Question

我有一个列表，如：

List_PatientState
    [[1]]
          BirthYear Hispanic tEye tStatus tKidney         
    12604         2        0    1       2       2 

    [[2]]
         BirthYear Hispanic tEye tStatus tKidney          
    9252         2        0    2       1       1 

    [[3]]
         BirthYear Hispanic tEye tStatus tKidney          
    6613         2        0    1       1       1 

    [[4]]
         BirthYear Hispanic tEye tStatus tKidney        
    6265         2        0    2       2       1 

    [[5]]
         BirthYear Hispanic tEye tStatus tKidney          
    6202         2        0    1       1       2

在一个循环中我创建了一个新列表，如果它在我当前的列表列表中可用，我想忽略该列表

有关更多信息，请考虑在我的循环中创建的此列表：

[[6]]
      BirthYear Hispanic tEye tStatus tKidney          
11773   2        0    1       1       1

这类似于：

 List_PatientState[[3]]

我想在我的列表列表中检测到这样的列表，所以，我可以拒绝它添加到我当前的列表列表中。有人帮我吗？我将不胜感激。 dput的输出（List_PatientState）：

> dput(List_PatientState)
list(structure(list(BirthYear = structure(2L, .Label = c("1", 
"2", "3"), class = "factor"), Hispanic = structure(1L, .Label = c("0", 
"1"), class = "factor"), tEye = structure(1L, .Label = c("1", 
"2"), class = "factor"), tStatus = structure(2L, .Label = c("1", 
"2"), class = "factor"), tKidney = structure(2L, .Label = c("1", 
"2"), class = "factor")), .Names = c("BirthYear", "Hispanic", 
"tEye", "tStatus", "tKidney"), row.names = 12604L, class = "data.frame"), 
    structure(list(BirthYear = structure(2L, .Label = c("1", 
    "2", "3"), class = "factor"), Hispanic = structure(1L, .Label = c("0", 
    "1"), class = "factor"), tEye = structure(2L, .Label = c("1", 
    "2"), class = "factor"), tStatus = structure(1L, .Label = c("1", 
    "2"), class = "factor"), tKidney = structure(1L, .Label = c("1", 
    "2"), class = "factor")), .Names = c("BirthYear", "Hispanic", 
    "tEye", "tStatus", "tKidney"), row.names = 9252L, class = "data.frame"))

Answer 1

首先我们创建一些示例数据：

dat <- lapply(1:5,function(x) mtcars[x,]) #List of the first 5 rows of mtcars set

然后，如果你必须使用for loop：

for (i in 3:6) {
   temp <- mtcars[i,]
   if(is.na(any(sapply(dat,function(y) all.equal(y,temp)[[1]])))) {
    dat[[i]] <- temp
 }
}

如果您在循环中创建的data frame已存在于列表中并且仅添加未列在列表中的data frames，则基本上会检查每次迭代。

Answer 2

您的数据

向量列表

List_PatientState <- list(c(Birthyear=2, Hispanic=0, tEye=1, tStatus=2, tKidney=2),
                          c(Birthyear=2, Hispanic=0, tEye=2, tStatus=1, tKidney=1),
                          c(Birthyear=2, Hispanic=0, tEye=1, tStatus=1, tKidney=1), 
                          c(Birthyear=2, Hispanic=0, tEye=2, tStatus=2, tKidney=1),
                          c(Birthyear=2, Hispanic=0, tEye=1, tStatus=1, tKidney=2))

可能添加的新向量

newv <- list(c(Birthyear=2, Hispanic=0, tEye=1, tStatus=1, tKidney=1))     # duplicate values
newv1 <- list(c(Birthyear=2, Hispanic=0, tEye=2, tStatus=1, tKidney=2))    # unique values

如果是唯一值

，则追加函数

myfun <- function(L, V) {
              if (any(sapply(L, function(x) all(V[[1]] == x)))) {
                    return(L)
              } else {
                    return(c(L,V))
              }
         }

测试

myfun(List_PatientState, newv)    # returns original list
myfun(List_PatientState, newv1)   # returns original list + newv1

如何检测R中列表列表中的列表

2 个答案:

您的数据

可能添加的新向量

如果是唯一值

测试