Question

我有每日数据(4011天)和指标(1周、2周),我想找到每周最高值和相应的指标。

mydat <- matrix(c(0.027,0.034,0.019,0.021,0.026,0.024,0.058,0.026,0.064,
     0.066,0.026,0.101,0.069,0.054,rep(2,2),rep(1,5),rep(2,2),rep(1,5)), ncol=2)

我尝试了以下的代码。我设法得到了最大序列( 在这种情况下, 每周最高数), 但我不想在指示器中看到最大序列。这是代码

week.max <- function(vec){    
   if(length(vec[is.na(vec)]) == 7){       
   return(NA)     
   } 
   else{
     return(max(vec, na.rm = T))
   } 
}

max.week.dat <- apply(mydat, 2, function(x) tapply(x, rep(1:(length(x)/7), 
  each=7, len=length(x)), week.max))

和结果

matrix(c(0.058,0.101,2,2),ncol=2)

我想要这样的输出:

matrix(c(0.058,0.101,1,1),ncol=2)

事先非常感谢。

Answer 1

以下是数据(第三周还有一天额外时间)

mydat <- data.frame(value = c(0.027,0.034,0.019,0.021,0.026,0.024,0.058,0.026,0.064,
     0.066,0.026,0.101,0.069,0.054,0.95), ind = c(rep(2,2),rep(1,5),rep(2,2),rep(1,5),2))

您的函数函数

week.max <- function(vec){    
  if(length(vec[is.na(vec)]) == 7){       
return(NA)     
 } 
 else{
return(max(vec, na.rm = T))
 } 
}

添加周信息

mydat$week <- c(rep(1:2,each=7),3)

使用与“https://stackoverflow.com/ questions/10705290/r-select-a-value-a-value-f-based-on-a-high-fest-value-in-another-clumn/10712545#10712545#10712545”相同的解决方案。

library(plyr)
ddply(mydat, .(week), subset, subset = value==week.max(value), select = -week)

友情链接