I am working with a huge dataset, but just to simplify what I'd like to do, I will use the following one.
testDF <- data.frame(v1 = rep(c('a', 'b', 'c', 'd', 'e', 'f'), 2),
v2 = rep(c(1,0),6))
Let's assume you could subset it like this.
v1 v2
1 a 1
2 b 0
3 c 1
4 d 0
5 e 1
6 f 0
7 a 1
8 b 0
9 c 1
10 d 0
11 e 1
12 f 0
When the first value of v1 assumes the same value (for example in the I would like to add a third column reporting sum of the second column values. The output will be like this:
testDF
v1 v2 tc
1 a 1 2
2 b 0 0
3 c 1 2
4 d 0 0
5 e 1 2
6 f 0 0
7 a 1 2
8 b 0 0
9 c 1 2
10 d 0 0
11 e 1 2
Which operation I could by perpetuating the dplyr code?
Thanks
CodePudding user response：
A similar approach to @jpsmith, except using map
testDF %>%
group_split(v1) %>%
map(~if(nrow(.x) > 1) mutate(.x, v3 = sum(v2)) else .x)
#> [[1]]
#> # A tibble: 1 x 2
#> v1 v2
#> <dbl> <dbl>
#> 1 1 1
#>
#> [[2]]
#> # A tibble: 2 x 3
#> v1 v2 v3
#> <dbl> <dbl> <dbl>
#> 1 3 0 0
#> 2 3 0 0
#>
#> [[3]]
#> # A tibble: 1 x 2
#> v1 v2
#> <dbl> <dbl>
#> 1 4 1
#>
#> [[4]]
#> # A tibble: 3 x 3
#> v1 v2 v3
#> <dbl> <dbl> <dbl>
#> 1 5 1 2
#> 2 5 1 2
#> 3 5 0 2
#>
#> [[5]]
#> # A tibble: 2 x 3
#> v1 v2 v3
#> <dbl> <dbl> <dbl>
#> 1 7 0 1
#> 2 7 1 1
#>
#> [[6]]
#> # A tibble: 1 x 2
#> v1 v2
#> <dbl> <dbl>
#> 1 8 1
#>
#> [[7]]
#> # A tibble: 1 x 2
#> v1 v2
#> <dbl> <dbl>
#> 1 9 0
#>
#> [[8]]
#> # A tibble: 1 x 2
#> v1 v2
#> <dbl> <dbl>
#> 1 NA 0
Ā
#> Error in eval(expr, envir, enclos): object 'A' not found
Created on 2022-11-06 with reprex v2.0.2
CodePudding user response：
If I am understanding you correctly, you could try this brute-force for loop, though I am sure there are more elegant solutions.
library(dplyr)
test_list <- testDF %>%
group_split(v1)
# Convert list elements to df
test_list_df <- lapply(test_list, as.data.frame)
# If more than one row, create new variable ("third_column") and sum v2
for(xx in seq_along(test_list_df)){
if(nrow(test_list[[xx]]) > 1){
test_list_df[[xx]]$third_column <- sum(test_list_df[[xx]]["v2"])
}}
(I am a bit unclear as to what "For each subset, when the first value of v1 is the same..." - I took that to mean if there was more than one row in the subset.)
Output
# [[1]]
# v1 v2
# 1 1 1
#
# [[2]]
# v1 v2 third_column
# 1 3 0 0
# 2 3 0 0
#
# [[3]]
# v1 v2
# 1 4 1
#
# [[4]]
# v1 v2 third_column
# 1 5 1 2
# 2 5 1 2
# 3 5 0 2
#
# [[5]]
# v1 v2 third_column
# 1 7 0 1
# 2 7 1 1
#
# [[6]]
# v1 v2
# 1 8 1
#
# [[7]]
# v1 v2
# 1 9 0
#
# [[8]]
# v1 v2
# 1 NA 0
CodePudding user response：
Is this what you need?
testDF %>%
group_by(v1) %>%
mutate(third_col = sum(v2, na.rm = TRUE)) %>%
arrange(v1)
# A tibble: 12 × 3
# Groups: v1 [8]
v1 v2 third_col
<dbl> <dbl> <dbl>
1 1 1 1
2 3 0 0
3 3 0 0
4 4 1 1
5 5 1 2
6 5 1 2
7 5 0 2
8 7 0 1
9 7 1 1
10 8 1 1
11 9 0 0
12 NA 0 0
If you need to split by group, then just replace arrange with group_split:
testDF %>%
group_by(v1) %>%
mutate(third_col = sum(v2, na.rm = TRUE)) %>%
group_split(v1)
<list_of<
tbl_df<
v1 : double
v2 : double
third_col: double
>
>[8]>
[[1]]
# A tibble: 1 × 3
v1 v2 third_col
<dbl> <dbl> <dbl>
1 1 1 1
[[2]]
# A tibble: 2 × 3
v1 v2 third_col
<dbl> <dbl> <dbl>
1 3 0 0
2 3 0 0
[[3]]
# A tibble: 1 × 3
v1 v2 third_col
<dbl> <dbl> <dbl>
1 4 1 1
[[4]]
# A tibble: 3 × 3
v1 v2 third_col
<dbl> <dbl> <dbl>
1 5 1 2
2 5 1 2
3 5 0 2
[[5]]
# A tibble: 2 × 3
v1 v2 third_col
<dbl> <dbl> <dbl>
1 7 0 1
2 7 1 1
[[6]]
# A tibble: 1 × 3
v1 v2 third_col
<dbl> <dbl> <dbl>
1 8 1 1
[[7]]
# A tibble: 1 × 3
v1 v2 third_col
<dbl> <dbl> <dbl>
1 9 0 0
[[8]]
# A tibble: 1 × 3
v1 v2 third_col
<dbl> <dbl> <dbl>
1 NA 0 0