基于两个或多个变量的所有可能组合的子数据表 [英] Subset data.table based on all possible combinations of two or more variables

问题描述

我想根据某些变量是正数，负数还是两者之间的组合来为data.frame子集.对于n变量，这应该导致2^n可能的组合.

I want to subset a data.frame based on if some variables are all positive, all negative or some combination in between. For n variables this should lead to 2^n possible combinations.

我认为combn可以用来实现这一目标，但我正在努力正确地做到这一点.

I think combn can be used to achieve this but I'm struggling to do it properly.

样本数据:

library(data.table)
dt <- data.table(x = runif(100, -1, 1), y = runif(100, -1, 1), z = runif(100, -1, 1))

我想要什么:

dt[x < 0 & y < 0 z < 0, ]
dt[x < 0 & y < 0 z > 0, ]
dt[x < 0 & y > 0 z < 0, ]
dt[x < 0 & y > 0 z > 0, ]
dt[x > 0 & y < 0 z < 0, ]
dt[x > 0 & y < 0 z > 0, ]
dt[x > 0 & y > 0 z < 0, ]
dt[x > 0 & y > 0 z > 0, ]

到目前为止我已经尝试过的:

What I've tried so far:

combinator <- function(z){
  cnames <- colnames(z)
  combinations <- t(combn(c(rep("<", ncol(z)), rep(">", ncol(z))),ncol(z)))

  retval <- t(sapply(1:nrow(combinations), function(p){
    sapply(1:ncol(z), function(q) paste(cnames[q], combinations[p,q], 0))
  }))

  return(apply(retval, 1, paste, collapse = " & "))
}

输出:

> l <- combinator(dt)
> l
 [1] "x < 0 & y < 0 & z < 0" "x < 0 & y < 0 & z > 0" "x < 0 & y < 0 & z > 0" "x < 0 & y < 0 & z > 0"
 [5] "x < 0 & y < 0 & z > 0" "x < 0 & y < 0 & z > 0" "x < 0 & y < 0 & z > 0" "x < 0 & y > 0 & z > 0"
 [9] "x < 0 & y > 0 & z > 0" "x < 0 & y > 0 & z > 0" "x < 0 & y < 0 & z > 0" "x < 0 & y < 0 & z > 0"
[13] "x < 0 & y < 0 & z > 0" "x < 0 & y > 0 & z > 0" "x < 0 & y > 0 & z > 0" "x < 0 & y > 0 & z > 0"
[17] "x < 0 & y > 0 & z > 0" "x < 0 & y > 0 & z > 0" "x < 0 & y > 0 & z > 0" "x > 0 & y > 0 & z > 0"

> l[1]
[1] "x < 0 & y < 0 & z < 0"

> subset(dt, eval(l[1]))
Error in subset.data.table(dt, eval(l[1])) : 
  'subset' must evaluate to logical

此外，如果以下内容表明我没有列出所有所需的组合:

Also if the following shows that I'm not listing all the desired combinations:

> unique(l)
[1] "x < 0 & y < 0 & z < 0" "x < 0 & y < 0 & z > 0" 
[3] "x < 0 & y > 0 & z > 0" "x > 0 & y > 0 & z > 0"

输出应具有8个唯一的结果，而不是上面显示的4个.

the output should have 8 unique results instead of the 4 shown above.

推荐答案

只需执行dt[, sign_combi := do.call(paste, lapply(dt, sign))]，您就可以根据需要split或by =该列，例如split(dt, dt$sign_combi).试图将代码粘贴在一起是一个坏主意.

Just do dt[, sign_combi := do.call(paste, lapply(dt, sign))] and you can split or by = that column as needed, e.g., split(dt, dt$sign_combi). Trying to paste together code is a Bad Idea.

例如:

set.seed(47) # setting seed for reproducibility
dt <- data.table(x = runif(100, -1, 1), y = runif(100, -1, 1), z = runif(100, -1, 1))

# create combination column (you could keep it separate if you prefer)
dt[, sign_combi := do.call(paste, lapply(dt, sign))]

# split original data by sign combinations
result = split(dt, dt$sign_combi)

# list of 8 resulting data tables
length(result)
# [1] 8

# peaking at the first three rows of the first three tables:
lapply(head(result, 3), head, 3)
# $`-1 -1 -1`
#             x          y          z sign_combi
# 1: -0.5713038 -0.7103555 -0.6873705   -1 -1 -1
# 2: -0.1407803 -0.8371153 -0.3686299   -1 -1 -1
# 3: -0.6478446 -0.7629461 -0.7458949   -1 -1 -1
# 
# $`-1 -1 1`
#             x          y         z sign_combi
# 1: -0.8070969 -0.3952283 0.9212030    -1 -1 1
# 2: -0.1190934 -0.4969318 0.8082232    -1 -1 1
# 3: -0.6536104 -0.3280965 0.6880454    -1 -1 1
# 
# $`-1 1 -1`
#              x         y          z sign_combi
# 1: -0.78789241 0.8577848 -0.7586369    -1 1 -1
# 2: -0.04442825 0.4736388 -0.3354734    -1 1 -1
# 3: -0.22105744 0.3012645 -0.4160631    -1 1 -1

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