我有一个非常基本的问题，尝试了一些已经在堆栈溢出中的东西，但不知何故它不起作用。
这是我的代码的头部：
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：
STATE - STATEFP - YEAR - GW_ratio - SW_ratio
我迷路了，如果有人能帮助我就太好了！
答案 0 :(得分：2)
我们可以使用 pivot_longer 和 names_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
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"))
答案 1 :(得分：1)
使用 reshape 的基本 R 选项
reshape(
df,
direction = "long",
idvar = c("STATE", "STATEFP"),
timevar = "YEAR",
varying = -(1:2)
)
给予
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