Reputation: 85
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
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