Question

我的数据框如下：

structure(list(symbol = c("u", "n", "v", "i", "a"), start = c(9L,
6L, 10L, 8L, 7L), end = c(14L, 15L, 12L, 13L, 11L)), .Names = c("symbol",
"start", "end"), class = "data.frame", row.names = c("1", "2",
"3", "4", "5"))

我想要与每个符号的（开始，结束）范围内的值一样多的行。因此，最终的数据框架将如下所示：

structure(list(symbol = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L,
4L, 4L, 5L, 5L, 5L, 5L, 5L), .Label = c("a", "l", "n", "v", "y"
), class = "factor"), value = c(7L, 8L, 9L, 10L, 11L, 6L, 7L,
8L, 9L, 10L, 11L, 12L, 13L, 14L, 8L, 9L, 10L, 11L, 12L, 10L,
11L, 12L, 13L, 14L, 15L, 9L, 10L, 11L, 12L, 13L)), class = "data.frame", row.names = c(NA,
-30L), .Names = c("symbol", "value"))

我原以为我可以简单地每行列出一个值列表，然后使用tidyr包unnest，如下所示：

df$value <- apply(df, 1, function(x) as.list(x[2]:x[3]))
dput(df)
structure(list(symbol = structure(c(4L, 3L, 5L, 2L, 1L), .Label = c("a",
"i", "n", "u", "v"), class = "factor"), start = c(9L, 6L, 10L,
8L, 7L), end = c(14L, 15L, 12L, 13L, 11L), value = structure(list(
    `1` = list(9L, 10L, 11L, 12L, 13L, 14L), `2` = list(6L, 7L,
        8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L), `3` = list(10L,
        11L, 12L), `4` = list(8L, 9L, 10L, 11L, 12L, 13L), `5` = list(
        7L, 8L, 9L, 10L, 11L)), .Names = c("1", "2", "3", "4",
"5"))), .Names = c("symbol", "start", "end", "value"), row.names = c("1",
"2", "3", "4", "5"), class = "data.frame")

df
  symbol start end                              value
1      u     9  14              9, 10, 11, 12, 13, 14
2      n     6  15 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
3      v    10  12                         10, 11, 12
4      i     8  13               8, 9, 10, 11, 12, 13
5      a     7  11                    7, 8, 9, 10, 11

然后做：

library(tidyr)
unnest(df, value)

但是，我想我正在处理这个待处理的功能/错误： https://github.com/tidyverse/tidyr/issues/278

Error: Each column must either be a list of vectors or a list of data frames [value]

有没有更好的方法来做到这一点，特别是避免申请家庭？

Answer 1

使用dplyr，我们可以将rowwise与do

一起使用

library(dplyr)
df1 %>% 
   rowwise() %>% 
   do(data.frame(symbol= .$symbol, value = .$start:.$end)) %>% 
   arrange(symbol)
# A tibble: 30 x 2
#   symbol value
#    <chr> <int>
# 1      a     7
# 2      a     8
# 3      a     9
# 4      a    10
# 5      a    11
# 6      i     8
# 7      i     9
# 8      i    10
# 9      i    11
#10      i    12
# ... with 20 more rows

Answer 2

您可以使用data.table并按所需行数复制df（基于每个start的{{1}}和end），然后在

之后为每行分配值

symbol

Answer 3

也许您可以使用map2将我们可以unnest的列添加到所需的结果中。

library(tidyverse)
df %>% 
  mutate(value = map2(start, end, ~ seq(from = .x, to = .y))) %>%
  select(symbol, value) %>%
  unnest()
#>    symbol    value
#> 1       u        9
#> 2       u       10
#> 3       u       11
#> 4       u       12
#> 5       u       13
#> 6       u       14
#> 7       n        6
#> 8       n        7
#> 9       n        8
#> 10      n        9
#> ...etc

将数据框展开为包含原始行中两列范围的行数

3 个答案: