有没有办法根据多列在单独的列中获取一系列月份值？

Question

我有 12 列，其中包含有关物种产卵的数据。在每一列中，1 表示该物种在该月产卵，0 表示它没有。我想创建另一个名为产卵期的列，它给出了几个月的范围。我有另一列包含调查日期，我想要另一列查看产卵期列和调查日期，并确定调查是否在产卵期间发生。

预期结果是：

Jan   Feb   Mar   Apr   May   Jun   Jul   Aug   Sep   Oct  Nov  Dec   Spawning_Period      Survey_Date   Survey_Sampling
1     1     1     1     1     1     1     1     1     1     1     1     Jan-Dec            17/01/2019        1
1     1     0     0     0     0     0     0     1     1     1     1     Jan-Feb, Sep-Dec   13/06/2019        0

我认为 for 循环可能是答案，因为有 500 个观察值。我看过类似的问题，但似乎找不到我要找的东西。

Answer 1

希望这是一个足够的解决方案，您可以在之后添加日期和采样：

A = matrix(c(rep(1,14),rep(0,6),rep(1,4),rep(0,12)),nrow=3,ncol=12,byrow=T)
A = data.frame(A)
colnames(A) = c("Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec")

# Change the colnames to numerics( we can handle them much easier)
colnames(A) = c( 1:12)

A$Spawning_Period=rep(0,nrow(A))

#a loop is inefficient but shows you what actually happens here:
for(i in 1:nrow(A)){

Month.Vector=c()

#Shows a vector of months pre row, where species spawn
Month.Vector=colnames(A)[A[i,]==1]

Spawning.Period=c(1,diff(as.numeric(Month.Vector)))

#new Code to determine when there is no spawning period
if(length(Month.Vector)==0){
A[i,"Spawning_Period"]=NA } else{
#Determines the Break point
Break.Points=which(Spawning.Period!=1)
if(length(Break.Points)==0){
A[i,"Spawning_Period"]="1-12"}else{
Collect = list()
for(j in 1:length(Break.Points)){
if(j==1){
Collect [[1]] = paste(Month.Vector[1:(Break.Points[j]-1)][1],Month.Vector[1:(Break.Points[j]-1)][length(Month.Vector[1:(Break.Points[j]-1)])],sep="-")

Collect[[length(Break.Points)+1]] = paste(Month.Vector[(Break.Points[j]):length(Month.Vector)][1],Month.Vector[(Break.Points[j]):length(Month.Vector)][length(Month.Vector[(Break.Points[j]):length(Month.Vector)])],sep="-")
}

if(j!=1){

Collect[[j]] = paste(Month.Vector[(Break.Points[j-1]):(Break.Points[j]-1)][1],Month.Vector[(Break.Points[j-1]):(Break.Points[j]-1)][length(Month.Vector[(Break.Points[j-1]):(Break.Points[j]-1)])],sep="-")

}


}

A[i,"Spawning_Period"]= paste(unlist(Collect), collapse =", ")
}
}
}

您将在此处将月份表示为日期，但如果您愿意，可以将它们转换为相应的值。希望这会有所帮助。

Answer 2

这是使用@RBeginner 的完整解决方案，以防万一有人想看到它。我注意到，当没有休息时，它仍然会发布 1-12，而它应该是 4,5，所以我添加了它，对于 1-12 中的 0，我添加了 NA。

spawning_period_ind <- spawning_period[,c(1:12)]
colnames(spawning_period_ind) = c( 1:12)
spawning_period_ind$Spawning_Period=c(0,0)

for(i in 1:nrow(spawning_period_ind)){

  Month.Vector=c()

  #Shows a vector of months pre row, where species spawn
  Month.Vector=colnames(spawning_period_ind)[spawning_period_ind[i,]==1]

  Spawning.Period=c(1,diff(as.numeric(Month.Vector)))

  #new Code to determine when there is no spawning period
  if(all(is.na(Month.Vector)==TRUE)){
    spawning_period_ind[i,"Spawning_Period"]=NA } 
  else if(all(is.na(Month.Vector)==FALSE & length(Spawning.Period)==12)) {
    spawning_period_ind[i,"Spawning_Period"]="1-12"
  }
  else
    {
      #Determines the Break point
      Break.Points=which(Spawning.Period!=1)
      if(length(Break.Points)==0 & length(Spawning.Period)!=12)
        {
        spawning_period_ind[i,"Spawning_Period"]= paste(Month.Vector, collapse = ",", sep = "")
        }
      else
        {
          Collect = list()
          for(j in 1:length(Break.Points)){
            if(j==1){
              Collect [[1]] = paste(Month.Vector[1:(Break.Points[j]-1)][1],Month.Vector[1:(Break.Points[j]-1)][length(Month.Vector[1:(Break.Points[j]-1)])],sep="-")

              Collect[[length(Break.Points)+1]] = paste(Month.Vector[(Break.Points[j]):length(Month.Vector)][1],Month.Vector[(Break.Points[j]):length(Month.Vector)][length(Month.Vector[(Break.Points[j]):length(Month.Vector)])],sep="-")
            }

            if(j!=1){

              Collect[[j]] = paste(Month.Vector[(Break.Points[j-1]):(Break.Points[j]-1)][1],Month.Vector[(Break.Points[j-1]):(Break.Points[j]-1)][length(Month.Vector[(Break.Points[j-1]):(Break.Points[j]-1)])],sep="-")

            }


          }

          spawning_period_ind[i,"Spawning_Period"]= paste(unlist(Collect), collapse =", ")
        }
    }
}