嵌套的ifelse（）是最糟糕的;什么是最好的？ [重复]

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How to implement coalesce efficiently in R 7 answers

如何在R 7答案中有效地实现合并

EDIT: this is a dupe of How to implement coalesce efficiently in R, agreed. I didn't realize how my problem was more general than my specific application, so this discussion has been great.

编辑:这是如何在R中有效实现合并的一个骗局,同意。我没有意识到我的问题比我的具体应用更普遍,所以这个讨论很棒。

Sometimes, the response variable in a randomized experiment is contained in a different column for each experimental group (Y_1 through Y_5 in the code below). It's often best to collect the response variable into a single column (Y_all). I end up doing it as in the example below. But I'm SURE there's a better way. thoughts?

有时,随机实验中的响应变量包含在每个实验组的不同列中(下面的代码中为Y_1到Y_5)。通常最好将响应变量收集到单个列(Y_all)中。我最终这样做,如下例所示。但我确定有更好的方法。想法?

set.seed(343)
N <- 1000
group <- sample(1:5, N, replace=TRUE)
Y_1 <- ifelse(group==1, rbinom(sum(group==1), 1, .5), NA)
Y_2 <- ifelse(group==2, rbinom(sum(group==2), 1, .5), NA)
Y_3 <- ifelse(group==3, rbinom(sum(group==3), 1, .5), NA)
Y_4 <- ifelse(group==4, rbinom(sum(group==4), 1, .5), NA)
Y_5 <- ifelse(group==5, rbinom(sum(group==5), 1, .5), NA)

## This is the part I want to make more efficient
Y_all <- ifelse(!is.na(Y_1), Y_1, 
                ifelse(!is.na(Y_2), Y_2, 
                       ifelse(!is.na(Y_3), Y_3, 
                              ifelse(!is.na(Y_4), Y_4, 
                                     ifelse(!is.na(Y_5), Y_5, 
                                            NA)))))

table(Y_all, Y_1, exclude = NULL)
table(Y_all, Y_2, exclude = NULL)

3 个解决方案

#1

I like to use a coalesce() function for this

我喜欢使用coalesce()函数

#available from https://gist.github.com/MrFlick/10205794
coalesce<-function(...) {
    x<-lapply(list(...), function(z) {if (is.factor(z)) as.character(z) else z})
    m<-is.na(x[[1]])
    i<-2
    while(any(m) & i<=length(x)) {
        if ( length(x[[i]])==length(x[[1]])) {
            x[[1]][m]<-x[[i]][m]
        } else if (length(x[[i]])==1) {
            x[[1]][m]<-x[[i]]
        } else {
            stop(paste("length mismatch in argument",i," - found:", length( x[[i]] ),"expected:",length( x[[1]] ) ))
        }
        m<-is.na(x[[1]])
        i<-i+1
    }
    return(x[[1]])
}

Then you can do

那你可以做

Y_all <- coalesce(Y_1,Y_2,Y_3,Y_4,Y_5)

Of course, this is very specific to getting the first non-NA value.

当然,这非常特定于获得第一个非NA值。

#2

I think in this case you can use the melt function to convert the data to long format and then get rid of the missing values:

我想在这种情况下你可以使用融合函数将数据转换为长格式,然后摆脱缺失的值:

library(reshape2)

set.seed(10)
N <- 1000
group <- sample(1:5, N, replace=TRUE)
Y_1 <- ifelse(group==1, rbinom(sum(group==1), 1, .5), NA)
Y_2 <- ifelse(group==2, rbinom(sum(group==2), 1, .5), NA)
Y_3 <- ifelse(group==3, rbinom(sum(group==3), 1, .5), NA)
Y_4 <- ifelse(group==4, rbinom(sum(group==4), 1, .5), NA)
Y_5 <- ifelse(group==5, rbinom(sum(group==5), 1, .5), NA)

Y_all = data.frame(group, Y_1, Y_2,Y_3,Y_4,Y_5)

Y_all.m = melt(Y_all, id.var="group")
Y_all.m = Y_all.m[!is.na(Y_all.m$value),]

#3

Store the vectors in a matrix and then select:

将矢量存储在矩阵中,然后选择:

Ymat    <- cbind(Y_1,Y_2,Y_3,Y_4,Y_5)
mycol   <- apply(!is.na(Ymat),1,which)

Y_all.f <- Ymat[cbind(1:nrow(Ymat),mycol)]

identical(Y_all,Y_all.f) # TRUE

#1