Question

我有一个向量列表。此向量由名称A1，A2 ..和数值组成：

a <- list()
a$`p1`["A1"] <- 1
a$`p1`["A2"] <- 0.5
a$`p1`["A3"] <- 0.3
a$`p2`["A3"] <- 2
a$`p2`["A4"] <- 2.5
a$`p2`["A5"] <- 2.3

我希望将其转换为具有以下结构的数据帧

P <- c("p1","p1","p1","p2","p2","p2")
A <- c("A1","A2","A3","A3","A4","A5")
V <- c(1,0.5,0.3,2,2.5,2.3)
out <- data.frame(P,A,V)

谢谢：）

Answer 1

df_list = lapply(a, function(x) data.frame(A = names(x), V = x, stringsAsFactors = FALSE))
dplyr::bind_rows(df_list, .id = "P")
#    P  A   V
# 1 p1 A1 1.0
# 2 p1 A2 0.5
# 3 p1 A3 0.3
# 4 p2 A3 2.0
# 5 p2 A4 2.5
# 6 p2 A5 2.3

我喜欢上面的方式，但这是一个选项，可能在较大的列表上可能更有效率：

data.frame(V = unlist(a)) %>% 
  tibble::rownames_to_column() %>%
  tidyr::separate(rowname, into = c("P", "A"))

Answer 2

一个purrr和tibble选项可以是：

map_dfr(a, ~ enframe(., name = "A", value = "V"), .id = "P")

  P     A         V
  <chr> <chr> <dbl>
1 p1    A1      1  
2 p1    A2      0.5
3 p1    A3      0.3
4 p2    A3      2  
5 p2    A4      2.5
6 p2    A5      2.3

Answer 3

在base-R

cbind(stack(a),A=unlist(lapply(a,names),use.names = F))

给予

  values ind  A
1    1.0  p1 A1
2    0.5  p1 A2
3    0.3  p1 A3
4    2.0  p2 A3
5    2.5  p2 A4
6    2.3  p2 A5

Answer 4

带有base R

的选项

do.call(rbind, Map(cbind, lapply(a, stack), P = names(a)))

将向量列表转换为数据帧

4 个答案: