Question

我从数据框中提取了以下向量

parent<-c(119.04832, 120.34873, 405.80245, 635.80934)
product<-c(119.04838, 800.56437, 361.87349, 582.09567, 444.65898, 286.08075)

我想删除所有元素＆＃34;几乎相等＆＃34;仅来自产品阵列。嵌套的for循环可以是一个选项。像这样：

limit<-0.0001
for(i in 1:length(parent)){
 for(j in 1:length(product)){
  if(abs(parent[i]-product[j]<=limit)){
    replicate<-rbind(replicate,product[j])
    } else {not_replicate<-rbind(not_replicate, product[j])
  }
 }
}

所以，我期望得到一个新的载体

>replicate
[1] 800.56437, 361.87349, 582.09567, 444.65898, 286.08075

有没有比这更聪明，更快的方法？例如暗示使用sapply或ifelse的东西？

P.S。一个更大的测试案例可能是：

parent<-c(225.08619,236.09266.16304,267.14706,270.10044,277.18305,282.15796,355.15658,496.23987,238.11061,239.06545,252.08988,252.08988,
,369.17223,413.25640,467.30356,551.39344,748.50853) 

product<-c(300.02900,227.10570,341.17340,305.17360,241.12110,369.18980,269.13730,347.10386,373.19980,357.18960,266.16307,233.19210,269.13720,357.18860,269.13740,406.24220,305.17370,405.22580,157.08490,157.08520,157.09490,169.08480,169.12130,181.12060,
187.09550,199.11050,209.11600,225.08620,227.12690)

Answer 1

您可以使用sapply代替for循环

product[!as.logical(apply(sapply(product, function(x) abs(x - parent) <= 0.0001), 2L, sum))]

Answer 2

以下是使用dplyr软件包near函数的示例：

parent<-c(119.04832, 120.34873, 405.80245, 635.80934)
product<-c(119.04838, 800.56437, 361.87349, 582.09567, 444.65898, 286.08075)

library(dplyr)
#create matrix of matches between parent and product
result<-sapply(parent, function(x){near(x, product, tol=0.0001)})
#find rows where there is at least one match
toremove<-which(rowSums(result)>0)
#remove unwanted values
product[-toremove]

Answer 3

假设：

parent <- c(119.04832, 120.34873, 405.80245, 635.80934)
product <- c(119.04838, 800.56437, 361.87349, 582.09567, 444.65898, 286.08075)
limit <- 0.0001

我会使用outer：

product[!as.logical(colSums(abs(outer(parent, product, "-")) <= limit))]
# [1] 800.5644 361.8735 582.0957 444.6590 286.0808

通过成对比较从数据框的单个列中删除元素

3 个答案: