lapply 对 svytable 的匿名函数调用导致找不到对象“x”

Question

我有一个正在为其创建列联表的调查数据集。数据框中的每一列都是一个问题，一般来说，问题倾向于组合在一起。所以为了让生活更轻松，我一直在使用 lapply 循环遍历部分并使用以下代码返回列联表：

> out <- lapply(dat[,162:170], function(x) round(prop.table(table(x,dat$seg_2),2),3)*100)
> out
$r3a_1
            
x               1    2
  Don't Know  1.9  1.4
  No         14.2  4.9
  Yes        83.9 93.7

$r3a_2
            
x               1    2
  Don't Know  2.7  1.7
  No         14.8  6.6
  Yes        82.4 91.6

etc...

如您所见，我正在遍历第 162:170 列并创建一个道具表，显示第 1 组和第 2 组之间的不同响应。

但是，我想对这些数据进行加权。所以我正在使用调查包创建一个名为 dat_weight 的简单加权调查设计对象，并使用 svytable() 而不是 table()。我可以手动在单个列上运行更新的代码：

> round(prop.table(svytable(~dat[,162] + dat$seg_2, dat_weight),2),3)*100 
            dat$seg_2
dat[, 162]      1    2
  Don't Know  2.5  2.7
  No         16.5  5.4
  Yes        80.9 91.9

但是，当我尝试使用 lapply 时，它不起作用：

> out <- lapply(dat[,162:170], function(x) round(prop.table(svytable(~x + dat$seg_2, dat_weight),2),3)*100)

Error in eval(expr, envir, enclos) : object 'x' not found 

显然，匿名函数调用和 svytable 不能很好地配合使用。我尝试创建一个也不起作用的 for 循环。我猜这与范围界定有关，但我不知道如何解决它。

当然，必须有一种方法可以循环遍历该调查的各个部分，并且避免为每一列创建唯一的代码行。任何帮助将不胜感激。

编辑以添加一些示例数据：

> library("survey")
> dat <- structure(list(r3a_1 = structure(c(3L, 2L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("Don't Know", 
"No", "Yes"), class = "factor"), r3a_2 = structure(c(3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L), .Label = c("Don't Know", "No", "Yes"), class = "factor"), 
    r3a_3 = structure(c(3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L), .Label = c("Don't Know", 
    "No", "Yes"), class = "factor"), r3a_4 = structure(c(3L, 
    2L, 2L, 2L, 3L, 2L, 2L, 3L, 3L, 2L, 2L, 3L, 2L, 3L, 2L, 2L, 
    3L, 3L, 3L, 1L), .Label = c("Don't Know", "No", "Yes"), class = "factor"), 
    r3a_5 = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 3L, 2L, 
    2L, 3L, 2L, 3L, 3L, 2L, 3L, 2L, 3L, 1L), .Label = c("Don't Know", 
    "No", "Yes"), class = "factor"), r3a_6 = structure(c(3L, 
    3L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 2L, 3L, 3L, 2L, 2L, 2L, 3L, 
    2L, 3L, 3L, 3L), .Label = c("Don't Know", "No", "Yes"), class = "factor"), 
    r3a_7 = structure(c(1L, 2L, 2L, 2L, 3L, 2L, 2L, 3L, 3L, 2L, 
    3L, 3L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L), .Label = c("Don't Know", 
    "No", "Yes"), class = "factor"), r3a_8 = structure(c(3L, 
    2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 2L, 3L, 3L, 2L, 3L, 3L, 2L, 
    2L, 2L, 3L, 3L), .Label = c("Don't Know", "No", "Yes"), class = "factor"), 
    r3a_9 = structure(c(1L, 3L, 2L, 2L, 3L, 2L, 2L, 3L, 3L, 3L, 
    3L, 3L, 2L, 2L, 2L, 3L, 2L, 2L, 3L, 3L), .Label = c("Don't Know", 
    "No", "Yes"), class = "factor"), weight = c(0.34, 0.34, 0.34, 
    0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.34, 0.43, 
    0.43, 0.43, 0.34, 0.34, 0.34, 0.34, 0.34), seg_2 = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L), .Label = c("1", "2"), class = "factor")), .Names = c("r3a_1", 
"r3a_2", "r3a_3", "r3a_4", "r3a_5", "r3a_6", "r3a_7", "r3a_8", 
"r3a_9", "weight", "seg_2"), row.names = c(NA, 20L), class = "data.frame")

> dat_weight <- svydesign(ids = ~1, weights = ~weight, data = dat)

从那里你可以得到加权和未加权的表格：

round(prop.table(table(dat[,1],dat$seg_2),2),3)*100  #unweighted

round(prop.table(svytable(~dat[,1] + dat$seg_2, dat_weight),2),3)*100   #weighted

但是，这是可行的：

lapply(dat[,1:9], function(x) round(prop.table(table(x,dat$seg_2),2),3)*100)

虽然不是这样：

lapply(dat[,1:9], function(x) round(prop.table(svytable(~x + dat$seg_2, dat_weight),2),3)*100)

Answer 1

好吧，看来svytable 函数很挑剔，只会在设计对象中查找数据。它似乎没有在封闭环境中寻找x。因此，另一种方法是动态构建公式。因此，我们不是传入数据列本身，而是从 data.frame 中传入列名。然后我们将它们代入公式，然后它们由指向原始 data.frame 的设计对象解析。这是使用您的示例数据的一些工作代码

lapply(names(dat)[1:9], function(x) round(prop.table(
    svytable(bquote(~.(as.name(x)) + seg_2), dat_weight),
2),3)*100)

所以这里我们使用bquote 来构建公式。 .() 允许我们插入表达式，在这里我们获取x 中的字符值并将其转换为专有名称对象。因此从"r3a_9" 变为r3a_9。