Question

我想将行名转换为值并保留相应的值。我的数据在第一列中包含国家/地区名称，在剩余的列中具有单元格中的值的年份为列名。我想将其转换为适当的表形式。参见以下示例。

表格表的示例：

国家| 2002 | 2003 | ...

加拿大| 2.2 | 2.4 | ...

美国| 4.2 | 7.4 | ...

。

我希望表格的格式为：国家（地区）|年份值

加拿大| 2002 | 2.2

加拿大| 2.2 | 2.4

...

我认为整洁的数据包应该在该国起作用。请参阅下面的示例。

数据：

ElectricCarStock_BEVandPHEV<- structure(list(Country = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 
7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 17L, 18L, 19L, 20L, 
22L, 23L, 16L, 21L), .Label = c("Australia", "Brazil", "Canada", 
"Chile", "China", "Finland", "France", "Germany", "India", "Japan", 
"Korea", "Mexico", "Netherlands", "New Zealand", "Norway", "Others", 
"Portugal", "South Africa", "Sweden", "Thailand", "Total", "United Kingdom", 
"United States"), class = "factor"), `2005` = c(NA, NA, NA, NA, 
NA, NA, 0.01, 0.02, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
0.22, 1.12, 0.53, 1.89), `2006` = c(NA, NA, NA, NA, NA, NA, 0.01, 
0.02, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.01, 0.55, 1.12, 
0.53, 2.23), `2007` = c(NA, NA, NA, NA, NA, NA, 0.01, 0.02, NA, 
NA, NA, NA, NA, NA, 0.01, NA, NA, NA, 0.01, 1, 1.12, 0.53, 2.69
), `2008` = c(NA, NA, NA, NA, NA, NA, 0.01, 0.09, 0.37, NA, NA, 
NA, 0.01, NA, 0.26, NA, NA, NA, 0.01, 1.22, 2.58, 0.61, 5.15), 
`2009` = c(NA, NA, NA, NA, 0.48, NA, 0.12, 0.1, 0.53, 1.08, 
NA, NA, 0.15, NA, 0.4, NA, NA, NA, 0.01, 1.4, 2.58, 0.64, 
7.48), `2010` = c(NA, NA, NA, NA, 1.91, NA, 0.3, 0.25, 0.88, 
3.52, 0.06, NA, 0.27, 0.01, 0.79, NA, NA, NA, 0.01, 1.68, 
3.77, 0.81, 14.26), `2011` = c(50, NA, 0.52, 0.01, 6.98, 
0.06, 3.03, 1.89, 1.33, 16.14, 0.34, NA, 1.14, 0.03, 2.63, 
NA, NA, 0.18, 0.01, 2.89, 21.5, 2.6, 61.33), `2012` = c(300, 
NA, 2.54, 0.01, 16.88, 0.24, 9.29, 5.26, 2.76, 40.58, 0.85, 
0.09, 6.26, 0.06, 7.15, NA, NA, 1.11, 0.02, 5.59, 74.74, 
5.31, 179.03), `2013` = c(600, NA, 5.66, 0.02, 32.22, 0.47, 
18.91, 12.19, 2.95, 69.46, 1.45, 0.1, 28.67, 0.09, 15.67, 
NA, 0.03, 2.66, 0.03, 9.34, 171.44, 9.35, 381.3), `2014` = c(1920, 
0.06, 10.73, 0.03, 105.39, 0.93, 31.54, 24.93, 3.35, 101.74, 
2.76, 0.15, 43.76, 0.41, 35.44, NA, 0.05, 7.32, 0.1, 24.08, 
290.22, 18.73, 703.65), `2015` = c(3690, 0.15, 17.69, 0.07, 
312.77, 1.59, 54.49, 48.12, 4.35, 126.4, 5.95, 0.25, 87.53, 
0.91, 69.17, NA, 0.29, 15.91, 0.37, 48.51, 404.09, 37.17, 
1239.45), `2016` = c(5060, 0.32, 29.27, 0.1, 648.77, 3.29, 
84, 72.73, 4.8, 151.25, 11.21, 0.66, 112.01, 2.41, 114.05, 
NA, 0.67, 29.33, 0.38, 86.42, 563.71, 61.63, 1982.04), `2017` = c(7340, 
0.68, 45.95, 0.25, 1227.77, 6.34, 118.77, 109.56, 6.8, 205.35, 
25.92, 0.92, 119.33, 5.88, 176.31, 1.78, 0.86, 49.6, 0.4, 
133.67, 762.06, 103.44, 3109.05)), .Names = c("Country", 
"2005", "2006", "2007", "2008", "2009", "2010", "2011", "2012", 
"2013", "2014", "2015", "2016", "2017"), class = "data.frame", row.names = c(NA, 
-23L))

代码：

library(tidyr)
library(dplyr)
 temp<-gather(ElectricCarStock_BEVandPHEV, Country, 2:13)
  head(temp)

Answer 1

这应该做到：

library(reshape2)
library(tidyverse)
ElectricCarStock_BEVandPHEV %>% 
  melt(id.vars="Country")

将行名转换为值并保留对应的值的列

1 个答案: