我想收集（）列表列以在我的数据框中创建新行。我正在使用repurrrsive包中的权力的游戏数据集。以下是我设置问题的代码：
library(tidyverse)
got_chars <- repurrrsive::got_chars
df <- got_chars %>%
{
tibble::tibble(
Name = map_chr(., 'name'),
Gender = map_chr(.,'gender'),
Culture = map_chr(.,'culture'),
Born = map_chr(.,'born'),
Alive = map_chr(.,'alive'),
Titles = map(.,'titles'),
Aliases = map(., "aliases"),
Allegiances = map(., "allegiances"),
Books = map(.,'books'),
POV_Books = map(.,'povBooks'),
TV_Series = map(.,'tvSeries'),
Actor = map(.,'playedBy')
)
}
我希望能够做到，但无法弄清楚是gather()列表列（例如Books，POV_Books等），以便为每条记录创建一个新行。例如：
姓名|图书
Theon Greyjoy |权力的游戏
Theon Greyjoy |剑的风暴
Theon Greyjoy |乌鸦的盛宴
我能得到的最接近的是：
df_books <- df %>%
separate_rows(Books,sep="\"")
这样可以工作，但是从向量中的c（）字符留下一堆垃圾。我可以过滤掉那些，但我觉得有一个更好的方法，我可能只是没有尝试正确的功能。任何建议都将非常感谢，谢谢！
答案 0 :(得分：2)
你的目标看起来像这样：
df
# # A tibble: 30 x 12
# Name Gender Culture Born Alive Titles Aliases Allegiances Books POV_Books TV_Series Actor
# <chr> <chr> <chr> <chr> <chr> <list> <list> <list> <lis> <list> <list> <lis>
# 1 Theon Greyjoy Male Ironbo… In 2… TRUE <chr … <chr [… <chr [1]> <chr… <chr [2]> <chr [6]> <chr…
# 2 Tyrion Lannister Male "" In 2… TRUE <chr … <chr [… <chr [1]> <chr… <chr [4]> <chr [6]> <chr…
# 3 Victarion Greyjoy Male Ironbo… In 2… TRUE <chr … <chr [… <chr [1]> <chr… <chr [2]> <chr [1]> <chr…
# 4 Will Male "" "" FALSE <chr … <chr [… <NULL> <chr… <chr [1]> <chr [1]> <chr…
# 5 Areo Hotah Male Norvos… In 2… TRUE <chr … <chr [… <chr [1]> <chr… <chr [2]> <chr [2]> <chr…
# 6 Chett Male "" At H… FALSE <chr … <chr [… <NULL> <chr… <chr [1]> <chr [1]> <chr…
# 7 Cressen Male "" In 2… FALSE <chr … <chr [… <NULL> <chr… <chr [1]> <chr [1]> <chr…
# 8 Arianne Martell Female Dornish In 2… TRUE <chr … <chr [… <chr [1]> <chr… <chr [1]> <chr [1]> <chr…
# 9 Daenerys Targaryen Female Valyri… In 2… TRUE <chr … <chr [… <chr [1]> <chr… <chr [4]> <chr [6]> <chr…
# 10 Davos Seaworth Male Wester… In 2… TRUE <chr … <chr [… <chr [2]> <chr… <chr [3]> <chr [5]> <chr…
# # ... with 20 more rows
unnest()将是一个显而易见的选择，但如果所有列表在扩展到的值数量方面相同，则不起作用。
library(tidyverse)
unnest(df)
