你是怎么抽象的aggregate
在函数中通过传递条件和值列表来总结?
# This works fine:
x <- data.frame(cond1 = sample(letters[1:3], 500, replace=TRUE),
cond2 = sample(LETTERS[1:7], 500, replace = TRUE),
cond3 = sample(LETTERS[1:4], 500, replace = TRUE),
value1 = rnorm(500),
value2 = rnorm(500))
aggregate(cbind(value1,value2) ~ cond1 + cond2, data = x, FUN=sum)
需要创建列名列表:(显示 3 个选项)然后调用函数:
c1 <- c("cond1","cond2","cond3"); v1 <- c("value1","value2")
c1 <- c("cond2","cond3"); v1 <- c("value2")
c1 <- c("cond3"); v1 <- c("value1")
aggregate(cbind(v1) ~ c1, data = x, FUN=sum)
我已经审查了许多替代方案,但还没有发现这种抽象的关键。
最佳答案
您可以使用 aggregate
的替代接口(interface),它不使用公式:
c1 <- c("cond1","cond2","cond3")
v1 <- c("value1","value2")
aggregate(x[v1],by=x[c1],FUN=sum)
cond1 cond2 cond3 value1 value2
1 a A A -3.3025839 -0.98304649
2 b A A 0.6326985 -3.08677485
3 c A A 3.6007853 2.23962265
4 a B A -0.5247620 -0.94644740
5 b B A 0.9242562 2.48268452
6 c B A 6.9215712 0.31512645
关于r - 如何将聚合与列名列表一起使用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18521811/