对不起,我可能使用了错误的搜索词,但我找不到解决办法。
给出一个由两个参与者(id)组成的实验,每个参与者在两个不同参数(par1、par2)下执行6次任务:
id <- c(rep(1,6),rep(2,6))
par1 <- c(rep("a",9),rep("b",3))
par2 <- c(rep("c",3),rep("d",9))
val <- rnorm(12)
data <- data.frame(id,par1,par2,val)如何将"id“、"par1”和"par2“的所有行替换为一行,其中"val”的值是被替换行的"val“值的平均值?
因此,结果是这样一张表格:
id par1 par2 val
1 a c (mean of row 1-3)
1 a d (mean of row 4-6)
2 a d (mean of row 7-9)
2 b d (mean of row 10-12)发布于 2017-10-12 23:37:58
对于dplyr方法:
library(dplyr)
set.seed(123) # for reproducibility
id <- c(rep(1, 6), rep(2, 6))
par1 <- c(rep("a", 9), rep("b", 3))
par2 <- c(rep("c", 3), rep("d", 9))
val <- rnorm(12)
data <- data.frame(id, par1, par2, val)
# group by all variables except `val`
data %>% group_by_at(vars(-val)) %>% summarize(val = mean(val))这意味着:
# A tibble: 4 x 4
# Groups: id, par1 [?]
id par1 par2 val
<dbl> <fctr> <fctr> <dbl>
1 1 a c 0.2560184
2 1 a d 0.6382870
3 2 a d -0.4969993
4 2 b d 0.3794112发布于 2017-10-13 02:24:11
下面是data.table的一个选项。将“data.frame”转换为“data.table”(setDT(data)),按“id”、“par1”、“par2 2”分组,得到“val”的mean
library(data.table)
setDT(data)[, .(val = mean(val)), by = .(id, par1, par2)]https://stackoverflow.com/questions/46720085
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