9

我的数据如下所示:

在此处输入图像描述

我试图让它看起来像这样:

在此处输入图像描述

我想在 tidyverse 中使用 %>%-chaining 来做到这一点。

df <- 
structure(list(id = c(2L, 2L, 4L, 5L, 5L, 5L, 5L), start_end = structure(c(2L, 
1L, 2L, 2L, 1L, 2L, 1L), .Label = c("end", "start"), class = "factor"), 
    date = structure(c(6L, 7L, 3L, 8L, 9L, 10L, 11L), .Label = c("1979-01-03", 
    "1979-06-21", "1979-07-18", "1989-09-12", "1991-01-04", "1994-05-01", 
    "1996-11-04", "2005-02-01", "2009-09-17", "2010-10-01", "2012-10-06"
    ), class = "factor")), .Names = c("id", "start_end", "date"
), row.names = c(3L, 4L, 7L, 8L, 9L, 10L, 11L), class = "data.frame")

我试过的:

data.table::dcast( df, formula = id ~ start_end, value.var = "date", drop = FALSE )  # does not work because it summarises the data

tidyr::spread( df, start_end, date )  # does not work because of duplicate values


df$id2 <- 1:nrow(df)
tidyr::spread( df, start_end, date ) # does not work because the dataset now has too many rows.

这些问题不回答我的问题:

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R:具有重复项的数据帧上的扩展函数 (因为它们将值粘贴在一起)

使用“登录”“注销”时间重塑 R 中的数据(因为没有使用 tidyverse 和链接专门要求/回答)

4

1 回答 1

16

我们可以使用tidyverse. 按 'start_end'、'id' 分组后,创建序列列 'ind' ,然后spread从 'long' 到 'wide' 格式

library(dplyr)
library(tidyr)
df %>%
   group_by(start_end, id) %>%
   mutate(ind = row_number()) %>%
   spread(start_end, date) %>% 
   select(start, end)
#     id      start        end
#* <int>     <fctr>     <fctr>
#1     2 1994-05-01 1996-11-04
#2     4 1979-07-18         NA
#3     5 2005-02-01 2009-09-17
#4     5 2010-10-01 2012-10-06

或使用tidyr_1.0.0

chop(df, date) %>%
     spread(start_end, date) %>%
     unnest(c(start, end))
于 2017-04-06T15:38:19.883 回答