我尝试了相关的解决方案，但它们不适合我的情况。我有一个数据框，在一列中有一个嵌套列表，我想拆分这个列表并将其放在列中。列表包含另一个列表，其中包含每个月的时间戳（ts）和每个月的消耗量（v）。数据框是：
id monthly_consum
1 112 list1
2 34 list2
3 54 list3
其中
list1<-list(list(ts = "2016-01-01T00:00:00+01:00", v = 466.6),list(ts = "2016-02-01T00:00:00+01:00", v = 565.6),
list(ts = "2016-03-01T00:00:00+01:00", v = 765.6),list(ts = "2016-04-01T00:00:00+01:00", v = 888.6),
list(ts = "2016-05-01T00:00:00+01:00", v = 465),list(ts = "2016-06-01T00:00:00+01:00", v = 465.6),
list(ts = "2016-07-01T00:00:00+01:00", v = 786),list(ts = "2016-08-01T00:00:00+01:00", v = 435),
list(ts = "2016-09-01T00:00:00+01:00", v = 568),list(ts = "2016-10-01T00:00:00+01:00", v = 678),
list(ts = "2016-11-01T00:00:00+01:00", v = 522),list(ts = "2016- 12-01T00:00:00+01:00", v = 555))
list2<-list(list(ts = "2016-01-01T00:00:00+01:00", v = 333.6),list(ts = "2016-02-01T00:00:00+01:00", v = 565.6),
list(ts = "2016-03-01T00:00:00+01:00", v = 765.6),list(ts = "2016-04-01T00:00:00+01:00", v = 333.6),
list(ts = "2016-05-01T00:00:00+01:00", v = 465),list(ts = "2016-06-01T00:00:00+01:00", v = 465.6),
list(ts = "2016-07-01T00:00:00+01:00", v = 786),list(ts = "2016-08-01T00:00:00+01:00", v = 435),
list(ts = "2016-09-01T00:00:00+01:00", v = 568),list(ts = "2016-10-01T00:00:00+01:00", v = 678),
list(ts = "2016-11-01T00:00:00+01:00", v = 522),list(ts = "2016-12-01T00:00:00+01:00", v = 555))
list3<-list(list(ts = "2016-01-01T00:00:00+01:00", v = 323.6),list(ts = "2016-02-01T00:00:00+01:00", v = 565.6),
list(ts = "2016-03-01T00:00:00+01:00", v = 333.6),list(ts = "2016-04-01T00:00:00+01:00", v = 888.6),
list(ts = "2016-05-01T00:00:00+01:00", v = 465),list(ts = "2016-06-01T00:00:00+01:00", v = 465.6),
list(ts = "2016-07-01T00:00:00+01:00", v = 786),list(ts = "2016-08-01T00:00:00+01:00", v = 435),
list(ts = "2016-09-01T00:00:00+01:00", v = 568),list(ts = "2016-10-01T00:00:00+01:00", v = 678),
list(ts = "2016-11-01T00:00:00+01:00", v = 522),list(ts = "2016-12-01T00:00:00+01:00", v = 555))
我想拆分列表并创建一个数据帧，该数据帧将具有以下两种格式之一：
id ts.1 cons.1 ts.2 cons.2 ts.3 etc..
1 112 2016-01-01T00:00:00+01:00 466.6 2016-02.. ... ...
2 34 2016-01-01T00:00:00+01:00 333.6 2016-02.. ... ...
3 54 2016-01-01T00:00:00+01:00 323.6 2016-02.. ... ...
OR
id ts consumption
112 2016-01-01T00:00:00+01:00 466.6
112 2016-02-01T00:00:00+01:00 565.6
112 2016-03-01T00:00:00+01:00 765.6
112 2016-04-01T00:00:00+01:00 888.6
112 2016-05-01T00:00:00+01:00 465
112 2016-06-01T00:00:00+01:00 465.6
112 2016-07-01T00:00:00+01:00 786
112 2016-08-01T00:00:00+01:00 435
112 2016-09-01T00:00:00+01:00 568
112 2016-10-01T00:00:00+01:00 678
112 2016-11-01T00:00:00+01:00 522
112 2016-12-01T00:00:00+01:00 555
34 2016-01-01T00:00:00+01:00 466.6
34 2016-02-01T00:00:00+01:00 333.6
34 2016-03-01T00:00:00+01:00 323.6
etc............