# Error: All nested columns must have the same number of elements.
一种方法是使用以下功能。 flatten()使数据“宽”，flattenLong()获取“广泛”数据并使其“长”。关于缺失数据的假设是，如果列表项中的向量比另一个列表项中的匹配向量短，则缺失的数据是最后的。
flatten <- function(indt, cols, drop = FALSE) {
require(data.table)
if (!is.data.table(indt)) indt <- as.data.table(indt)
x <- unlist(indt[, lapply(.SD, function(x) max(lengths(x))), .SDcols = cols])
nams <- paste(rep(cols, x), sequence(x), sep = "_")
indt[, (nams) := unlist(lapply(.SD, data.table::transpose), recursive = FALSE), .SDcols = (cols)]
if (isTRUE(drop)) indt[, (cols) := NULL]
indt[]
}
flattenLong <- function(indt, cols) {
ob <- setdiff(names(indt), cols)
x <- flatten(indt, cols, TRUE)
mv <- lapply(cols, function(y) grep(sprintf("^%s_", y), names(x)))
setorderv(melt(x, measure.vars = mv, value.name = cols), ob)[]
}
以下是使用它的一种方法，将其应用于所有list列。
flattenLong(df, names(df)[sapply(df, is.list)])
# Name Gender Culture Born Alive variable
# 1: Aeron Greyjoy Male Ironborn In or between 269 AC and 273 AC, at Pyke TRUE 1
# 2: Aeron Greyjoy Male Ironborn In or between 269 AC and 273 AC, at Pyke TRUE 2
# 3: Aeron Greyjoy Male Ironborn In or between 269 AC and 273 AC, at Pyke TRUE 3
# 4: Aeron Greyjoy Male Ironborn In or between 269 AC and 273 AC, at Pyke TRUE 4
# 5: Aeron Greyjoy Male Ironborn In or between 269 AC and 273 AC, at Pyke TRUE 5
# ---
# 476: Will Male FALSE 12
# 477: Will Male FALSE 13
# 478: Will Male FALSE 14
# 479: Will Male FALSE 15
# 480: Will Male FALSE 16
# Titles Aliases Allegiances Books
# 1: Priest of the Drowned God The Damphair House Greyjoy of Pyke A Game of Thrones
# 2: Captain of the Golden Storm (formerly) Aeron Damphair NA A Clash of Kings
# 3: NA NA NA A Storm of Swords
# 4: NA NA NA A Dance with Dragons
# 5: NA NA NA NA
# ---
# 476: NA NA NA NA
# 477: NA NA NA NA
# 478: NA NA NA NA
# 479: NA NA NA NA
# 480: NA NA NA NA
# POV_Books TV_Series Actor
# 1: A Feast for Crows Season 6 Michael Feast
# 2: NA NA NA
# 3: NA NA NA
# 4: NA NA NA
# 5: NA NA NA
# ---
# 476: NA NA NA
# 477: NA NA NA
# 478: NA NA NA
# 479: NA NA NA
# 480: NA NA NA
您还可以执行以下任何操作来处理单个列：
flattenLong(df[c(names(df)[!sapply(df, is.list)], "Books")], "Books")
flattenLong(df[c("Name", "Gender", "Culture", "Born", "Alive", "Books")], "Books")
df %>%
select(Name, Gender, Culture, Born, Alive, Books) %>%
flattenLong("Books")
这完全不等同于“tidyverse”方法。它以不同的方式处理NULL，并且unnest每个组的长度相同。请考虑以下数据集：
mydf <- data.frame(V1 = c("a", "b", "c"),
V2 = I(list(c(10, 20), NA_real_, c(20, 40, 60))),
V3 = I(list(NULL, c("x", "y", "z"), c("BA", "BB"))))
mydf
# V1 V2 V3
# 1 a 10, 20
# 2 b NA x, y, z
# 3 c 20, 40, 60 BA, BB
差异＃1：每组的值数：
# Note the resulting number of values per group
# Equivalent of
# as.data.table(mydf)[, list(unlist(V2)), V1]
mydf %>% select(V1, V2) %>% unnest()
# V1 V2
# 1 a 10
# 2 a 20
# 3 b NA
# 4 c 20
# 5 c 40
# 6 c 60
flattenLong(mydf[c("V1", "V2")], "V2")
# V1 variable V2
# 1: a V2_1 10
# 2: a V2_2 20
# 3: a V2_3 NA
# 4: b V2_1 NA
# 5: b V2_2 NA
# 6: b V2_3 NA
# 7: c V2_1 20
# 8: c V2_2 40
# 9: c V2_3 60
差异＃2：处理NULL值
mydf %>% select(V1, V3) %>% unnest()
# Error: Each column must either be a list of vectors or a list of data frames [V3]
flattenLong(mydf[c("V1", "V2")], "V2")
# V1 variable V3
# 1: a V3_1 NA
# 2: a V3_2 NA
# 3: a V3_3 NA
# 4: b V3_1 x
# 5: b V3_2 y
# 6: b V3_3 z
# 7: c V3_1 BA
# 8: c V3_2 BB
# 9: c V3_3 NA
答案 1 :(得分：1)
您可以使用unnest，但首先您必须以tidyr理解的方式格式化列。
这意味着：
NULL个元素data.frames而不是vectors
library(tidyverse)
df %>%
select(Name,Books) %>% # skip this line to keep all columns
slice(which(lengths(Books)>0)) %>%
mutate(Books = map(Books,~tibble(Book=.x))) %>%
unnest(Books)
# # A tibble: 77 x 2
# Name Book
# <chr> <chr>
# 1 Theon Greyjoy A Game of Thrones
# 2 Theon Greyjoy A Storm of Swords
# 3 Theon Greyjoy A Feast for Crows
# 4 Tyrion Lannister A Feast for Crows
# 5 Tyrion Lannister The World of Ice and Fire
# 6 Victarion Greyjoy A Game of Thrones
# 7 Victarion Greyjoy A Clash of Kings
# 8 Victarion Greyjoy A Storm of Swords
# 9 Will A Clash of Kings
# 10 Areo Hotah A Game of Thrones
# # ... with 67 more rows
如果我们过滤输出（与我的解决方案相同的输出），您尝试过的解决方案工作正常：
df %>%
select(Name, Books) %>%
separate_rows(Books,sep="\"") %>%
filter(!Books %in% c("c(",", ",")") & lengths(Books)>0)