Converting nested list to dataframe

问题:

The goal is to convert a nested list which sometimes contain missing records into a data frame. An example of the structure when there are missing records is:

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"

When there are no missing records the list can be converted into a data frame using data.frame(do.call(rbind.data.frame, mylist)). However, when records are missing this results in a column mismatch. I know there are functions to merge data frames of non-matching columns but I'm yet to find one that can be applied to lists. The ideal outcome would keep record 2 with NA for all variables. Hoping for some help.

Edit to add 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")))

回答1:

You can also use (at least v1.9.3) of rbindlist in the data.table package:

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

回答2:

You could create a list of data.frames:

dfs <- lapply(mylist, data.frame, stringsAsFactors = FALSE)

Then use one of these:

library(plyr)
rbind.fill(dfs)

or the faster

library(dplyr)
rbind_all(dfs)

In the case of dplyr::rbind_all, I am surprised that it chooses to use "" instead of NA for missing data. If you remove stringsAsFactors = FALSE, you will get NA but at the cost of a warning... So suppressWarnings(rbind_all(lapply(mylist, data.frame))) would be an ugly but fast solution.

回答3:

I just developed a solution for this question that is applicable here, so I'll provide it here as well:

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,...));

Execution:

## 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>

My function is more powerful than data.table::rbindlist() as of 1.9.6, in that it can handle any number of nesting levels and different vector lengths across branches. In the linked question, my function correctly flattens the OP's list to a data.frame, but data.table::rbindlist() fails with "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".