我试图找出如何将values的{{1}}和属于category的{{1}}乘以因子file，同时还要保留原始数据。
library(dplyr)
df <- data.frame(ID = 1:20, values = runif(20), category = rep(letters[1:5], 4), file = as.factor(sort(rep(1:5, 4))))
ID values category file
1 1 0.65699229 a 1
2 2 0.70506478 b 1
3 3 0.45774178 c 1
4 4 0.71911225 d 1
5 5 0.93467225 e 1
6 6 0.25542882 a 2
7 7 0.46229282 b 2
8 8 0.94001452 c 2
9 9 0.97822643 d 2
10 10 0.11748736 e 2
11 11 0.47499708 a 3
12 12 0.56033275 b 3
13 13 0.90403139 c 3
14 14 0.13871017 d 3
15 15 0.98889173 e 3
16 16 0.94666823 a 4
17 17 0.08243756 b 4
18 18 0.51421178 c 4
19 19 0.39020347 d 4
20 20 0.90573813 e 4
这样
df[1,2]将被添加到df[2,2]的文件1的类别'ab' df[6,2]将被添加到df[7,2]的文件2的类别'ab'上到目前为止，我有这个：
df %>%
filter(category %in% c('a' , 'b')) %>%
group_by(file) %>%
summarise(values = sum(values))
我想将求和值的类别更改为“ ab”，并将其附加到同一管道中的原始数据帧中。
所需的输出：
ID values category file
1 1 0.65699229 a 1
2 2 0.70506478 b 1
3 3 0.45774178 c 1
4 4 0.71911225 d 1
5 5 0.93467225 e 1
6 6 0.25542882 a 2
7 7 0.46229282 b 2
8 8 0.94001452 c 2
9 9 0.97822643 d 2
10 10 0.11748736 e 2
11 11 0.47499708 a 3
12 12 0.56033275 b 3
13 13 0.90403139 c 3
14 14 0.13871017 d 3
15 15 0.98889173 e 3
16 16 0.94666823 a 4
17 17 0.08243756 b 4
18 18 0.51421178 c 4
19 19 0.39020347 d 4
20 20 0.90573813 e 4
21 21 1.25486225 ab 1
22 22 1.87216325 ab 2
23 23 1.36548126 ab 3
答案 0 :(得分：1)
这将为您带来结果
function setValue(val) { a = val; websocket.send(a) }
代码示例中生成数据帧的代码是这个：
df %>% bind_rows(
df %>%
filter(category %in% c('a' , 'b')) %>%
group_by(file) %>%
mutate(values = sum(values), category = paste0(category,collapse='')) %>%
filter(row_number() == 1 & n() > 1)
) %>% mutate(ID = row_number())
现在让我们说您要对多个列求和，需要以向量形式提供列表：
df <- data.frame(ID = 1:20, values = runif(20), category = rep(letters[1:5], 4), file = as.factor(sort(rep(1:4, 5))))
答案 1 :(得分：1)
library(dplyr)
df1 %>%
filter(category %in% c('a' , 'b')) %>%
group_by(file) %>%
filter(n_distinct(category) > 1) %>%
summarise(values = sum(values)) %>%
mutate(category="ab",
ID=max(df1$ID)+1:n()) %>%
bind_rows(df1, .)
#> Warning in bind_rows_(x, .id): binding factor and character vector,
#> coercing into character vector
#> Warning in bind_rows_(x, .id): binding character and factor vector,
#> coercing into character vector
#> ID values category file
#> 1 1 0.62585921 a 1
#> 2 2 0.61865851 b 1
#> 3 3 0.05274456 c 1
#> 4 4 0.68156961 d 1
.
.
.
#> 19 19 0.43239411 d 5
#> 20 20 0.85886314 e 5
#> 21 21 1.24451773 ab 1
#> 22 22 0.99001810 ab 2
#> 23 23 1.25331943 ab 3
答案 2 :(得分：1)
此data.table方法使用自连接来获取所有可能的两个字符的组合。
library(data.table)
setDT(df)
df_self_join <- df[df, on = .(file), allow.cartesian = T
][category != i.category,
.(category = paste0(i.category, category), values = values + i.values, file)
][order(category), .(ID = .I + nrow(df), values, category, file)]
rbindlist(list(df, df_self_join))
ID values category file
1: 1 0.76984382 a 1
2: 2 0.54311583 b 1
3: 3 0.23462016 c 1
4: 4 0.60179043 d 1
...
20: 20 0.03534223 e 5
21: 21 1.31295965 ab 1
22: 22 0.51666175 ab 2
23: 23 1.02305754 ab 3
24: 24 1.00446399 ac 1
25: 25 0.96910373 ac 2
26: 26 0.87795389 ac 4
#total of 80 rows
dplyr翻译非常接近
library(dplyr)
tib <- as_tibble(df)
inner_join(tib, tib, by = 'file')%>%
filter(ID.x != ID.y)%>%
transmute(category = paste0(category.x, category.y)
, values = values.x + values.y
, file)%>%
arrange(category)%>%
bind_rows(tib, .)%>%
mutate(ID = row_number())%>%
filter(category == 'ab') #filter added to show the "ab" files
# A tibble: 3 x 4
ID values category file
<int> <dbl> <chr> <fct>
1 21 1.31 ab 1
2 22 0.517 ab 2
3 23 1.02 ab 3