＆＃34; rbindlist（....）出错： 列表输入的第1项不是data.frame，data.table或list＆＃34;
提前谢谢！
更新 使用dput我会得到（在真正的问题中）：
>dput(locs_total[9:12,1:5])
structure(list(X.dep_id. = c("34", "34", "34", "34"), X.loc_id. = c("17761",
"17406", "23591", "27838"), X.surface. = c("200", "1250", "54",
"150"), X.sector. = c("HOUSING", "SMALL-STORE-FOOD", "LIBRARY",
"OFFICE-BUILDING"),
X.avg_cons_main. = list(list(structure(list(
ts = "2016-01-01T00:00:00+01:00", v = 466.65), .Names = c("ts",
"v")), structure(list(ts = "2016-02-01T00:00:00+01:00", v = 406.45),
.Names = c("ts",
"v")), structure(list(ts = "2016-03-01T00:00:00+01:00", v = 483.35),
.Names = c("ts",
"v")), structure(list(ts = "2016-04-01T00:00:00+02:00", v = 79.45), .
Names = c("ts",
"v"))), NULL, NULL, NULL)), .Names = c("X.dep_id.", "X.loc_id.",
"X.surface.", "X.sector.", "X.avg_cons_main."
), row.names = c("9", "10", "11", "12"), class = "data.frame")
答案 0 :(得分：0)
我们可以遍历list
res <- do.call(rbind, Map(cbind, id = df1$id, lapply(mget(df1$monthly_consum),
function(x) do.call(rbind.data.frame, x))))
names(res)[3] <- "consumption"
row.names(res) <- NULL
head(res, 14)
# id ts consumption
#1 112 2016-01-01T00:00:00+01:00 466.6
#2 112 2016-02-01T00:00:00+01:00 565.6
#3 112 2016-03-01T00:00:00+01:00 765.6
#4 112 2016-04-01T00:00:00+01:00 888.6
#5 112 2016-05-01T00:00:00+01:00 465.0
#6 112 2016-06-01T00:00:00+01:00 465.6
#7 112 2016-07-01T00:00:00+01:00 786.0
#8 112 2016-08-01T00:00:00+01:00 435.0
#9 112 2016-09-01T00:00:00+01:00 568.0
#10 112 2016-10-01T00:00:00+01:00 678.0
#11 112 2016-11-01T00:00:00+01:00 522.0
#12 112 2016- 12-01T00:00:00+01:00 555.0
#13 34 2016-01-01T00:00:00+01:00 333.6
#14 34 2016-02-01T00:00:00+01:00 565.6
df1 <- structure(list(id = c(112L, 34L, 54L), monthly_consum = c("list1",
"list2", "list3")), .Names = c("id", "monthly_consum"),
class = "data.frame", row.names = c("1", "2", "3"))
答案 1 :(得分：0)
如果ID也在列表中，您可以使用dplyr::bind_rows
dplyr::bind_rows(list1, list2, list3)
# A tibble: 36 × 2
ts v
<chr> <dbl>
1 2016-01-01T00:00:00+01:00 466.6
2 2016-02-01T00:00:00+01:00 565.6
3 2016-03-01T00:00:00+01:00 765.6
4 2016-04-01T00:00:00+01:00 888.6
5 2016-05-01T00:00:00+01:00 465.0
6 2016-06-01T00:00:00+01:00 465.6
7 2016-07-01T00:00:00+01:00 786.0
8 2016-08-01T00:00:00+01:00 435.0
9 2016-09-01T00:00:00+01:00 568.0
10 2016-10-01T00:00:00+01:00 678.0
# ... with 26 more rows
从其他df添加ID
library(dplyr)
ids <- data_frame(list_id = c(112, 34, 54),
monthly_consum = c("list1", "list2", "list3"))
如果我们考虑嵌套列表，您可以使用purrr：map，如下所示：
- 将三个列表合并到一个列表中
k <- list(list1, list2, list3)
- 使用map独立地映射到每列中的bind_rows
k1 <- purrr:: map(k, bind_rows)
- 使用ID作为列表的名称
names(k1) <- ids$list_id
-bind_rows使用.id
bind_rows(k1, .id = "id")
# A tibble: 36 × 3
id ts v
<chr> <chr> <dbl>
1 112 2016-01-01T00:00:00+01:00 466.6
2 112 2016-02-01T00:00:00+01:00 565.6
3 112 2016-03-01T00:00:00+01:00 765.6
4 112 2016-04-01T00:00:00+01:00 888.6
5 112 2016-05-01T00:00:00+01:00 465.0
6 112 2016-06-01T00:00:00+01:00 465.6
7 112 2016-07-01T00:00:00+01:00 786.0
8 112 2016-08-01T00:00:00+01:00 435.0
9 112 2016-09-01T00:00:00+01:00 568.0
10 112 2016-10-01T00:00:00+01:00 678.0