我有一个包含连续变量和两个分类 ID 变量的数据集。我想缩放我的数据,使得连续变量对于每个 ID 的某个值具有 1 的值。我尝试将它实现为 R 中的一个函数,这样我就可以避免重复编码。
此外,我有一个在 Stata 中实现的重新缩放的示例:
gen value_var_i_k= value_var if ID1=="15t16" & ID2 =="AUS"
egen value_var_i_k_m = mean(value_var_i_k)
drop value_var_i_k
rename value_var_i_k_m value_var_i_k
gen value_var_k= value_var if ID1 =="15t16"
bys ID2: egen value_var_k_m = mean(value_var_k)
drop value_var_k
rename value_var_k_m value_var_k
gen value_var_i = value_var if ID2=="AUS"
bys ID1: egen value_var_i_m = mean(value_var_i)
drop value_var_i
rename value_var_i_m value_var_i
gen value_var_i_k_norm= value_var * value_var_i_k/(value_var_i*value_var_k)
我在 R 中尝试使用新变量创建value_var_i value_var_k
andvalue_var_i_k
变量,该变量因错误而退出(不适用于类“字符”的方法):
library(dplyr)
library(magrittr)
normalize<-function(var,data,i,k) {
varname <- paste("value", var , sep="_")
#Id columns and define variables to select
col1<-"ID1"
col2<-"ID2"
select_variables<-c(col2,col1,varname)
#name of the output variables
name_ik<-paste(i,k,sep="_")
name.ik<-paste(name_ik, "df", sep="_")
name.i<-paste(i, "df", sep="_")
name.k<-paste(k, "df", sep="_")
#my attempt to replicate the Stata code with dplyr
data %>% filter_(as.name(col1)==as.name(i) & as.name(col2)==as.name(k)) %>% select_( .dots=select_variables) %$% as.name( name.ik)
data %>% filter_(as.name(col1)==as.name(i)) %>% select_( .dots = select_variables ) %>% group_by_(as.name(col2)) %>%transform( interp(~mean(b, na.rm = TRUE),b=as.name(varname) )) %$% as.name(name.i)
data %>% filter_(as.name(col2)==as.name(k)) %>% select_( .dots = select_variables ) %>% group_by_(as.name(col1)) %>%transform( interp(~mean(b, na.rm = TRUE),b=as.name(varname) )) %$% as.name(name.k)
norm <- data[eval(substitute(varname)]*as.name(name.ik)/ ( as.name(name.i) * as.name(name.k) )
}
更新二:计算步骤的最小工作示例:value_var
两个表中的变量相同。Stata 代码将value_var_k
andvalue_var_i
变量替换为变量value_var_i_m
and value_var_k_m
。然后value_var
变身。
value_var ID1 ID2 value_var_i_k value_var_k value_var_k_m
1.154662 15t16 AUS 1.154662 1.154662 1.154662
1.070471 17t18 AUS 1.154662 . 1.154662
0.9643197 19 AUS 1.154662 . 1.154662
1.036398 20 AUS 1.154662 . 1.154662
1.084701 21t22 AUS 1.154662 . 1.154662
1.463215 15t16 AUT 1.154662 1.463215 1.463215
1.431824 17t18 AUT 1.154662 . 1.463215
1.276983 19 AUT 1.154662 . 1.463215
1.441925 20 AUT 1.154662 . 1.463215
1.506117 21t22 AUT 1.154662 . 1.463215
1.589491 15t16 BEL 1.154662 1.589491 1.589491
1.540076 17t18 BEL 1.154662 . 1.589491
1.188218 19 BEL 1.154662 . 1.589491
1.386074 20 BEL 1.154662 . 1.589491
1.48204 21t22 BEL 1.154662 . 1.589491
value_var ID1 ID2 value_var_i value_var_i_m
1.154662 15t16 AUS 1.154662 1.154662
1.589491 15t16 BEL . 1.154662
1.463215 15t16 AUT . 1.154662
1.070471 17t18 AUS 1.070471 1.070471
1.540076 17t18 BEL . 1.070471
1.431824 17t18 AUT . 1.070471
0.9643197 19 AUS 0.9643197 0.9643197
1.276983 19 AUT . 0.9643197
1.188218 19 BEL . 0.9643197
1.036398 20 AUS 1.036398 1.036398
1.441925 20 AUT . 1.036398
1.386074 20 BEL . 1.036398
1.084701 21t22 AUS 1.084701 1.084701
1.506117 21t22 AUT . 1.084701
1.48204 21t22 BEL . 1.084701
归一化值 var 如下(对 ID1 级别“15t16”和 ID2 级别“AUS”进行归一化):
ID1 ID2 value_var_i_k_norm
AUS 15t16 1
AUS 17t18 1
AUS 19 1
AUS 20 1
AUS 21t22 1
AUT 15t16 1
AUT 17t18 1.055508
AUT 19 1.044988
AUT 20 1.097901
AUT 21t22 1.09571
BEL 15t16 1
BEL 17t18 1.045116
BEL 19 .8951011
BEL 20 .9715319
BEL 21t22 .9925373
更新:为了使标准化(或缩放)步骤更清晰,我在这里以宽格式显示标准化数据的 pre after。
首先我从以下广泛的数据开始
Row-/Colnames 15t16 17t18 19t 20t 21t22
AUS 1.154662 1.070471 0.9643197 1.036398 1.084701
AUT 1.463215 1.431824 1.276983 1.441925 1.506117
BEL 1.589491 1.540076 1.188218 1.386074 1.48204
我将矩阵规范化为具有 AUS 值的行和具有 15t16 值的列。所以,我会得到
Row-/Colnames 15t16 17t18 19t 20t 21t22
AUS 1 1 1 1 1
AUT 1 1.055508 1.044988 1.097901 1.09571
BEL 1 1.045116 .8951011 .9715319 .9925373