrow(df),df$f,function(i) i[which.min(df$v1[i])]),]
+ )
   user  system elapsed 
   0.17    0.00    0.19 

> system.time( ddply(df, .var = "f", .fun = function(x) {
+     return(subset(x, v1 %in% min(v1)))
+     }
+ )
+ )
   user  system elapsed 
   0.28    0.00    0.28 

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