3

我有这个大数据，其中包含几个具有常见列名的数据框的汇编。为了说明，它看起来有点像这样：

df <- data_frame(A = c(runif(3), "A", runif(4), "A", runif(5)),
                 B = c(runif(3), "B", runif(4), "B", runif(5)),
                 C = c(runif(3), "C", runif(4), "C", runif(5)),
                 D = c(runif(3), "D", runif(4), "D", runif(5)))

如何为包含原始列名的每个新行创建一个数据框，这样我会得到一个看起来像这样的结果：

df1:
A     B     C     D     id
0.668 0.411 0.553 0.477 1
0.794 0.821 0.530 0.732 1
0.108 0.647 0.789 0.693 1

df2:
A     B     C     D     id
0.724 0.783 0.023 0.478 2
0.861 0.099 0.407 0.332 2
0.438 0.316 0.913 0.651 2
0.245 0.519 0.294 0.258 2
0.070 0.662 0.459 0.479 2
0.766 0.839 0.892 0.961 2

df3:
A     B     C     D     id
0.084 0.347 0.864 0.435 3
0.875 0.334 0.390 0.713 3
0.339 0.476 0.777 0.400 3
0.084 0.347 0.864 0.435 3

谢谢！

4

2 回答 2

2

这是使用apply()+的基本 R 解决方案split()

idx <- apply(df, 1, function(v) all(v==names(df)))
df <- within(df[-which(idx),],id <- (cumsum(idx)+1)[-which(idx)])
res <- split(df,df$id)

这样

> res
$`1`
      A     B     C     D id
1 0.668 0.411 0.553 0.477  1
2 0.794 0.821 0.530 0.732  1
3 0.108 0.647 0.789 0.693  1

$`2`
       A     B     C     D id
5  0.724 0.783 0.023 0.478  2
6  0.861 0.099 0.407 0.332  2
7  0.438 0.316 0.913 0.651  2
8  0.245 0.519 0.294 0.258  2
9  0.070 0.662 0.459 0.479  2
10 0.766 0.839 0.892 0.961  2

$`3`
       A     B     C     D id
12 0.084 0.347 0.864 0.435  3
13 0.875 0.334 0.390 0.713  3
14 0.339 0.476 0.777 0.400  3
15 0.084 0.347 0.864 0.435  3
于 2020-01-19T07:40:02.427 回答
1

一种方法是在数据框中找到列的名称并以此为基础进行拆分。我们可以group_split用来将数据拆分为数据框，我们用来map_at从每个列表中删除第一行，因为它是列名，并type.convert用来将列转换为各自的类。

library(dplyr)
library(purrr)

temp <- df %>%
          group_split(id = cumsum(A == names(.)[1]) + 1) %>%
          map_at(-1, tail, -1) %>%
          map(type.convert)
temp

#[[1]]
# A tibble: 3 x 5
#     A     B     C     D     id
#   <dbl> <dbl> <dbl> <dbl> <int>
#1 0.668 0.411 0.553 0.477     1
#2 0.794 0.821 0.530 0.732     1
#3 0.108 0.647 0.789 0.693     1

#[[2]]
# A tibble: 6 x 5
#    A     B     C     D    id
#  <dbl> <dbl> <dbl> <dbl> <int>
#1 0.724 0.783 0.023 0.478     2
#2 0.861 0.099 0.407 0.332     2
#3 0.438 0.316 0.913 0.651     2
#4 0.245 0.519 0.294 0.258     2
#5 0.07  0.662 0.459 0.479     2
#6 0.766 0.839 0.892 0.961     2

#[[3]]
# A tibble: 4 x 5
#      A     B     C     D    id
#  <dbl> <dbl> <dbl> <dbl> <int>
#1 0.084 0.347 0.864 0.435     3
#2 0.875 0.334 0.39  0.713     3
#3 0.339 0.476 0.777 0.4       3
#4 0.084 0.347 0.864 0.435     3

在基础 R 中使用相同的逻辑，我们可以做

df$id <- cumsum(df$A == names(df)[1]) + 1
temp <- split(df, df$id)
temp[-1] <- lapply(temp[-1], tail, -1)
temp <- lapply(temp, type.convert)

如果您需要它们作为单独的数据框，

names(temp) <- paste0("df", seq_along(temp))
list2env(temp, .GlobalEnv)

数据

df <- structure(list(A = structure(c(7L, 10L, 3L, 13L, 8L, 11L, 6L, 
4L, 1L, 9L, 13L, 2L, 12L, 5L, 2L), .Label = c("0.070", "0.084", 
"0.108", "0.245", "0.339", "0.438", "0.668", "0.724", "0.766", 
"0.794", "0.861", "0.875", "A"), class = "factor"), B = structure(c(5L, 
11L, 8L, 13L, 10L, 1L, 2L, 7L, 9L, 12L, 13L, 4L, 3L, 6L, 4L), .Label = c("0.099", 
"0.316", "0.334", "0.347", "0.411", "0.476", "0.519", "0.647", 
"0.662", "0.783", "0.821", "0.839", "B"), class = "factor"), 
 C = structure(c(7L, 6L, 9L, 13L, 1L, 4L, 12L, 2L, 5L, 11L, 
13L, 10L, 3L, 8L, 10L), .Label = c("0.023", "0.294", "0.390", 
"0.407", "0.459", "0.530", "0.553", "0.777", "0.789", "0.864", 
"0.892", "0.913", "C"), class = "factor"), D = structure(c(5L, 
11L, 9L, 13L, 6L, 2L, 8L, 1L, 7L, 12L, 13L, 4L, 10L, 3L, 
4L), .Label = c("0.258", "0.332", "0.400", "0.435", "0.477", 
"0.478", "0.479", "0.651", "0.693", "0.713", "0.732", "0.961", 
"D"), class = "factor")), class = "data.frame", row.names = c(NA, -15L))
于 2020-01-19T07:18:51.557 回答