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Question

我有一个具有以下结构的列表列表：

a = list("t" = list('abc'), "c" = list('def','ghi')) 
b = list("t" = list('jk','kl'), "c" = list('lmn')) 
c = list("t" = list('op','pq','z'), "c" = list('qrs','tuv','wxy'))
ls = list('one'=a,'two'=b,'three'=c)

我需要将其存储在文本文件中。由于并非所有元素的长度都相同，因此在将其转换为数据框并写入文件之前，我需要填充nan。我想要一个看起来像这样的文件（此处大致显示为.csv，但.txt可以）：

 t1, t2, t3, c1, c2, c3
one,abc,nan,nan,def,ghi,nan
two,jk,kl,nan,lmn,nan,nan
three,op,pq,z,qrs,tuv,wxy

我是R的新手，所以我对如何执行此操作只有一般的认识，无法弄清语法。伪代码中是这样的：

lapply(ls, fill nans)
lapply(ls, unlist)
lapply(ls, names=[t1,t2,t3,c1,c2,c3])
df=data.frame(ls)
write.table(df)

有人可以引导我完成这个过程吗？

编辑：我能够在以下方面取得一些进展

：

ellength <- function(ls,i) {return(length(ls[[i]]))}

fillna <- function(ls,i,m) {
if (length(ls[[i]])<m) {
return(append(ls[[i]],vector('list',length=m-length(ls[[i]]))))
}
else {return(ls[[i]])}
}

make_col <- function(ls,i){
return(lapply(ls,fillna,i=i,m=max(unlist(lapply(ls,ellength,i=i)))))
}

> matrix(list(make_col(ls,'t'),make_col(ls,'c')))
     [,1]  
[1,] List,3
[2,] List,3

但是我仍然无法以任何连贯的方式将其写入文件。这在Python中是如此简单。我肯定错过了什么。帮助吗？

Answer 1

这完全有帮助吗？

library(tidyr)
library(dplyr)
a = list("t" = 'abc', "c" = list('def','ghi')) 
b = list("t" = 'jk', "c" = list('lmn')) 
c = list("t" = 'op', "c" = list('qrs','tuv','wxy'))
ls = list('one'=a,'two'=b,'three'=c)
#unlist and turn into a df
lx <- as.data.frame( unlist(ls),stringsAsFactors = FALSE)
#make rownames as column
lx$nms <- rownames(lx)

#split nms column so you can transpose your data
lx <- separate(lx, nms, c("v1","v2"), sep = "[.]")
lx <- mutate(lx, v3 = `unlist(ls)`) %>% 
      select(-`unlist(ls)`)
#transpose your data - it fills with NA
#NaN is a numeric field so you can't use it to fill character variables
lx2 <- spread(lx,v2,v3)

更新

如果您希望对变量进行折叠，可以使用ifelse替换某些列中的NA：

lx2 <- mutate(lx2, c_1 = ifelse(is.na(c),c1,c))
lx2 <- mutate(lx2, t_1 = ifelse(is.na(t),t1,t))
lx3 <- lx2[c('v1','c_1','c2','c3','t_1','t2','t3')]

Answer 2

考虑在不列出嵌套列表之后构建数据帧列表：

char_vec <- unlist(ls)

df_list <- lapply(names(ls), function(x) {
  tmp <- data.frame(t(char_vec[names(char_vec)[grep(x, names(char_vec))]]),
                    stringsAsFactors = FALSE)
  names(tmp) <- gsub(".*\\.", "", names(tmp))

  return(tmp)
})

df_list
# [[1]]
#     t  c1  c2
# 1 abc def ghi

# [[2]]
#   t1 t2   c
# 1 jk kl lmn

# [[3]]
#   t1 t2 t3  c1  c2  c3
# 1 op pq  z qrs tuv wxy

要将所有数据框项目绑定在一起，可以使用dplyr，data.table甚至base R：

dplyr （带有bind_rows）

final_df1 <- bind_rows(df_list)

# CLEAN UP AND RE-ORDER COLUMNS
final_df1 <- transform(final_df1, 
                       t1 = ifelse(is.na(t), t1, t),
                       t = NULL,
                       c1 = ifelse(is.na(c), c1, c),
                       c = NULL
              )

final_df1 <- final_df1[order(names(final_df1))]

data.table （带有rbindlist）

final_df2 <- data.frame(rbindlist(df_list, fill=TRUE))

# CLEAN UP AND RE-ORDER COLUMNS
final_df2 <- transform(final_df2, 
                       t1 = ifelse(is.na(t), t1, t),
                       t = NULL,
                       c1 = ifelse(is.na(c), c1, c),
                       c = NULL
              )

final_df2 <- final_df2[order(names(final_df2))]

基本 R （带有do.call）

# RETRIEVE ALL COLUMN NAMES
nms <- names(unlist(df_list))

df_list <- lapply(df_list, function(df){
  # CREATE BLANK COLUMNS FOR rbind
  for(i in nms) {
    if(!i %in% names(df)) {
      df[[i]] <- NA
    }
  }

  # CLEAN UP AND RE-ORDER COLUMNS
  df <- within(df, {t1 <- ifelse(is.na(t), t1, t)
                    t <- NULL
                    c1 <- ifelse(is.na(c), c1, c)
                    c <- NULL
        })

  return(df[order(names(df))])
})

final_df3 <- do.call(rbind, df_list)

输出

final_df1
#    c1   c2   c3  t1   t2   t3
# 1 def  ghi <NA> abc <NA> <NA>
# 2 lmn <NA> <NA>  jk   kl <NA>
# 3 qrs  tuv  wxy  op   pq    z

identical(final_df1, final_df2)
# [1] TRUE
identical(final_df1, final_df3)
# [1] TRUE
identical(final_df2, final_df3)
# [1] TRUE

R：将列表列表的列表写到带有名称的文本文件中

2 个答案:

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