如何解决如何在R中的表中的估计值下方或旁边移动标准误差
我想创建有关调查中健康数据的表格摘要统计信息。我希望对表格进行操作，以便每个变量的标准误差出现在均值/中值正下方或正侧的括号中。我已经看到了仅包含1个或2个变量的表的更简单的解释，但是不确定如何应用许多变量。这是MRE：
#Here is the data frame
#the data frame
structure(list(AGE = c(40L,23L,24L,18L,30L,33L,32L,63L,22L,24L),FAMSIZE = c(2L,2L,3L,6L,1L,1L
),HYPERTEN = c(0,1,0),ALC = c(0,2,3,2),region_group = c("Region 4","Region 3","Region 4","Region 1","Region 2","Region 4"),PSU = c(2L,2L),PERWEIGHT_MERGE = c(31.2,615.2,37.6,1626,44,149.8,745.2,984.2,1512,399.6),SAMPWEIGHT_MERGE = c(65,860.4,94.4,9146,170.8,310.4,755.2,1053.4,3964.4,706.2),STRATA = c(6296L,6165L,6296L,6224L,6045L,6083L,6029L,6073L,6287L,6247L
)),row.names = c(NA,10L),class = "data.frame")
#weighting data frame so accounts for sample design
sample_survey<- as_survey_design(A,ids=PSU,weights=SAMPWEIGHT_MERGE,strata=STRATA,nest=TRUE)
options(survey.lonely.psu="remove")
#producing desired table
out1<-sample_survey %>%
group_by(region_group) %>%
summarise("Number of drinks (mean)"=survey_mean(ALC),"Number of drinks (median)"=survey_median(ALC),"Hypertension"=survey_mean(HYPERTEN),"Family Size"=survey_mean(FAMSIZE),"Age"=survey_median(AGE))
out1=t(out1)
out1
[,1] [,2] [,3] [,4]
region_group "Region 1" "Region 2" "Region 3" "Region 4"
Number of drinks (mean) "1.663778" "2.131566" "1.744107" "2.009594"
Number of drinks (mean)_se "0.1375124" "0.1245772" "0.0957500" "0.1199982"
Number of drinks (median) "1" "2" "1" "2"
Number of drinks (median)_se "0.0000000" "0.2531528" "0.0000000" "0.2533324"
Hypertension "0.1340147" "0.1685102" "0.1834528" "0.1225418"
Hypertension_se "0.01623974" "0.01529678" "0.01463019" "0.01475651"
Family \n Size "3.121062" "2.883905" "3.107202" "3.265012"
Family \n Size_se "0.11668906" "0.07435704" "0.08004129" "0.11138869"
Age "30" "27" "30" "28"
Age_se "1.3615690" "1.0126110" "0.7616152" "0.7599972"
我该如何修改此表，以使标准错误不是它自己的行，而是与其相关变量一起出现在同一行中（在下面还是在旁边）？谢谢！
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