目标是将有时包含缺失记录的嵌套列表转换为数据框。缺少记录时的结构示例如下：
str(mylist)
List of 3
$ :List of 7
..$ Hit : chr "True"
..$ Project: chr "Blue"
..$ Year : chr "2011"
..$ Rating : chr "4"
..$ Launch : chr "26 Jan 2012"
..$ ID : chr "19"
..$ Dept : chr "1, 2, 4"
$ :List of 2
..$ Hit : chr "False"
..$ Error: chr "Record not found"
$ :List of 7
..$ Hit : chr "True"
..$ Project: chr "Green"
..$ Year : chr "2004"
..$ Rating : chr "8"
..$ Launch : chr "29 Feb 2004"
..$ ID : chr "183"
..$ Dept : chr "6, 8"
当没有丢失记录时，可以使用data.frame(do.call(rbind.data.frame, mylist))将列表转换为数据框。但是，如果缺少记录，则会导致列不匹配。我知道有一些功能可以合并非匹配列的数据帧，但我还没有找到一个可以应用于列表的数据帧。对于所有变量，理想的结果将保留记录2和NA。希望得到一些帮助。
修改以添加dput(mylist)：
list(structure(list(Hit = "True", Project = "Blue", Year = "2011",
Rating = "4", Launch = "26 Jan 2012", ID = "19", Dept = "1, 2, 4"), .Names = c("Hit",
"Project", "Year", "Rating", "Launch", "ID", "Dept")), structure(list(
Hit = "False", Error = "Record not found"), .Names = c("Hit",
"Error")), structure(list(Hit = "True", Project = "Green", Year = "2004",
Rating = "8", Launch = "29 Feb 2004", ID = "183", Dept = "6, 8"), .Names = c("Hit",
"Project", "Year", "Rating", "Launch", "ID", "Dept")))
答案 0 :(得分：29)
您还可以在rbindlist包中使用data.table（至少v1.9.3）：
library(data.table)
rbindlist(mylist, fill=TRUE)
## Hit Project Year Rating Launch ID Dept Error
## 1: True Blue 2011 4 26 Jan 2012 19 1, 2, 4 NA
## 2: False NA NA NA NA NA NA Record not found
## 3: True Green 2004 8 29 Feb 2004 183 6, 8 NA
答案 1 :(得分：9)
您可以创建data.frames列表：
dfs <- lapply(mylist, data.frame, stringsAsFactors = FALSE)
然后使用其中一个：
library(plyr)
rbind.fill(dfs)
或更快
library(dplyr)
rbind_all(dfs)
对于dplyr::rbind_all，我很惊讶它选择使用""代替NA来删除数据。如果您删除stringsAsFactors = FALSE，则会收到NA，但会以警告为代价......所以suppressWarnings(rbind_all(lapply(mylist, data.frame)))将是一个丑陋但快速的解决方案。
答案 2 :(得分：9)
我刚刚开发了适用于this question的解决方案，所以我也会在这里提供：
tl <- function(e) { if (is.null(e)) return(NULL); ret <- typeof(e); if (ret == 'list' && !is.null(names(e))) ret <- list(type='namedlist') else ret <- list(type=ret,len=length(e)); ret; };
mkcsv <- function(v) paste0(collapse=',',v);
keyListToStr <- function(keyList) paste0(collapse='','/',sapply(keyList,function(key) if (is.null(key)) '*' else paste0(collapse=',',key)));
extractLevelColumns <- function(
nodes, ## current level node selection
..., ## additional arguments to data.frame()
keyList=list(), ## current key path under main list
sep=NULL, ## optional string separator on which to join multi-element vectors; if NULL, will leave as separate columns
mkname=function(keyList,maxLen) paste0(collapse='.',if (is.null(sep) && maxLen == 1L) keyList[-length(keyList)] else keyList) ## name builder from current keyList and character vector max length across node level; default to dot-separated keys, and remove last index component for scalars
) {
cat(sprintf('extractLevelColumns(): %s\n',keyListToStr(keyList)));
if (length(nodes) == 0L) return(list()); ## handle corner case of empty main list
tlList <- lapply(nodes,tl);
typeList <- do.call(c,lapply(tlList,`[[`,'type'));
if (length(unique(typeList)) != 1L) stop(sprintf('error: inconsistent types (%s) at %s.',mkcsv(typeList),keyListToStr(keyList)));
type <- typeList[1L];
if (type == 'namedlist') { ## hash; recurse
allKeys <- unique(do.call(c,lapply(nodes,names)));
ret <- do.call(c,lapply(allKeys,function(key) extractLevelColumns(lapply(nodes,`[[`,key),...,keyList=c(keyList,key),sep=sep,mkname=mkname)));
} else if (type == 'list') { ## array; recurse
lenList <- do.call(c,lapply(tlList,`[[`,'len'));
maxLen <- max(lenList,na.rm=T);
allIndexes <- seq_len(maxLen);
ret <- do.call(c,lapply(allIndexes,function(index) extractLevelColumns(lapply(nodes,function(node) if (length(node) < index) NULL else node[[index]]),...,keyList=c(keyList,index),sep=sep,mkname=mkname))); ## must be careful to translate out-of-bounds to NULL; happens automatically with string keys, but not with integer indexes
} else if (type%in%c('raw','logical','integer','double','complex','character')) { ## atomic leaf node; build column
lenList <- do.call(c,lapply(tlList,`[[`,'len'));
maxLen <- max(lenList,na.rm=T);
if (is.null(sep)) {
ret <- lapply(seq_len(maxLen),function(i) setNames(data.frame(sapply(nodes,function(node) if (length(node) < i) NA else node[[i]]),...),mkname(c(keyList,i),maxLen)));
} else {
## keep original type if maxLen is 1, IOW don't stringify
ret <- list(setNames(data.frame(sapply(nodes,function(node) if (length(node) == 0L) NA else if (maxLen == 1L) node else paste(collapse=sep,node)),...),mkname(keyList,maxLen)));
}; ## end if
} else stop(sprintf('error: unsupported type %s at %s.',type,keyListToStr(keyList)));
if (is.null(ret)) ret <- list(); ## handle corner case of exclusively empty sublists
ret;
}; ## end extractLevelColumns()
## simple interface function
flattenList <- function(mainList,...) do.call(cbind,extractLevelColumns(mainList,...));
执行：
## define data
mylist <- list(structure(list(Hit='True',Project='Blue',Year='2011',Rating='4',Launch='26 Jan 2012',ID='19',Dept='1, 2, 4'),.Names=c('Hit','Project','Year','Rating','Launch','ID','Dept')),structure(list(Hit='False',Error='Record not found'),.Names=c('Hit','Error')),structure(list(Hit='True',Project='Green',Year='2004',Rating='8',Launch='29 Feb 2004',ID='183',Dept='6, 8'),.Names=c('Hit','Project','Year','Rating','Launch','ID','Dept')));
## run it
df <- flattenList(mylist);
## extractLevelColumns():
## extractLevelColumns(): Hit
## extractLevelColumns(): Project
## extractLevelColumns(): Year
## extractLevelColumns(): Rating
## extractLevelColumns(): Launch
## extractLevelColumns(): ID
## extractLevelColumns(): Dept
## extractLevelColumns(): Error
df;
## Hit Project Year Rating Launch ID Dept Error
## 1 True Blue 2011 4 26 Jan 2012 19 1, 2, 4 <NA>
## 2 False <NA> <NA> <NA> <NA> <NA> <NA> Record not found
## 3 True Green 2004 8 29 Feb 2004 183 6, 8 <NA>
我的函数比1.9.6中的data.table::rbindlist()更强大，因为它可以处理任意数量的嵌套级别和跨分支的不同向量长度。在链接的问题中，我的功能正确地将OP列表展平为data.frame，但data.table::rbindlist()失败并显示"Error in rbindlist(jsonRList, fill = T) : Column 4 of item 16 is length 2, inconsistent with first column of that item which is length 1. rbind/rbindlist doesn't recycle as it already expects each item to be a uniform list, data.frame or data.table"。
答案 3 :(得分：0)
这是一个将任何嵌套/不均匀列表转换为数据框的解决方案。 rbindlist 不适用于许多情况，尤其是列表列表。所以我必须创造出比 rbindlist 更好的东西。
rbindlist.v2 <- function(l)
{
l <- l[lapply(l, class) == "list"]
df <- foreach(element = l, .combine = bind_rows, .errorhandling = 'remove') %do%
{df = unlist(element); df = as.data.frame(t(df)); rm(element); return(df)}
rm(l)
return(df)
}
对于大型列表，您可以通过将 %do% 替换为 %dopar% 来加快流程。这也是我的案例所需要的东西。