控制嵌套列表的结构时适用FUN返回列表或NA

Question

我有一个向量v和矩阵m并使用适用于从cor.test函数（相关性v和m列之间）提取的结果的子集。

set.seed(1) 
m <- matrix(runif(12), nrow = 3) 
v <- 3:1 

res <- apply(m, 2, function(x) { 
    cor.test(x, v, method = 'spearman', exact = F)[c(1,3,4)] 
}) 

这与长度等于列的数目的列表中m嵌套列表 - 和在结构我想作为输出（2级表）。

> str(res) 
List of 4 
$ :List of 3 
    ..$ statistic: Named num 8 
    .. ..- attr(*, "names")= chr "S" 
    ..$ p.value : num 0 
    ..$ estimate : Named num -1 
    .. ..- attr(*, "names")= chr "rho" 
$ :List of 3 
    ..$ statistic: Named num 2 
    .. ..- attr(*, "names")= chr "S" 
    ..$ p.value : num 0.667 
    ..$ estimate : Named num 0.5 
    .. ..- attr(*, "names")= chr "rho" 
$ :List of 3 
    ..$ statistic: Named num 0 
    .. ..- attr(*, "names")= chr "S" 
    ..$ p.value : num 0 
    ..$ estimate : Named num 1 
    .. ..- attr(*, "names")= chr "rho" 
$ :List of 3 
    ..$ statistic: Named num 6 
    .. ..- attr(*, "names")= chr "S" 
    ..$ p.value : num 0.667 
    ..$ estimate : Named num -0.5 
    .. ..- attr(*, "names")= chr "rho" 

我要筛选的每种cor.test结果，说p.value，内环路申请并返回NA指示过滤的结果（保留结果的长度，这里四个）。

res <- apply(m, 2, function(x) { 
    tmp <- cor.test(x, v, method = 'spearman', exact = F)[c(1,3,4)] 
    ifelse(tmp$p.value < 0.1, list(tmp), NA) 
}) 

我的问题是，我们现在得到一个3级列表结构

res2 <- apply(m, 2, function(x) { 
    tmp <- cor.test(x, v, method = 'spearman', exact = F)[c(1,3,4)] 
    ifelse(tmp$p.value < 0.1, list(tmp), NA) 
    }) 

> str(res2) 
List of 4 
$ :List of 1 
    ..$ :List of 3 
    .. ..$ statistic: Named num 8 
    .. .. ..- attr(*, "names")= chr "S" 
    .. ..$ p.value : num 0 
    .. ..$ estimate : Named num -1 
    .. .. ..- attr(*, "names")= chr "rho" 
$ : logi NA 
$ :List of 1 
    ..$ :List of 3 
    .. ..$ statistic: Named num 0 
    .. .. ..- attr(*, "names")= chr "S" 
    .. ..$ p.value : num 0 
    .. ..$ estimate : Named num 1 
    .. .. ..- attr(*, "names")= chr "rho" 
$ : logi NA 

只有从apply的第一个结果是NA结果的结构非常像期望的，显然因为apply则可以容纳未过滤导致结构。

res3 <- apply(m, 2, function(x) { 
    tmp <- cor.test(x, v, method = 'spearman', exact = F)[c(1,3,4)] 
    ifelse(tmp$p.value > 0.1, list(tmp), NA) #'invert' the test 
}) 

>res3 
List of 4 
$ : logi NA 
$ :List of 3 
    ..$ statistic: Named num 2 
    .. ..- attr(*, "names")= chr "S" 
    ..$ p.value : num 0.667 
    ..$ estimate : Named num 0.5 
    .. ..- attr(*, "names")= chr "rho" 
$ : logi NA 
$ :List of 3 
    ..$ statistic: Named num 6 
    .. ..- attr(*, "names")= chr "S" 
    ..$ p.value : num 0.667 
    ..$ estimate : Named num -0.5 
    .. ..- attr(*, "names")= chr "rho" 

我试图徒然返回ifelse(tmp$p.value < 0.1, tmp, NA)和ifelse(tmp$p.value < 0.1, list(tmp), list(NA))。

我发现的唯一的解决办法是分配NA的apply外：

res4 <- apply(m, 2, function(x) { 
    cor.test(x, v, method = 'spearman', exact = F)[c(1,3,4)] 
}) 
res4[sapply(res4, "[[", 2) > 0.1] <- NA 

很显然，我错过了什么关于适用的内部运作。

Answer 1

您的问题不在apply，而是在ifelse。如果您使用if() {} else {}相反，它的工作方式，你打算从ifelse

ifelse形状相同返回一个值作为测试

res3 <- apply(m, 2, function(x) { 
     tmp <- cor.test(x, v, method = 'spearman', exact = F)[c(1,3,4)] 
     if (tmp$p.value < 0.1) { return(tmp) } else { return(NA) } 
    }) 

str(res3) 
# List of 4 
# $ :List of 3 
    # ..$ statistic: Named num 8 
    # .. ..- attr(*, "names")= chr "S" 
    # ..$ p.value : num 0 
    # ..$ estimate : Named num -1 
    # .. ..- attr(*, "names")= chr "rho" 
# $ : logi NA 
# $ :List of 3 
    # ..$ statistic: Named num 0 
    # .. ..- attr(*, "names")= chr "S" 
    # ..$ p.value : num 0 
    # ..$ estimate : Named num 1 
    # .. ..- attr(*, "names")= chr "rho" 
# $ : logi NA 

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