Question

我有矢量列表，并希望将矢量分配给它的一个位置（覆盖）。以下是示例代码：

for (nodeId in names(chains)) {
    chains[nodeId] <- unlist(chains[nodeId])[-1]
}

分配后，我收到很多警告，告诉我列表的长度不一样。我知道发生的任务不是我想要的。

有没有办法只用对象chains[nodeId]替换unlist(chains[nodeId])[-1]中的元素？

当我执行str(chains)，str(chains[nodeId])和str(unlist(chains[nodeId])[-1])时，我会得到以下输出：

$str(chains)
List of 15
 $ 4  : chr [1:3] "root" "alcohol< 9.85" "totalSulfurDioxide>=60.5"
 $ 10 : chr [1:4] "root" "alcohol< 9.85" "totalSulfurDioxide< 60.5" "sulphates< 0.575"
 $ 22 : chr [1:5] "root" "alcohol< 9.85" "totalSulfurDioxide< 60.5" "sulphates>=0.575" ...
 (...) lots more

$str(chains[nodeId])
List of 1
 $ 4: chr [1:3] "root" "alcohol< 9.85" "totalSulfurDioxide>=60.5"

$str(unlist(chains[nodeId])[-1])
Named chr [1:2] "alcohol< 9.85" "totalSulfurDioxide>=60.5"
 - attr(*, "names")= chr [1:2] "42" "43"

更新： str已替换为dput;已添加dput(chains[nodeId])

$ dput(chains)
structure(list(`4` = "alcohol< 9.85", `10` = "alcohol< 9.85", 
    `22` = "alcohol< 9.85", `92` = "alcohol< 9.85", `93` = "alcohol< 9.85", 
    `47` = "alcohol< 9.85", `24` = "alcohol>=9.85", `50` = "alcohol>=9.85", 
    `102` = "alcohol>=9.85", `103` = "alcohol>=9.85", `26` = "alcohol>=9.85", 
    `27` = "alcohol>=9.85", `28` = "alcohol>=9.85", `29` = "alcohol>=9.85", 
    `15` = c("root", "alcohol>=9.85", "alcohol>=11.55", "sulphates>=0.685"
    )), .Names = c("4", "10", "22", "92", "93", "47", "24", "50", 
"102", "103", "26", "27", "28", "29", "15"))

$ dput(chains[nodeId])
structure(list(`15` = c("root", "alcohol>=9.85", "alcohol>=11.55", 
"sulphates>=0.685")), .Names = "15")

$ dput(unlist(chains[nodeId])[-1))
structure(c("alcohol>=9.85", "alcohol>=11.55", "sulphates>=0.685"
), .Names = c("152", "153", "154"))

$ dput(chains[nodeId])
structure(list(`15` = "alcohol>=9.85"), .Names = "15")

我想要实现的是从链[vectorId]

中的向量中删除第一个元素

Answer 1

如果chains是列表而nodeId是字符串，则chains[nodeId]将是长度为1的列表。您需要chains[[nodeId]]，其中包含该列表的内容。

Answer 2

这是你想要的吗？

# make a list of vectors since no data provided
origlist<-lapply(1:3,function(x)c("a",paste0("b",x),"c"))
names(origlist)<-c("_1","_2","_3")

$`_1`
[1] "a"  "b1" "c" 

$`_2`
[1] "a"  "b2" "c" 

$`_3`
[1] "a"  "b3" "c" 

# remove first item from each as per your example
lapply(origlist, tail, n = -1)

$`_1`
[1] "b1" "c" 

$`_2`
[1] "b2" "c" 

$`_3`
[1] "b3" "c"

R - 替换向量列表中的元素

2 个答案: