Question

是否有一个看起来像这样的数据框

Forest Grass Shrub Water Binary
0.6    0.5   0.3   0.2   1
0.2    0.3   0.4   0.5   0
0.3    0.5   0.2   0.6   1
0.2    0.6   0.3   0.2   0
0.6    0.5   0.3   0.2   1

我希望R查看所有行和写一个新的专栏，写下最大数量的土地覆盖名称，这样我就有了这样一张桌子。

Forest Grass Shrub Water Binary Most
0.6    0.5   0.3   0.2   1      Forest
0.2    0.3   0.4   0.5   0      Water
0.3    0.5   0.2   0.6   1      Water
0.2    0.6   0.3   0.2   0      Grass
0.6    0.5   0.3   0.2   1      Forest

然后我希望R查看二进制列并计算森林-1森林-0，水-1，水0组合出现的频率

不幸的是，我对如何做到这一点没有任何线索，非常感谢你的帮助！

Answer 1

更新

您可以使用apply代替max.col，而不是names(mydf)[max.col(mydf[-length(mydf)])] # [1] "Forest" "Water" "Water" "Grass" "Forest"：

## Create your "Most" column. 
## This assumes "Binary" to be the last column. 
## Use names along with `setdiff`, index positions, or other approaches
##     if this is not the case with your actual data.
mydf$Most <- apply(mydf[-length(mydf)], 1, function(x) names(x)[which.max(x)])
mydf
#   Forest Grass Shrub Water Binary   Most
# 1    0.6   0.5   0.3   0.2      1 Forest
# 2    0.2   0.3   0.4   0.5      0  Water
# 3    0.3   0.5   0.2   0.6      1  Water
# 4    0.2   0.6   0.3   0.2      0  Grass
# 5    0.6   0.5   0.3   0.2      1 Forest

## This is the tabulation step.
## I'm assuming this is separate from the original data.frame
##     as it isn't shown to be as part of your desired output.
table(mydf[c("Most", "Binary")])
#         Binary
# Most     0 1
#   Forest 0 2
#   Grass  1 0
#   Water  1 1

原始答案

通过它的声音，你可能正在寻找这样的东西：

{{1}}

Answer 2

library(data.table)
library(reshape2)
setDT(dt)
dt[,RowNo := .I]

dt2 <- melt(dt[,c(setdiff(colnames(dt),'Binary')), with = F], id.vars = c('RowNo'))
dt2 <- unique(dt2[, Most := as.character(variable)[which.max(value)], by = RowNo][,list(RowNo,Most)])

dt <- merge(dt2,dt, by = 'RowNo')

dt[,list(.N), by = list(Binary,Most)]

输出 -

   Binary   Most N
1:      1 Forest 2
2:      0  Water 1
3:      1  Water 1
4:      0  Grass 1

使用其他列的标题在数据框中创建新列

2 个答案:

更新

原始答案