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r - 循环遍历子列表，其中“子子”列表保持不变

问题描述

我想遍历一个子列表，其中“子子”列表保持不变。我提到的所有代码都只是可重现的示例（请注意实际数据非常大）并且应该可以正常工作。

我有一个列表列表，每个列表都有 2 个子列表，如下所示：

library(data.table)  
library(mice)
df <- fread(
    "A   B  C  D  E  F  iso   year   
     0   A   NA  1  NA  NA  NLD   2009   
     1   Y   NA  2  NA  NA  NLD   2009   
     0   Q   NA  3  NA  NA  AUS   2011   
     1   NA  NA  4  NA  NA  AUS   2011   
     0   0   NA  7  NA  NA  NLD   2008   
     1   1   NA  1  NA  NA  NLD   2008   
     0   1   NA  3  NA  NA  AUS   2012   
     0   NA  1   NA  1  NA  ECU   2009   
     1   NA  0   NA  2  0   ECU   2009   
     0   NA  0   NA  3  0   BRA   2011   
     1   NA  0   NA  4  0   BRA   2011   
     0   NA  1   NA  7  NA  ECU   2008   
     1   NA  0   NA  1  0   ECU   2008   
     0   NA  0   NA  3  2   BRA   2012   
     1   NA  0   NA  4  NA  BRA   2012",
   header = TRUE
)

# Creates a list
df_iso <- split(df, df$iso) # Creates a list of dataframes

# Creates a list of lists
mylist.names <- names(df_iso)
df_iso_list <- vector("list", length(mylist.names))
names(df_iso_list) <- mylist.names

f <- function(x) return(list(a = list(), b = list()))
new_nested <- lapply(df_iso, f)

现在new_nested$AUS$a访问a子列表的列表AUS。到目前为止，一切都很好。

我想将df_iso_1, df_iso_2下面的两个列表 ( ) 重新分配到我刚刚创建的列表结构中。

df_iso_1 = list()
for (i in 1:length(df_iso))  {
  tryCatch({
    df_iso_1 [[i]] <- mice(df_iso[[i]], m=1, maxit = 5, seed=1)
    if (i==1000) stop("stop")
  }, error=function(e){cat("ERROR :",conditionMessage(e), "\n")})
}

df_iso_2 = list()
for (i in 1:length(df_iso))  {
  tryCatch({
    df_iso_2 [[i]] <- mice(df_iso[[i]], m=1, maxit = 5, seed=2)
    if (i==1000) stop("stop")
  }, error=function(e){cat("ERROR :",conditionMessage(e), "\n")})
}
names(df_iso_1) <- names(df_iso)
names(df_iso_2) <- names(df_iso)

虽然new_nested$AUS$a访问列表，但我想iso用索引循环代码而不是按名称引用它们：

for (n in length(df_iso)) {
new_nested$[i]$a <- df_iso_1[n]
}

然而，这不起作用。遍历这些列表的正确语法是什么？

期望的输出：

从df_iso_1和开始df_iso_2，mids对象按iso代码池化。换句话说，所有iso代码都放在新结构中：

因此new_nested，它的列表NLD填充了来自and的 NLDmids对象，其列表列表填充了来自and的 AUS对象。df_iso_1df_iso_2AUSmidsdf_iso_1df_iso_2

标签： rlistfor-looplapply

解决方案

这里有两种方法。第一个是修复你的循环seq_len(length(df_iso))并匹配你的输出，将你的更改df_iso_1[n]为df_iso1[[n]]：

for (n in seq_len(length(df_iso))) {
  new_nested[[names(df_iso)[n]]]$a <- df_iso_1[[n]] 
  new_nested[[names(df_iso)[n]]]$b <- df_iso_2[[n]]
}

new_nested$AUS$a

Class: mids
Number of multiple imputations:  1 
Imputation methods:
   A    B    C    D    E    F  iso year 
  ""   ""   ""   ""   ""   ""   ""   "" 
PredictorMatrix:
  A B C D E F iso year
A 0 0 0 0 0 0   0    1
B 0 0 0 0 0 0   0    0
C 0 0 0 0 0 0   0    0
D 1 0 0 0 0 0   0    1
E 0 0 0 0 0 0   0    0
F 0 0 0 0 0 0   0    0
Number of logged events:  6 
  it im dep      meth out
1  0  0      constant   B
2  0  0      constant   C
3  0  0      constant   E
4  0  0      constant   F
5  0  0      constant iso
6  0  0     collinear   D

