1

我有一个非常基本的问题，尝试了一些已经在堆栈溢出的东西，但不知何故它不起作用。

这是我的代码的负责人：

  STATE STATEFP GW_ratio.2000 GW_ratio.2005 GW_ratio.2010 GW_ratio.2015 SW_ratio.2000 SW_ratio.2005 SW_ratio.2010 SW_ratio.2015
1    AL       1    0.04247763    0.04742443    0.04685309    0.05630994     0.9575224     0.9525756     0.9531469     0.9436901
2    AR       5    0.62710731    0.64621860    0.67762854    0.68590438     0.3728927     0.3537814     0.3223715     0.3140956
3    AZ       4    0.50738010    0.48620606    0.41704896    0.45948459     0.4926199     0.5137939     0.5829510     0.5405154
4    CA       6    0.39589058    0.32538360    0.39230956    0.67370799     0.6041094     0.6746164     0.6076904     0.3262920
5    CO       8    0.18116832    0.18174957    0.13833921    0.14408643     0.8188317     0.8182504     0.8616608     0.8559136
6    CT       9    0.13035722    0.18318935    0.25172450    0.20938673     0.8696428     0.8168107     0.7482755     0.7906133

我想要一个看起来像这样的框架：

州 - STATEFP - 年 - GW_ratio - SW_ratio

我迷路了，如果有人能帮助我，那就太好了！

4

2 回答 2

2

我们可以使用pivot_longerwithnames_sep作为分隔符.

library(dplyr)
library(tidyr)
df1 %>%
   pivot_longer(cols = -c(STATE, STATEFP),
         names_to = c(".value", "YEAR"), names_sep = "\\.")

-输出

# A tibble: 24 x 5
#   STATE STATEFP YEAR  GW_ratio SW_ratio
#   <chr>   <int> <chr>    <dbl>    <dbl>
# 1 AL          1 2000    0.0425    0.958
# 2 AL          1 2005    0.0474    0.953
# 3 AL          1 2010    0.0469    0.953
# 4 AL          1 2015    0.0563    0.944
# 5 AR          5 2000    0.627     0.373
# 6 AR          5 2005    0.646     0.354
# 7 AR          5 2010    0.678     0.322
# 8 AR          5 2015    0.686     0.314
# 9 AZ          4 2000    0.507     0.493
#10 AZ          4 2005    0.486     0.514
# … with 14 more rows

数据

df1 <- structure(list(STATE = c("AL", "AR", "AZ", "CA", "CO", "CT"), 
    STATEFP = c(1L, 5L, 4L, 6L, 8L, 9L), GW_ratio.2000 = c(0.04247763, 
    0.62710731, 0.5073801, 0.39589058, 0.18116832, 0.13035722
    ), GW_ratio.2005 = c(0.04742443, 0.6462186, 0.48620606, 0.3253836, 
    0.18174957, 0.18318935), GW_ratio.2010 = c(0.04685309, 0.67762854, 
    0.41704896, 0.39230956, 0.13833921, 0.2517245), GW_ratio.2015 = c(0.05630994, 
    0.68590438, 0.45948459, 0.67370799, 0.14408643, 0.20938673
    ), SW_ratio.2000 = c(0.9575224, 0.3728927, 0.4926199, 0.6041094, 
    0.8188317, 0.8696428), SW_ratio.2005 = c(0.9525756, 0.3537814, 
    0.5137939, 0.6746164, 0.8182504, 0.8168107), SW_ratio.2010 = c(0.9531469, 
    0.3223715, 0.582951, 0.6076904, 0.8616608, 0.7482755), 
    SW_ratio.2015 = c(0.9436901, 
    0.3140956, 0.5405154, 0.326292, 0.8559136, 0.7906133)), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6"))
于 2021-03-05T16:35:31.070 回答
1

使用基本 R 选项reshape

reshape(
  df,
  direction = "long",
  idvar = c("STATE", "STATEFP"),
  timevar = "YEAR",
  varying = -(1:2)
)

给

          STATE STATEFP YEAR   GW_ratio  SW_ratio
AL.1.2000    AL       1 2000 0.04247763 0.9575224
AR.5.2000    AR       5 2000 0.62710731 0.3728927
AZ.4.2000    AZ       4 2000 0.50738010 0.4926199
CA.6.2000    CA       6 2000 0.39589058 0.6041094
CO.8.2000    CO       8 2000 0.18116832 0.8188317
CT.9.2000    CT       9 2000 0.13035722 0.8696428
AL.1.2005    AL       1 2005 0.04742443 0.9525756
AR.5.2005    AR       5 2005 0.64621860 0.3537814
AZ.4.2005    AZ       4 2005 0.48620606 0.5137939
CA.6.2005    CA       6 2005 0.32538360 0.6746164
CO.8.2005    CO       8 2005 0.18174957 0.8182504
CT.9.2005    CT       9 2005 0.18318935 0.8168107
AL.1.2010    AL       1 2010 0.04685309 0.9531469
AR.5.2010    AR       5 2010 0.67762854 0.3223715
AZ.4.2010    AZ       4 2010 0.41704896 0.5829510
CA.6.2010    CA       6 2010 0.39230956 0.6076904
CO.8.2010    CO       8 2010 0.13833921 0.8616608
CT.9.2010    CT       9 2010 0.25172450 0.7482755
AL.1.2015    AL       1 2015 0.05630994 0.9436901
AR.5.2015    AR       5 2015 0.68590438 0.3140956
AZ.4.2015    AZ       4 2015 0.45948459 0.5405154
CA.6.2015    CA       6 2015 0.67370799 0.3262920
CO.8.2015    CO       8 2015 0.14408643 0.8559136
CT.9.2015    CT       9 2015 0.20938673 0.7906133
于 2021-03-05T16:59:55.320 回答