# Only keep min value for each factor level

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

I got the following data frame

```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 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
```

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

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

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

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 == min(x))
+             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
```

A data.table solution.

```library(data.table)
DT <- as.data.table(df)
DT[,.SD[which.min(v1)], by = f]

##   f  v1 v2
## 1: a 1.3  1
## 2: b 2.0  3
## 3: c 1.1  6
## 4: d 3.1  8
```

Or, more efficiently

```DT[DT[,.I[which.min(v1)],by=f][['V1']]]
```

# some benchmarking

```f <- rep(letters[1:20],100000)
v1 <- rnorm(20*100000)
v2 <- 1:(20*100000)
df <- data.frame(f,v1,v2)
DT <- as.data.table(df)
f1<-function(){df2<-df[order(df\$f,df\$v1),]
df2[!duplicated(df2\$f),]}

f2<-function(){df2<-df[order(df\$v1),]
df2[!duplicated(df2\$f),]}

f3<-function(){df[ df\$v1 == ave(df\$v1, df\$f, FUN=min), ]}

f4 <- function(){DT[,.SD[which.min(v1)], by = f]}

f5 <- function(){DT[DT[,.I[which.min(v1)],by=f][['V1']]]}

library(microbenchmark)
microbenchmark(f1(),f2(),f3(),f4(), f5(),times = 5)
# Unit: milliseconds
# expr       min        lq    median        uq       max neval
# f1() 3254.6620 3265.4760 3286.5440 3411.4054 3475.4198     5
# f2() 1630.8572 1639.3472 1651.5422 1721.4670 1738.6684     5
# f3()  172.2639  174.0448  177.4985  179.9604  184.7365     5
# f4()  206.1837  209.8161  209.8584  210.4896  210.7893     5
# f5()  105.5960  106.5006  107.9486  109.7216  111.1286     5
```

The .I approach is the winner (FR #2330 will hopefully render the elegance of the .SD approach similarly fast when implemented).

With plyr, I'd use:

```ddply(df, .var = "f", .fun = function(x) {
return(subset(x, v1 %in% min(v1)))
}
)
```

Give that a try and see if it returns what you want.

Another tapply solution, with no unnecessary scanning of vector with %in%:

```df[tapply(1:nrow(df),df\$f,function(i) i[which.min(df\$v1[i])]),]
```

EDIT: This will left only first row in case of a tie.

```df[sapply(split(1:nrow(df),df\$f),function(x) x[which.min(df\$v1[x])]),]
```

On my machine (using Joris' benchmark data):

```> system.time(df[ df\$v1 == ave(df\$v1, df\$f, FUN=min), ])
user  system elapsed
0.022   0.000   0.021
> system.time(df[sapply(split(1:nrow(df),df\$f),function(x) x[which.min(df\$v1[x])]),])
user  system elapsed
0.006   0.000   0.007
```

Here's a tapply solution;

```> df[ df\$v1 %in% tapply(df\$v1, df\$f, min), ]

f  v1 v2
1 a 1.3  1
3 b 2.0  3
6 c 1.1  6
8 d 3.1  8
```

In your example it only picks out one per group, but if there were ties this method would show them all. (As would Parker's and Luštrik's I suspect.)

I'm sorry, my thinking power is depleted, and this ugly solution is all I can come up with at almost 1 am.

```lapply(split(df, df\$f), FUN = function(x) {
vec <- which(x == min(x))
return(x[vec, ])
})
```

Another way is to use order and !duplicated, but you would only get the first on ties.

```df2 <- df[order(df\$f,df\$v1),]
df2[!duplicated(df2\$f),]

f  v1 v2
1 a 1.3  1
3 b 2.0  3
6 c 1.1  6
8 d 3.1  8
```

Timings

```f1<-function(){df2<-df[order(df\$f,df\$v1),]
df2[!duplicated(df2\$f),]}

f2<-function(){df2<-df[order(df\$v1),]
df2[!duplicated(df2\$f),]}

f3<-function(){df[ df\$v1 == ave(df\$v1, df\$f, FUN=min), ]}

library(rbenchmark)
> benchmark(f1(),f2(),f3())
test replications elapsed relative user.self sys.self user.child sys.child
1 f1()          100   38.16 7.040590     36.66     1.48         NA        NA
2 f2()          100   20.54 3.789668     19.30     1.23         NA        NA
3 f3()          100    5.42 1.000000      4.96     0.46         NA        NA
```

Here is a solution with by

```do.call(rbind, unname(by(df, df\$f, function(x) x[x\$v1 == min(x\$v1),])))
##   f  v1 v2
## 1 a 1.3  1
## 3 b 2.0  3
## 6 c 1.1  6
## 8 d 3.1  8
```

This is the dplyr-way to filter for the minimum v1 values by groups of f:

```require(dplyr)
df %>%
group_by(f) %>%
filter(v1 == min(v1))

#Source: local data frame [4 x 3]
#Groups: f
#
#  f  v1 v2
#1 a 1.3  1
#2 b 2.0  3
#3 c 1.1  6
#4 d 3.1  8
```

In cases of ties in v1, this would result in multiple rows per group of f. If you want to avoid that, you can use:

```df %>%
group_by(f) %>%
filter(rank(v1, ties.method= "first") == 1)
```

This way, you'll only get the first row in case of ties. You could alternatively use ties.method = "random" or others as described in the help file.