第二种方法是使用mapply循环遍历列表的每个元素，df_iso_n将它们组合成一个新的列表矩阵：

mapply(list, df_iso_1, df_iso_2)

#     AUS     BRA     ECU     NLD    
#[1,] List,21 List,21 List,21 List,21
#[2,] List,21 List,21 List,21 List,21

mapply(list, df_iso_1, df_iso_2)[, 'AUS']

[[1]]
Class: mids
Number of multiple imputations:  1 
Imputation methods:
   A    B    C    D    E    F  iso year 
  ""   ""   ""   ""   ""   ""   ""   "" 
PredictorMatrix:
  A B C D E F iso year
A 0 0 0 0 0 0   0    1
B 0 0 0 0 0 0   0    0
C 0 0 0 0 0 0   0    0
D 1 0 0 0 0 0   0    1
E 0 0 0 0 0 0   0    0
F 0 0 0 0 0 0   0    0
Number of logged events:  6 
  it im dep      meth out
1  0  0      constant   B
2  0  0      constant   C
3  0  0      constant   E
4  0  0      constant   F
5  0  0      constant iso
6  0  0     collinear   D

[[2]]
Class: mids
Number of multiple imputations:  1 
Imputation methods:
   A    B    C    D    E    F  iso year 
  ""   ""   ""   ""   ""   ""   ""   "" 
PredictorMatrix:
  A B C D E F iso year
A 0 0 0 0 0 0   0    1
B 0 0 0 0 0 0   0    0
C 0 0 0 0 0 0   0    0
D 1 0 0 0 0 0   0    1
E 0 0 0 0 0 0   0    0
F 0 0 0 0 0 0   0    0
Number of logged events:  6 
  it im dep      meth out
1  0  0      constant   B
2  0  0      constant   C
3  0  0      constant   E
4  0  0      constant   F
5  0  0      constant iso
6  0  0     collinear   D

此外，考虑重构代码仍然是一个好主意。这主要通过 3 行完成所有操作：

seeds = c(1,2)

by(data = df, INDICES = df$iso,
   FUN = function(ISO) lapply(seeds, function(seed) mice(ISO, m = 1, maxit = 5, seed = seed)))

df$iso: AUS
[[1]]
Class: mids
Number of multiple imputations:  1 
Imputation methods:
   A    B    C    D    E    F  iso year 
  ""   ""   ""   ""   ""   ""   ""   "" 
PredictorMatrix:
  A B C D E F iso year
A 0 0 0 0 0 0   0    1
B 0 0 0 0 0 0   0    0
C 0 0 0 0 0 0   0    0
D 1 0 0 0 0 0   0    1
E 0 0 0 0 0 0   0    0
F 0 0 0 0 0 0   0    0
Number of logged events:  6 
  it im dep      meth out
1  0  0      constant   B
2  0  0      constant   C
3  0  0      constant   E
4  0  0      constant   F
5  0  0      constant iso
6  0  0     collinear   D

[[2]]
Class: mids
Number of multiple imputations:  1 
Imputation methods:
   A    B    C    D    E    F  iso year 
  ""   ""   ""   ""   ""   ""   ""   "" 
PredictorMatrix:
  A B C D E F iso year
A 0 0 0 0 0 0   0    1
B 0 0 0 0 0 0   0    0
C 0 0 0 0 0 0   0    0
D 1 0 0 0 0 0   0    1
E 0 0 0 0 0 0   0    0
F 0 0 0 0 0 0   0    0
Number of logged events:  6 
  it im dep      meth out
1  0  0      constant   B
2  0  0      constant   C
3  0  0      constant   E
4  0  0      constant   F
5  0  0      constant iso
6  0  0     collinear   D

---------------------------------------------------- 
df$iso: BRA
[[1]]
...

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