Question

df2 <- structure(list(location = c("Dayton", "Toledo"), total_voters = c(236L, 
332L), candidate_1 = c(49L, 61L), candidate_2 = c(33L, 78L), 
    candidate_3 = c(19L, 71L), candidate_5 = c(42L, 52L)), row.names = c(NA, 
-2L), class = "data.frame")

我有来自SQL查询的数据，其形状如下：

+----------+--------------+-------------+-------------+-------------+-------------+-------------+
| location | total_voters | candidate_1 | candidate_2 | candidate_3 | candidate_4 | candidate_5 |
+----------+--------------+-------------+-------------+-------------+-------------+-------------+
| Dayton   |          236 |          49 |          33 |          19 |          93 |          42 |
| Toledo   |          332 |          61 |          78 |          71 |          70 |          52 |
+----------+--------------+-------------+-------------+-------------+-------------+-------------+

数字代表每个候选人的票数。那我想做什么？我想使用R（我想通过dplyr或tidyr）来旋转数据，使其看起来像这样：

+-------------+-------+----------+--------------+
|  candidate  | votes | location | total_voters |
+-------------+-------+----------+--------------+
| candidate_1 |    49 | Dayton   |          236 |
| candidate_2 |    33 | Dayton   |          236 |
| candidate_3 |    19 | Dayton   |          236 |
| candidate_4 |    93 | Dayton   |          236 |
| candidate_5 |    42 | Dayton   |          236 |
| candidate_1 |    61 | Toledo   |          332 |
| candidate_2 |    78 | Toledo   |          332 |
| candidate_3 |    71 | Toledo   |          332 |
| candidate_4 |    70 | Toledo   |          332 |
| candidate_5 |    52 | Toledo   |          332 |
+-------------+-------+----------+--------------+

在R中完成此操作的最有效方法是什么？

Answer 1

这是pivot_longer

的一个选项

library(dplyr)
library(tidyr)
df1 %>%
   pivot_longer(cols = everything(), names_to = 'candidate', values_to = 'votes')
# A tibble: 5 x 2
#  candidate   votes
#  <chr>       <dbl>
#1 candidate_1    49
#2 candidate_2    33
#3 candidate_3    19
#4 candidate_4    93
#5 candidate_5    42

使用更新的数据，

df2 %>%
   pivot_longer(cols = -c(location, total_voters),
                 names_to = 'candidate', values_to = 'votes')
# A tibble: 8 x 4
#  location total_voters candidate   votes
#  <chr>           <int> <chr>       <int>
#1 Dayton            236 candidate_1    49
#2 Dayton            236 candidate_2    33
#3 Dayton            236 candidate_3    19
#4 Dayton            236 candidate_5    42
#5 Toledo            332 candidate_1    61
#6 Toledo            332 candidate_2    78
#7 Toledo            332 candidate_3    71
#8 Toledo            332 candidate_5    52

或者在base R中，可以使用stack

stack(df1)[2:1]

或通过转换为table

as.data.frame.table(as.matrix(df1))[,-1]

或按照@markus的建议

reshape2::melt(df1)

数据

df1 <- data.frame(candidate_1 = 49, candidate_2 = 33, 
         candidate_3 = 19, candidate_4 = 93, candidate_5 = 42)

df2 <- structure(list(location = c("Dayton", "Toledo"), total_voters = c(236L, 
332L), candidate_1 = c(49L, 61L), candidate_2 = c(33L, 78L), 
    candidate_3 = c(19L, 71L), candidate_5 = c(42L, 52L)), row.names = c(NA, 
-2L), class = "data.frame")

Answer 2

实际上，您可以使用data.frame + t进行修改，即

dflong <- data.frame(t(dfwide))

Answer 3

如果候选人_1候选人_2等是列名，则可以使用reshape2软件包中的melt函数。

a=data.frame(candidate_1=49,
             candidate_2=33,
             candidate_3=19,
             candidate_4=93,
             candidate_5=42)

b=reshape2::melt(a)

将列透视成行

3 个答案:

数据