在R（表決）

我有5個矢量的每個項目在這些載體多於2個向量之間進行比較或者爲「是」或「否」因此，我希望這些5個載體（逐行）之間進行比較，並計算出對每一行進行多數投票並將結果添加到新的向量中。如何以有效的方式執行此操作。在R（表決）

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"

來源

2017-05-06 Nobel

首先把數據的形式是基於字符的：

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)

然後拉出的最大值的名稱爲表對象：

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

來源

2017-05-07 00:03:21

它的工作原理也如果數據存儲爲因素，因爲這將是一個「標準」 data.frame的情況下 – Gilles

另一種方法是使用mapply與==返回真假對比一個矩陣，其中向量的元素相等的東西（在這裏，「是」）。然後rowMeans計算跨行的比例，並且> 0.5檢查多數。我們加1，轉換爲數字位置，然後用這個作爲位置從c("no", "yes")元素進行選擇。

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

使用矩陣乘法另一種是，你通過把向量在列表如下開始

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

注意。

數據

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"))

來源

2017-05-07 00:41:44 lmo

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