CODEWITHSUNDEEP
rlistiterationsurvey
edo
edo

Reputation: 85

Iterate through groups of variables in a survey - R

As indicated here, if you want to calculate proportions of a categorical variable in the amazing srvyr package you first have to group over the variables as factors and then use an empty srvyr::survey_mean, as in this example.

My goal is to iterate over the second variables cname and sch.wide while keeping the first grouping variable stype to avoid duplicating the code.

library(survey)
library(srvyr)

data(api)

df <- apiclus1 %>% 
  mutate(cname=as.factor(cname)) %>% 
  select(pw,stype, cname,sch.wide) %>% 
  as_survey_design(weights=pw) 

# proportions of sch.wide
df %>% 
  group_by(stype,sch.wide) %>% 
  summarise(prop=srvyr::survey_mean())
#> # A tibble: 6 x 4
#>   stype sch.wide   prop prop_se
#>   <fct> <fct>     <dbl>   <dbl>
#> 1 E     No       0.0833  0.0231
#> 2 E     Yes      0.917   0.0231
#> 3 H     No       0.214   0.110 
#> 4 H     Yes      0.786   0.110 
#> 5 M     No       0.32    0.0936
#> 6 M     Yes      0.68    0.0936

# proportions of cname
df %>% 
  group_by(stype,cname) %>% 
  summarise(prop=srvyr::survey_mean())
#> # A tibble: 33 x 4
#>    stype cname          prop prop_se
#>    <fct> <fct>         <dbl>   <dbl>
#>  1 E     Alameda     0.0556  0.0191 
#>  2 E     Fresno      0.0139  0.00978
#>  3 E     Kern        0.00694 0.00694
#>  4 E     Los Angeles 0.0833  0.0231 
#>  5 E     Mendocino   0.0139  0.00978
#>  6 E     Merced      0.0139  0.00978
#>  7 E     Orange      0.0903  0.0239 
#>  8 E     Plumas      0.0278  0.0137 
#>  9 E     San Diego   0.347   0.0398 
#> 10 E     San Joaquin 0.208   0.0339 
#> # ... with 23 more rows
Created on 2019-11-28 by the reprex package (v0.3.0)

Maybe the way to go here is creating lists that keep the first grouping variable and divide the data by a another group of variables, and then calculate the proportions.

I would like to find a solution that involves purrr:map or tidyverse.

Thanks in advance for the help, or for pointing to the answer!

Upvotes: 1

Views: 419

Answers (1)

akrun
akrun

Reputation: 887991

There are multiple ways. If we pass as string, one option is to make use of group_by_at which takes strings as argument

library(purrr)
library(dplyr)
library(survey)
library(srvyr)
map(c('sch.wide', 'cname'), ~
        df %>%
           group_by_at(vars("stype", .x)) %>%
           summarise(prop = srvyr::survey_mean()))
#[[1]]
# A tibble: 6 x 4
#  stype sch.wide   prop prop_se
#  <fct> <fct>     <dbl>   <dbl>
#1 E     No       0.0833  0.0231
#2 E     Yes      0.917   0.0231
#3 H     No       0.214   0.110 
#4 H     Yes      0.786   0.110 
#5 M     No       0.32    0.0936
#6 M     Yes      0.68    0.0936

#[[2]]
# A tibble: 30 x 4
#   stype cname          prop prop_se
#   <fct> <fct>         <dbl>   <dbl>
# 1 E     Alameda     0.0556  0.0191 
# 2 E     Fresno      0.0139  0.00978
# 3 E     Kern        0.00694 0.00694
# 4 E     Los Angeles 0.0833  0.0231 
# 5 E     Mendocino   0.0139  0.00978
# 6 E     Merced      0.0139  0.00978
# 7 E     Orange      0.0903  0.0239 
# 8 E     Plumas      0.0278  0.0137 
# 9 E     San Diego   0.347   0.0398 
#10 E     San Joaquin 0.208   0.0339 
# … with 20 more rows

Or another option is to wrap with quos to create a quosure list and evaluate (!!) it in group_by

map(quos(sch.wide, cname), ~  
        df %>%
          group_by(stype, !!.x) %>% 
          summarise(prop = srvyr::survey_mean()))

Upvotes: 1

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