在 R 中，如何更改行和列的位置？

Question

我有这个原始数据集，下面是示例数据集：

 X1         X2
 1      Born 1946-05-27
 2  bioguide    A000370
 3      Born 1979-06-19
 4  bioguide    A000371
 5      Born 1980-04-18
 6  bioguide    A000367
 7      Born 1958-06-12
 8  bioguide    A000369
 9      Born 1948-03-23
 10 bioguide    B001291

使用这个，我想要的输出如下：

       Born        biouguide
1     1946-05-27     A000370
2     1979-06-19     A000371
3     1980-04-18     A000367
4     1958-06-12     A000369
5     1980-04-18     A000367

此外，以下是原始数据集的 dput：

structure(list(X1 = c("Born", "bioguide", "Born", "bioguide", 
"Born", "bioguide", "Born", "bioguide", "Born", "bioguide"), 
    X2 = c("1946-05-27", "A000370", "1979-06-19", "A000371", 
    "1980-04-18", "A000367", "1958-06-12", "A000369", "1948-03-23", 
    "B001291")), row.names = c(NA, 10L), class = "data.frame")

你能帮我做出想要的输出吗？

Answer 1

我们可以用 pivot_wider

library(dplyr)
library(tidyr)
df1 %>%
  group_by(X1) %>%
  mutate(rn = row_number()) %>%
  pivot_wider(names_from = X1, values_from = X2) %>%
  select(-rn)
# A tibble: 5 x 2
#  Born       bioguide
#  <chr>      <chr>   
#1 1946-05-27 A000370 
#2 1979-06-19 A000371 
#3 1980-04-18 A000367 
#4 1958-06-12 A000369 
#5 1948-03-23 B001291 

或者在 base R

unstack(df1, X2 ~ X1)

Answer 2

一种base R选择可能是：

data.frame(Born = df[c(TRUE, FALSE), 2],
           biouguide = df[c(FALSE, TRUE), 2])

        Born biouguide
1 1946-05-27   A000370
2 1979-06-19   A000371
3 1980-04-18   A000367
4 1958-06-12   A000369
5 1948-03-23   B001291