Question

I'm trying to analyse some results I obtained from an experience. I have a large dataframe formatted the following way :

Name        V1      V2      V3

X1247       1       9       0
X1247.1     1       7       0
X1247.10    2       2       4
X1247.100   8       3       7
X874        3       7       1
X874.1      4       0       1
X66         8       9       1   
X66.1       0       1       8

I'm looking to get a dataframe that will merge the rows that have the same name before the ".", so that it would look like this :

Name        V1      V2      V3

X1247       12      21      11
X874        7       7       2
X66         8       10      9

How could this be done? This may be trivial but I'm more a biologist than a computer scientist. Thanks

Answer 1

We group by the substring of 'Name' (str_remove from stringr) and use summarise_all to get the output

library(tidyverse)
df1 %>%
   group_by(Name = str_remove(Name, "\\..*")) %>%
   summarise_all(sum)
# A tibble: 3 x 4
#  Name     V1    V2    V3
#  <chr> <int> <int> <int>
#1 X1247    12    21    11
#2 X66       8    10     9
#3 X874      7     7     2

Or using base R with aggregate and sub

aggregate(.~ Name, transform(df1, Name = sub("\\..*", "", Name)), FUN = sum)

data

df1 <- structure(list(Name = c("X1247", "X1247.1", "X1247.10", "X1247.100", 
"X874", "X874.1", "X66", "X66.1"), V1 = c(1L, 1L, 2L, 8L, 3L, 
4L, 8L, 0L), V2 = c(9L, 7L, 2L, 3L, 7L, 0L, 9L, 1L), V3 = c(0L, 
0L, 4L, 7L, 1L, 1L, 1L, 8L)), class = "data.frame", row.names = c(NA, 
-8L))

Answer 2

data.table解决方案：

data.table::setDT(df1)[,as.list(colSums(.SD)), by=sub("\\.[^.]*", "", Name),][]
#     sub V1 V2 V3
#1: X1247 12 21 11
#2:  X874  7  7  2
#3:   X66  8 10  9

数据：（从akrun借来）

基准：

df1 <- do.call(rbind,rep(list(mtcars),1000))
df1 <- cbind(Name = paste0(rownames(df1),".HAHA"), df1)

f1 <- function(df1) {
    setDT(df1)[,as.list(colSums(.SD)), by=sub("\\.[^.]*", "", Name),][]
}

f2 <- function(df1) { 
    df1 %>% 
        group_by(Name = str_remove(Name, "\\..*")) %>%
        summarise_all(sum)
}
f3 <- function(df1) { 
    aggregate(.~ Name, transform(df1, Name = sub("\\..*", "", Name)), FUN = sum)
    colSums(df1[,-1])
}

microbenchmark::microbenchmark(f1(df1),f2(df1),f3(df1),times=5)

#Unit: milliseconds
#    expr      min        lq      mean    median        uq       max neval cld
 #f1(df1) 2982.691 3012.7485 3098.9819 3102.7779 3175.1407 3221.5515     5   c
 #f2(df1)  292.278  295.4829  299.7829  296.5712  301.3588  313.2236     5 a  
 #f3(df1) 2244.550 2254.7699 2318.5398 2256.9791 2274.2181 2562.1817     5  b

令人惊讶的是，@ akrun dplyr解决方案的速度快了10倍……至少让我感到惊讶。

Merge rows with same pattern name in dataframe R

2 个答案:

data