我有一个实验，我需要从对照（基准）中减去两种不同处理的值，但是这些减法必须对应于其他列（称为块和采样年份）。
虚拟数据框：
df <- data.frame("Treatment" = c("Control","Treat1", "Treat2"),
"Block" = rep(1:3, each=3), "Year" = rep(2011:2013, each=3),
"Value" = c(6,12,4,3,9,5,6,3,1));df
Treatment Block Year Value
1 Control 1 2011 6
2 Treat1 1 2011 12
3 Treat2 1 2011 4
4 Control 2 2012 3
5 Treat1 2 2012 9
6 Treat2 2 2012 5
7 Control 3 2013 6
8 Treat1 3 2013 3
9 Treat2 3 2013 1
所需的输出：
Treatment Block Year Value
1 Control-Treat1 1 2011 -6
2 Control-Treat2 1 2011 2
3 Control-Treat1 2 2012 -6
4 Control-Treat2 2 2012 -2
5 Control-Treat1 3 2013 3
6 Control-Treat2 3 2013 5
任何建议，最好使用dplyr？
我发现了类似的问题，但没有一个解决这个特定的问题。
答案 0 :(得分：1)
我们可以使用dplyr，group_by Block并从每个Value中减去Treatment == "Control"的{{1}}，然后删除“控件”行。
Value不确定，是否仅出于演示计算目的显示预期输出（library(dplyr)
df %>%
group_by(Block) %>%
mutate(Value = Value[which.max(Treatment == "Control")] - Value) %>%
filter(Treatment != "Control")
# Treatment Block Year Value
# <fct> <int> <int> <dbl>
#1 Treat1 1 2011 -6
#2 Treat2 1 2011 2
#3 Treat1 2 2012 -6
#4 Treat2 2 2012 -2
#5 Treat1 3 2013 3
#6 Treat2 3 2013 5
，Treatment）中Control-Treat1列中的值，或者OP确实希望将其用作输出。如果需要将其作为输出，我们可以使用
Control-Treat2答案 1 :(得分：1)
某种不同的tidyverse可能是：
df %>%
spread(Treatment, Value) %>%
gather(var, val, -c(Block, Year, Control)) %>%
mutate(Value = Control - val,
Treatment = paste("Control", var, sep = " - ")) %>%
select(Treatment, Block, Year, Value) %>%
arrange(Block)
Treatment Block Year Value
1 Control - Treat1 1 2011 -6
2 Control - Treat2 1 2011 2
3 Control - Treat1 2 2012 -6
4 Control - Treat2 2 2012 -2
5 Control - Treat1 3 2013 3
6 Control - Treat2 3 2013 5
答案 2 :(得分：1)
这可以通过如下的SQL自连接来完成：
library(sqldf)
sqldf("select a.Treatment || '-' || b.Treatment as Treatment,
a.Block,
a.Year,
a.Value - b.Value as Value
from df a
join df b on a.block = b.block and
a.Treatment = 'Control' and
b.Treatment != 'Control'")
给予：
Treatment Block Year Value
1 Control-Treat1 1 2011 -6
2 Control-Treat2 1 2011 2
3 Control-Treat1 2 2012 -6
4 Control-Treat2 2 2012 -2
5 Control-Treat1 3 2013 3
6 Control-Treat2 3 2013 5
答案 3 :(得分：0)
另一种dplyr-tidyr方法：您可以使用select删除不需要的列：
library(tidyr)
library(dplyr)
dummy_df %>%
spread(Treatment,Value) %>%
gather(key,value,Treat1:Treat2) %>%
group_by(Block,Year,key) %>%
mutate(Val=Control-value)
# A tibble: 6 x 6
# Groups: Block, Year, key [6]
Block Year Control key value Val
<int> <int> <dbl> <chr> <dbl> <dbl>
1 1 2011 6 Treat1 12 -6
2 2 2012 3 Treat1 9 -6
3 3 2013 6 Treat1 3 3
4 1 2011 6 Treat2 4 2
5 2 2012 3 Treat2 5 -2
6 3 2013 6 Treat2 1 5
精确的输出：
dummy_df %>%
spread(Treatment,Value) %>%
gather(key,value,Treat1:Treat2) %>%
mutate(Treatment=paste0("Control-",key)) %>%
group_by(Block,Year,Treatment) %>%
mutate(Val=Control-value) %>%
select(Treatment,everything(),-value,-key)%>%
arrange(Year)
结果：
# A tibble: 6 x 5
# Groups: Block, Year, Treatment [6]
Treatment Block Year Control Val
<chr> <int> <int> <dbl> <dbl>
1 Control-Treat1 1 2011 6 -6
2 Control-Treat2 1 2011 6 2
3 Control-Treat1 2 2012 3 -6
4 Control-Treat2 2 2012 3 -2
5 Control-Treat1 3 2013 6 3
6 Control-Treat2 3 2013 6 5
答案 4 :(得分：0)
另一个tidyverse解决方案。我们可以使用filter将“控制”和“处理”分离到不同的数据帧，使用left_join通过Block和Year组合它们，然后处理数据帧
library(tidyverse)
df2 <- df %>%
filter(!Treatment %in% "Control") %>%
left_join(df %>% filter(Treatment %in% "Control"),
.,
by = c("Block", "Year")) %>%
mutate(Value = Value.x - Value.y) %>%
unite(Treatment, Treatment.x, Treatment.y, sep = "-") %>%
select(names(df))
# Treatment Block Year Value
# 1 Control-Treat1 1 2011 -6
# 2 Control-Treat2 1 2011 2
# 3 Control-Treat1 2 2012 -6
# 4 Control-Treat2 2 2012 -2
# 5 Control-Treat1 3 2013 3
# 6 Control-Treat2 3 2013 5