mean for subsets using survey

Question

I have two variables "c" and "q" in a data.frame. "c" is a number between zero and one (a level of poverty) and "q" indicate if the household (or subject) is poor with 1 or non-poverty with zero. How can I calculate the mean of "c" only of the poor households (q=1).

What I need

Important detail: I have a database for a coutry and I want this result for regions. I am using the svyby like this:

svyby( ~q , ~region , design = base2015_pos , na.rm=TRUE, svytotal)

so in that way the R give me the number of poor by region and I don't need this now. I need the mean of a subset (see image above) by region.

structure(list(domicilio = c(11000015001, 11000015003, 11000015004), agua = c(0, 0, 6), ind_agua = c(0, 0, 1), esgoto = c(1, 1, 6), ind_cond_sanitaria = c(1, 1, 1), lixo = c(0, 0, 0), ind_lixo = c(0, 0, 0), luz = c(0, 0, 0), ind_iluminacao = c(0, 0, 0), ativos = c(0, 0, 0), ind_ativos = c(0, 0, 0), emprego = c(0, 0, 0), ind_emprego = c(0, 0, 0), renda = c(0, 0, 0), ind_renda = c(0, 0, 0), casa = c(1, 1, 0), ind_riqueza = c(1, 1, 0), anos = c(0, 0, 0), ind_estudo = c(0, 0, 0), ler = c(0, 0, 0), ind_alfabetizado = c(0, 0, 0), peso = c(270, 270, 270), sexo = c(0, 1, 1), uf = c("11", "11", "11"), v4609 = c("001772940", "001772940", "001772940"), v4617 = c(110001, 110001, 110001), v4618 = c(1, 1, 1), pre_wgt = c(200, 200, 200), one = c(1L, 1L, 1L), region = c("1", "1", "1"), c = c(0.2, 0.2, 0.2), q = c(0, 0, 0)), .Names = c("domicilio", "agua", "ind_agua", "esgoto", "ind_cond_sanitaria", "lixo", "ind_lixo", "luz", "ind_iluminacao","ativos", "ind_ativos", "emprego", "ind_emprego", "renda", "ind_renda", "casa", "ind_riqueza", "anos", "ind_estudo", "ler", "ind_alfabetizado","peso", "sexo", "uf", "v4609", "v4617", "v4618", "pre_wgt", "one", "region", "c", "q"), row.names = c(NA, 3L), class = "data.frame")

Answer 1

# complex sample survey design
library(survey)

# your data.frame
x <- structure(list(domicilio = c(11000015001, 11000015003, 11000015004), agua = c(0, 0, 6), ind_agua = c(0, 0, 1), esgoto = c(1, 1, 6), ind_cond_sanitaria = c(1, 1, 1), lixo = c(0, 0, 0), ind_lixo = c(0, 0, 0), luz = c(0, 0, 0), ind_iluminacao = c(0, 0, 0), ativos = c(0, 0, 0), ind_ativos = c(0, 0, 0), emprego = c(0, 0, 0), ind_emprego = c(0, 0, 0), renda = c(0, 0, 0), ind_renda = c(0, 0, 0), casa = c(1, 1, 0), ind_riqueza = c(1, 1, 0), anos = c(0, 0, 0), ind_estudo = c(0, 0, 0), ler = c(0, 0, 0), ind_alfabetizado = c(0, 0, 0), peso = c(270, 270, 270), sexo = c(0, 1, 1), uf = c("11", "11", "11"), v4609 = c("001772940", "001772940", "001772940"), v4617 = c(110001, 110001, 110001), v4618 = c(1, 1, 1), pre_wgt = c(200, 200, 200), one = c(1L, 
1L, 1L), region = c("1", "1", "1"), c = c(0.2, 0.2, 0.2), q = c(0, 0, 0)), .Names = c("domicilio", "agua", "ind_agua", "esgoto", "ind_cond_sanitaria", "lixo", "ind_lixo", "luz", "ind_iluminacao","ativos", "ind_ativos", "emprego", "ind_emprego", "renda", "ind_renda", "casa", "ind_riqueza", "anos", "ind_estudo", "ler", "ind_alfabetizado","peso", "sexo", "uf", "v4609", "v4617", "v4618", "pre_wgt", "one", "region", "c", "q"), row.names = c(NA, 3L), class = "data.frame")

# your survey.design (this is not the correct svydesign statement, please follow the directions specific to your data set)
y <- svydesign( ~ 1 , data = x , weights = ~ pre_wgt )

# your desired subset
z <- subset( y , q == 1 )

# your desired mean
svyby( ~ c , ~ region , z , svymean )

Answer 2

Here's another possibility. To illustrate, create a dataset per your parameters:

set.seed(787)
dat.a <-runif(n=10,min=0,max=1)
dat.b <-rbinom(n=10, size=1, prob=0.5)
dat.1 <-data.frame(matrix(c(dat.a, dat.b), ncol=2, nrow=10))
colnames(dat.1) <-c("c","q")

dat.1
            c q
1  0.35326234 1 
2  0.45277055 0
3  0.29505270 0
4  0.78723105 1
5  0.95915348 1
6  0.17505284 0
7  0.79693672 0
8  0.01648420 1
9  0.02706417 0
10 0.93996311 1

Now subset by extracting all rows that match q=1 and compute mean for column c in resulting output:

dat.1.subset <-dat.1[dat.1$q==1,]

mean(dat.1.subset$c)
[1] 0.6112188

Answer 3

aggregate(df$c, by=list(df$q), FUN=mean)