根据R中的ID将列值拆分为多个列

Question

我有一个数据集，每个 ID 都有不同的时间点。我希望每个 ID 有一条记录，我希望将时间点分成不同的列。我不想使用传播，因为我想要列的实际值。一些 ID 每个 ID 有 14 条记录，我希望将 14 条记录分成 14 列。我怎样才能在 R 中实现这一点？

示例数据

ID	时间点	价值
一个	1	是
一个	2	是
一个	3	是
一个	4	是
B	7	是
B	11	是
C	4	是
C	5	是
D	7	是

ID	时间点 1	时间点 2	时间点 3	时间点 4	价值
一个	1	2	3	4	是
B	7	11			是
C	4	5			是
D	7				是

Answer 1

我们可以用dcast

library(data.table)
dcast(setDT(df1), ID + Value ~ paste0("Timepoint",
       rowid(ID)), value.var = 'Timepoint')

-输出

  ID Value Timepoint1 Timepoint2 Timepoint3 Timepoint4
1:  A   yes          1          2          3          4
2:  B   yes          7         11         NA         NA
3:  C   yes          4          5         NA         NA
4:  D   yes          7         NA         NA         NA

数据

df1 <- structure(list(ID = c("A", "A", "A", "A", "B", "B", "C", "C", 
"D"), Timepoint = c(1L, 2L, 3L, 4L, 7L, 11L, 4L, 5L, 7L), Value = c("yes", 
"yes", "yes", "yes", "yes", "yes", "yes", "yes", "yes")),
 class = "data.frame", row.names = c(NA, 
-9L))

Answer 2

tidyverse

df1 %>%
  group_by(ID) %>%
  mutate(id_rows = row_number()) %>%
  pivot_wider( 
    id_cols = c(ID, Value),
    names_from = id_rows,
    values_from = Timepoint,
    names_prefix = "Timepoint"
  ) %>% 
  ungroup()

输出

# A tibble: 4 x 6
  ID    Value Timepoint1 Timepoint2 Timepoint3 Timepoint4
  <chr> <chr>      <int>      <int>      <int>      <int>
1 A     yes            1          2          3          4
2 B     yes            7         11         NA         NA
3 C     yes            4          5         NA         NA
4 D     yes            7         NA         NA         NA

Answer 3

在基础 R 中：

 reshape(transform(df1, time = ave(ID, ID, FUN = seq)), 
          dir = 'wide', idvar = c('ID', 'Value'), sep='')

  ID Value Timepoint1 Timepoint2 Timepoint3 Timepoint4
1  A   yes          1          2          3          4
5  B   yes          7         11         NA         NA
7  C   yes          4          5         NA         NA
9  D   yes          7         NA         NA         NA

Answer 4

不是最好的方法，但它有效：splitstackshape 和 data.table 的组合。其他解决方案已经提出：

library(splitstackshape)
library(data.table)

df <- dcast(getanID(df1, 'ID'), ID~.id, value.var='Timepoint')
colnames(df)[2:5] <- paste("Timepoint", colnames(df)[2:5], sep = "")

输出：

   ID Timepoint1 Timepoint2 Timepoint3 Timepoint4
1:  A          1          2          3          4
2:  B          7         11         NA         NA
3:  C          4          5         NA         NA
4:  D          7         NA         NA         NA

根据R中的ID将列值拆分为多个列

Split a column value into Multiple columns based on an ID in R

r

dplyr

split

multiple-columns

data-transform

数据