如何将列表列表保存到R中的CSV文件?

Question

我有一个包含864项的列表列表存储在一个变量中，如下所示:

→allpermsGlob

[[1]]
[[1]][[1]]
[1] "Agent is selling "

[[1]][[2]]
[1] "Yes, if asked to subscribe "

[[1]][[3]]
[1] "Yes, if offered the two-year option "

[[1]][[4]]
[1] "No, Agent should offer"

[[1]][[5]]
[1] "No, it is never ethical"
.
.
.
.
.
[[864]]
[[864]][[1]]
[1] "Agent is selling "
[[864]][[2]]
[1] "Yes, if asked to subscribe"
[[864]][[3]]
[1] "Yes, if offered the two-year "
[[864]][[4]]
[1] "No, Agent should offer "
[[864]][[5]]
[1] "No, it is never ethical 

我需要输出存储在一个csv文件与2列．第一列将方括号替换为Stem &4个选项，如下例所示&第二列应包含所有864项的对应值&对于所有864项，应该如下所示

'Stem',"Agent is selling  subscriptions "

"Option 1",
"Yes, if asked to subscribe"

"Option 2",
"Yes, if offered the two-year option "

"Option 3",
"No, Agent should offer "

"Option 4",
"No, it is never ethical 

我如何实现这一点?

执行dput(head(yourlist))给出-

list(list("Agent is selling subscriptions ", 
"Yes, if offered the two-year option", 
"Yes, if asked to subscribe ", 
"No, Agent should offer ", 
"No, it is never ethical"), 
list("Agent is selling subscriptions", 
"Yes, if offered the two-year option", 
"Yes, if asked to subscribe ", 
"No, it is never ethical ", 
"No, Agent should offer "))

Answer 1

下面是一个小例子:

> x <- list(as.list(letters[1:5])) #same structure as OP, different text
> print(x)
[[1]]
[[1]][[1]]
[1] "a"
[[1]][[2]]
[1] "b"
[[1]][[3]]
[1] "c"
[[1]][[4]]
[1] "d"
[[1]][[5]]
[1] "e"
> y <- as.data.frame(x)
> y
X.a. X.b. X.c. X.d. X.e.
1    a    b    c    d    e
> names(y) <- c("stem", "opt1", "opt2", "opt3", "opt4")
> y
stem opt1 opt2 opt3 opt4
1    a    b    c    d    e

则可以使用write.csv()将y导出为CSV文件。不确定如何将y作为数据帧和仍然像您所示的那样打印。也许你只是想要一个名字列表?也许这会让你更接近你想要的输出:

> sapply(y, function(i) list(i))
$stem
[1] "a"
$opt1
[1] "b"
$opt2
[1] "c"
$opt3
[1] "d"
$opt4
[1] "e"

根据你的项目有多大，你也可以考虑创建x作为list类的一个特例，它有自己的print方法。

Answer 2

假设这个示例模拟了您的数据结构，下面是一个可能的解决方案:

# the structure of your object
ex_list <- list(list("a", "b", "c"),
list("a", "b", "c"))
print(ex_list)
#> [[1]]
#> [[1]][[1]]
#> [1] "a"
#> 
#> [[1]][[2]]
#> [1] "b"
#> 
#> [[1]][[3]]
#> [1] "c"
#> 
#> 
#> [[2]]
#> [[2]][[1]]
#> [1] "a"
#> 
#> [[2]][[2]]
#> [1] "b"
#> 
#> [[2]][[3]]
#> [1] "c"
# give names to each otion a, b and c
ex_list <- lapply(ex_list, function(x) setNames(x, c("stem", "opt1", "opt2")))
# merge everything in a data.frame
temp <- unlist(ex_list)
out <- data.frame(ID = rep(1:length(ex_list), lengths(ex_list)), 
Options = names(temp), 
Values = unlist(temp))
print(out)
#>   ID Options Values
#> 1  1    stem      a
#> 2  1    opt1      b
#> 3  1    opt2      c
#> 4  2    stem      a
#> 5  2    opt1      b
#> 6  2    opt2      c

^{创建于2022-12-20与reprex v2.0.2}

那么您可以使用write.csv()和对象out来获得CSV文件。

创意来自:https://www.r-bloggers.com/2021/12/an-easy-to-convert-list-to-long-table/

Answer 3

如果所有列表都是完整的，我们可以使用cbind的回收能力:

cbind(Options = c("Stem1", "Option1", "Option2", "Option3", "Option4"), Values = unlist(allpermsGlob)) |>
write.csv("test.xls", row.names = FALSE)

输出(csv):

"Options","Values"
"Stem1","Agent is selling subscriptions "
"Option1","Yes, if offered the two-year option"
"Option2","Yes, if asked to subscribe "
"Option3","No, Agent should offer "
"Option4","No, it is never ethical"
"Stem1","Agent is selling subscriptions"
"Option1","Yes, if offered the two-year option"
"Option2","Yes, if asked to subscribe "
"Option3","No, it is never ethical "
"Option4","No, Agent should offer "