在 R 中处理一个复杂的列表

Question

我在 R 中有一个复杂的列表结构，我想将其放入数据框格式中。数据如下。该列表包含五个不同音乐专辑的八列数据。每个列表元素中的行数等于为相册贡献的人员数量，这是我的主要兴趣。

将列表转换为 12 行 8 列的数据框的有效方法是什么？理想情况下，我想创建一个函数，它接受一个参数（具有相同 8 列的任意数量的音乐专辑的列表）和 returns 一个数据框。

我尝试过使用 flatten、unnest 和 map 的各种方法，但无法破解这个。

df2<-list(NULL, structure(list(join = c(\"/\", \"/\", \"/\", \"/\", \"/\", \"\"\r\n), name = c(\"Art Tatum\", \"Lionel Hampton\", \"Harry Edison\", \"Buddy Rich\", \r\n\"Red Callender\", \"Barney Kessel\"), anv = c(\"\", \"\", \"\", \"\", \"\", \r\n\"\"), tracks = c(\"\", \"\", \"\", \"\", \"\", \"\"), role = c(\"\", \"\", \"\", \r\n\"\", \"\", \"\"), resource_url = c(\"https://api.discogs.com/artists/265634\", \r\n\"https://api.discogs.com/artists/136133\", \"https://api.discogs.com/artists/258469\", \r\n\"https://api.discogs.com/artists/57620\", \"https://api.discogs.com/artists/272014\", \r\n\"https://api.discogs.com/artists/253476\"), id = c(265634L, 136133L, \r\n258469L, 57620L, 272014L, 253476L), `frset$title` = c(\"The Tatum Group Masterpieces, Vol. 5\", \r\n\"The Tatum Group Masterpieces, Vol. 5\", \"The Tatum Group Masterpieces, Vol. 5\", \r\n\"The Tatum Group Masterpieces, Vol. 5\", \"The Tatum Group Masterpieces, Vol. 5\", \r\n\"The Tatum Group Masterpieces, Vol. 5\")), class = \"data.frame\", row.names = c(NA, \r\n6L)), structure(list(join = \"\", name = \"Art Tatum\", anv = \"\", \r\n    tracks = \"\", role = \"\", resource_url = \"https://api.discogs.com/artists/265634\", \r\n    id = 265634L, `frset$title` = \"This Is...Art Tatum  - Vol 1\"), class = \"data.frame\", row.names = 1L), \r\n    structure(list(join = c(\"/\", \"/\", \"\"), name = c(\"Oscar Peterson\", \r\n    \"Erroll Garner\", \"Art Tatum\"), anv = c(\"\", \"\", \"\"), tracks = c(\"\", \r\n    \"\", \"\"), role = c(\"\", \"\", \"\"), resource_url = c(\"https://api.discogs.com/artists/254394\", \r\n    \"https://api.discogs.com/artists/262816\", \"https://api.discogs.com/artists/265634\"\r\n    ), id = c(254394L, 262816L, 265634L), `frset$title` = c(\"Great Jazz Pianists\", \r\n    \"Great Jazz Pianists\", \"Great Jazz Pianists\")), class = \"data.frame\", row.names = c(NA, \r\n    3L)), structure(list(join = \"\", name = \"Art Tatum\", anv = \"\", \r\n        tracks = \"\", role = \"\", resource_url = \"https://api.discogs.com/artists/265634\", \r\n        id = 265634L, `frset$title` = \"Art Tatum\"), class = \"data.frame\", row.names = 1L), \r\n    structure(list(join = \"\", name = \"Art Tatum\", anv = \"\", tracks = \"\", \r\n        role = \"\", resource_url = \"https://api.discogs.com/artists/265634\", \r\n        id = 265634L, `frset$title` = \"Art!\"), class = \"data.frame\", row.names = 1L))
df2<-unlist(lapply(df, function(x) if (length(x)==8) list(x) else x), recursive=FALSE) # remove NULL lists or those with wrong dimensions

albumdata<-list()
personnel<-function(df){
 for(i in 1:length(df))
  albumdata[[i]]<-as.data.frame(unlist(df[i+1],recursive=FALSE))
  return(albumdata)
  message("Album #",dim(albumdata[[2]])[1]) 
      }

这个函数returns又是一个列表。为什么？

Answer 1

清理数据后，效果很好

jsonlite::rbind_pages(df2)

在 R 中处理一个复杂的列表

Wrangling a Complex List in R

r

list

discogs-api