Df_01a
Name re1 re2 re3 parameter
a 144 39.7 0.012 fed
b 223 31.2 5 fed
c 304 6.53 100 fed
d 187 51.3 25 fed
e 110 2.94 100 fed
f 151 4.23 75 fed
g 127 36.7 0.012 fed
Df_01b
Name re1 re2 re3 parameter
a 142 39.3 0.042 feh
b 221 31.0 4 feh
c 301 6.13 90 feh
d 185 41.3 15 feh
e 107 2.44 940 feh
f 143 2.23 75 feh
g 121 31.7 0.012 feh
Df_02
parameter c1 c2 c3
1 fed 5 4 3
2 feh 3 4 2
3 fea 5 4 3
4 few 2 4 3
期望的结果：
c-value re-value name
5 142 a_fed
4 39.3 a_fed
3 0.042 a_fed
5 221 b_fed
4 31.0 b_fed
3 4 b_fed
5 304 c_fed
4 6.53 c_fed
3 100 c_fed
....
3 0.012 g_fed
3 142 a_feh
4 39.3 a_feh
2 0.042 a_feh
3 221 b_feh
4 31.0 b_feh
2 4 b_feh
....
我有Df_01a, Df_01b, Df_01c, Df_01d。这些参数有
第5列：fed, feh, fea, few（见Df_02）。
每个参数都有3个值，由c1中的c2，c3和Df_02给出。
如何获得上面显示的所需data.frame？
答案 0 :(得分：2)
<强>码
(?=.*[[:alpha:]])[[:alnum:]]{1}[[:alnum:] -]*[[:alnum:]]{1}
<强>结果
library(dplyr)
library(tidyr)
rbind(Df_01a,Df_01b) %>% gather("re-col","re-value",c("re1","re2","re3")) %>%
inner_join(Df_02 %>% rename(re1=c1,re2=c2,re3=c3) %>% gather("re-col","c-value",c("re1","re2","re3"))) %>%
arrange(parameter,Name) %>%
unite(name,Name,parameter) %>%
select(`c-value`,`re-value`,`name`)
数据
# c-value re-value name
# 1 5 144.000 a_fed
# 2 4 39.700 a_fed
# 3 3 0.012 a_fed
# 4 5 223.000 b_fed
# 5 4 31.200 b_fed
# 6 3 5.000 b_fed
# 7 5 304.000 c_fed
# 8 4 6.530 c_fed
# 9 3 100.000 c_fed
# 10 5 187.000 d_fed
# 11 4 51.300 d_fed
# 12 3 25.000 d_fed
# 13 5 110.000 e_fed
# 14 4 2.940 e_fed
# 15 3 100.000 e_fed
# 16 5 151.000 f_fed
# 17 4 4.230 f_fed
# 18 3 75.000 f_fed
# 19 5 127.000 g_fed
# 20 4 36.700 g_fed
# 21 3 0.012 g_fed
# 22 3 142.000 a_feh
# 23 4 39.300 a_feh
# 24 2 0.042 a_feh
# 25 3 221.000 b_feh
# 26 4 31.000 b_feh
# 27 2 4.000 b_feh
# 28 3 301.000 c_feh
# 29 4 6.130 c_feh
# 30 2 90.000 c_feh
# 31 3 185.000 d_feh
# 32 4 41.300 d_feh
# 33 2 15.000 d_feh
# 34 3 107.000 e_feh
# 35 4 2.440 e_feh
# 36 2 940.000 e_feh
# 37 3 143.000 f_feh
# 38 4 2.230 f_feh
# 39 2 75.000 f_feh
# 40 3 121.000 g_feh
# 41 4 31.700 g_feh
# 42 2 0.012 g_feh