我有一个要分组的变量。那很容易。但是,我希望结果表按随机顺序列出其行。我实际上想要做的是有点复杂。但请允许我向您展示简化版本。
mydf = data.table(
x = rep(1:4, each = 5),
y = rep(c('A', 'B','c','D', 'E'), times = 2),
v = rpois(20, 30)
)
mydf[,list(sum(x),sum(v)), by=y]
mydf[,list(sum(x),sum(v)), by=list(y=sample(y))]
#to list all the raw data in order of y
mydf[,list(x,v), by=y]
mydf[,list(x,v), by=list(y=sample(y))]
如果查看结果输出,您会注意到y确实是随机顺序的,但是它已经与行中的数据无关。
我能做什么?
最佳答案
我会进行操作,然后随机订购:
mydf[,list(x,v),by=y][sample(seq_len(nrow(mydf)),replace=FALSE)]
编辑:分组后随机重新排序:
mydf[,list(sum(x),sum(v)), by=y][sample(seq_len(length(y)),replace=FALSE)]
您可以在分组之前进行如下操作来分组和随机排序,看起来它确实保留了更改后的顺序:
mydf[order(setNames(sample(unique(y)),unique(y))[y])]
mydf[order(setNames(sample(unique(y)),unique(y))[y]),list(sum(x),sum(v)),by=y]
#perhaps more readable:
mydf[{z <- unique(y); order(setNames(sample(z),z)[y])}]
mydf[{z <- unique(y); order(setNames(sample(z),z)[y])},list(sum(x),sum(v)),by=y]
通过在订购前先添加一列可以更加透明。
mydf[,new.y := setNames(sample(unique(y)),unique(y))[y]][order(new.y)]
分解:
##a random ordering of the elements of y
##(set.seed is used here to get consistent results)
set.seed(1); mydf[,{z <- unique(y);sample(z)}]
# [1] "B" "E" "D" "c" "A"
##assigning names to the elements of y
##creating a 1-1 bijective function between the elements of y
set.seed(1); mydf[,{z <- unique(y);setNames(sample(z),z)}]
# A B c D E
#"B" "E" "D" "c" "A"
##subsetting by y puts y through the map
##in effect every element of y is posing as an element of y, picked at random
##notice that the names (top row) are the original y
##the values (bottom row) are the mapped-to values
# A B c D E A B c D E A B c D E A B c D E
#"B" "E" "D" "c" "A" "B" "E" "D" "c" "A" "B" "E" "D" "c" "A" "B" "E" "D" "c" "A"
##ordering by this now orders by the mapped-to values
set.seed(1); mydf[{z <- unique(y);order(setNames(sample(z),z)[y])}]
编辑:在注释中加入Arun的建议以使用
setattr
设置名称:mydf[{z <- unique(y); order(setattr(sample(z),'names',z)[y])}]
mydf[{z <- unique(y); order(setattr(sample(z),'names',z)[y])},list(sum(x),sum(v)),by=y]