如何根據 r 中的不同值組合行？

Question

我想總結某些行的值。 例如，雖然 ID 相同，但我們將對字母 a 和 b 時的值求和。

  ID <- c(1,1,1,8,8,8,2,2,2,5,5,5,7,7,7)
  letter <- c("a","b","c","a","b","c","a","b","c","a","b","c","a","b","c")
  value <- c(111,75,94,37,85,86,76,55,72,62,56,35,85,44,65)
  dataframe <- data.frame(ID,letter,value)

Output如下

   ID letter value
1   1      a   111
2   1      b    75
3   1      c    94
4   8      a    37
5   8      b    85
6   8      c    86
7   2      a    76
8   2      b    55
9   2      c    72
10  5      a    62
11  5      b    56
12  5      c    35
13  7      a    85
14  7      b    44
15  7      c    65

Answer 1

使用dplyr -

library(dplyr)

dataframe %>%
  group_by(ID) %>%
  summarise(total = sum(value[letter %in% c('a', 'b')]))

#    ID total
#  <dbl> <dbl>
#1     1   186
#2     2   131
#3     5   118
#4     7   129
#5     8   122

Answer 2

data.table

library(data.table)
setDT(dataframe)
dataframe[letter %in% c("a","b"), .(ab_sum = sum(value)), by=ID]

或 tidyverse/dplyr：

library(magrittr)
library(dplyr)

dataframe %>% 
  filter(letter %in% c("a","b")) %>%
  group_by(ID) %>%
  summarize(ab_sum = sum(value))

Answer 3

使用dplyr你可以做到這一點

library(dplyr)

dataframe %>% 
  group_by(ID) %>% 
  mutate(sum=sum(ifelse(letter=="a"|letter=="b",value,NA),na.rm=T)) %>% 
  ungroup()
# A tibble: 15 × 4
      ID letter value   sum
   <dbl> <chr>  <dbl> <dbl>
 1     1 a        111   186
 2     1 b         75   186
 3     1 c         94   186
 4     8 a         37   122
 5     8 b         85   122
 6     8 c         86   122
 7     2 a         76   131
 8     2 b         55   131
 9     2 c         72   131
10     5 a         62   118
11     5 b         56   118
12     5 c         35   118
13     7 a         85   129
14     7 b         44   129
15     7 c         65   129

或更緊湊

dataframe %>% 
  group_by(ID) %>% 
  summarise(sum=sum(ifelse(letter=="a"|letter=="b",value,NA),na.rm=T))
# A tibble: 5 × 2
     ID   sum
  <dbl> <dbl>
1     1   186
2     2   131
3     5   118
4     7   129
5     8   122