比较 R 中超过 2 个向量（投票）答案

【问题标题】：compare between more than 2 vectors in R (Voting)比较 R 中超过 2 个向量（投票）
【发布时间】：2017-05-07 09:14:05
【问题描述】：

我有 5 个向量，这些向量中的每个项目要么是“是”，要么是“否” 所以我想比较这 5 个向量（逐行）并计算每一行的多数投票并将结果添加到一个新向量中。我怎样才能有效地做到这一点。

v1=c("yes","no","no","yes")
v2=c("no","no","yes","yes")
v3=c("yes","yes","no","yes")
v4=c("yes","no","yes","yes")
v5=c("yes","yes","yes","no")
#The expected output is "yes", "no", "yes", "yes"

【问题讨论】：

标签： r

【解决方案1】：

首先将数据放入基于字符的形式中：

dat <- data.frame( v1=c("yes","no","no","yes"),
                  v2=c("no","no","yes","yes"),
                  v3=c("yes","yes","no","yes"),
                  v4=c("yes","no","yes","yes"),
                  v5=c("yes","yes","yes","no"), stringsAsFactors=FALSE)

然后拉出一个表对象的最大值的名称：

 apply(dat, 1, function(x) names(which.max(table(x))) )
[1] "yes" "no"  "yes" "yes"

【讨论】：

如果将数据存储为因子，它也可以工作，就像“标准”data.frame 的情况一样

【解决方案2】：

另一种方法是使用 mapply 和 == 来返回一个 TRUE 和 FALSE 矩阵，比较向量的元素是否等于某个值（此处为“是”）。然后rowMeans 计算跨行的比例，> 0.5 检查多数。我们加 1 以转换为数字位置，然后将其用作从c("no", "yes") 中的元素中选择的位置。

c("no", "yes")[(rowMeans(mapply("==", moreArgs=list("yes"), myList)) > 0.5) + 1L]
[1] "yes" "no"  "yes" "yes"

使用矩阵乘法的替代方法是

c("no", "yes")[((do.call(cbind, myList) == "yes") %*%
               rep(1, length(myList)) > (length(myList) / 2)) + 1L]
[1] "yes" "no"  "yes" "yes"

请注意，首先将向量放在下面的列表中。

数据

myList <- list(v1=c("yes","no","no","yes"),
               v2=c("no","no","yes","yes"),
               v3=c("yes","yes","no","yes"),
               v4=c("yes","no","yes","yes"),
               v5=c("yes","yes","yes","no"))

【讨论】：