Looping over all possible column combinations with some contstraints - r

Question

xx is a sample data. it contains variables dep1,dep2,dep3,bet1,bet2,bet3. I want to select all possible 2 columns combinations but not the ones with the same name (except,number) . In this examples there are
9 such combos {dep1:bet1,dep1:bet2,dep1:bet3,dep2:bet1...........}
Below is code which I want to run for all combinations ( I did it just for one) also in last line
I added a code to keep track which variables were included in calculations. I believe the regex will help
to understand. help appreciated !
xx<-data.frame(id=1:10,
category=c(rep("A",5),rep("B",5)),
dep1=sample(1:5,10,replace = T),
dep2=sample(1:5,10,replace = T),
dep3=sample(1:5,10,replace = T),
bet1=sample(1:5,10,replace = T),
bet2=sample(1:5,10,replace = T),
bet3=sample(1:5,10,replace = T))
xx%>%select(2,dep1,bet1)%>%
mutate(vdep=if_else(dep1>3,1,0),
vbet=if_else(bet1>3,1,0))%>%
group_by(category)%>%
summarise(vdep=mean(vdep),
vbet=mean(vbet))%>%ungroup()%>%
gather(variable,value,-category)%>%
mutate(variable=as.factor(variable))%>%
unite(variable,category,col = "new")%>%
spread(new,value)%>%
mutate(first="dep1",second="bet1")

Answer 1

If I understand you correctly, something like the following should do it:
# the data
xx<-data.frame(id=1:10,
category=c(rep("A",5),rep("B",5)),
dep1=sample(1:5,10,replace = T),
dep2=sample(1:5,10,replace = T),
dep3=sample(1:5,10,replace = T),
bet1=sample(1:5,10,replace = T),
bet2=sample(1:5,10,replace = T),
bet3=sample(1:5,10,replace = T))
# Getting the column names with "dep" or "bet"
cols = names(xx)[grepl("dep|bet", names(xx))]
deps = cols[grepl("dep", cols)]
bets = cols[grepl("bet", cols)]
# Getting all possible combinations of these columns
comb = expand.grid(deps, bets)
comb
# Var1 Var2
# 1 dep1 bet1
# 2 dep2 bet1
# 3 dep3 bet1
# 4 dep1 bet2
# 5 dep2 bet2
# 6 dep3 bet2
# 7 dep1 bet3
# 8 dep2 bet3
# 9 dep3 bet3
# Transposing the dataframe containing these combinations, so that
# we can directly use sapply / lapply on the columns latter
comb = data.frame(t(comb), stringsAsFactors = FALSE)
# For each combination, subset the dataframe xx
result = sapply(comb, function(x){
xx[, x]
}, simplify = FALSE)
result
# $X1
# dep1 bet1
# 1 1 5
# 2 1 5
# 3 2 2
# 4 2 2
# 5 1 5
# 6 3 3
# 7 1 1
# 8 2 2
# 9 3 2
# 10 1 5
#
# $X2
# dep2 bet1
# 1 1 5
# 2 2 5
# 3 4 2
# 4 5 2
# 5 1 5
# 6 5 3
# 7 2 1
# 8 1 2
# 9 4 2
# 10 4 5
#
# $X3
# dep3 bet1
# 1 3 5
# 2 2 5
# 3 4 2
# 4 3 2
# 5 3 5
# 6 2 3
# 7 1 1
# 8 4 2
# 9 5 2
# 10 5 5
#
# $X4
# dep1 bet2
# 1 1 5
# 2 1 1
# 3 2 1
# 4 2 2
# 5 1 2
# 6 3 2
# 7 1 3
# 8 2 3
# 9 3 5
# 10 1 1
#
# $X5
# dep2 bet2
# 1 1 5
# 2 2 1
# 3 4 1
# 4 5 2
# 5 1 2
# 6 5 2
# 7 2 3
# 8 1 3
# 9 4 5
# 10 4 1
#
# $X6
# dep3 bet2
# 1 3 5
# 2 2 1
# 3 4 1
# 4 3 2
# 5 3 2
# 6 2 2
# 7 1 3
# 8 4 3
# 9 5 5
# 10 5 1
#
# $X7
# dep1 bet3
# 1 1 3
# 2 1 2
# 3 2 5
# 4 2 1
# 5 1 3
# 6 3 2
# 7 1 4
# 8 2 1
# 9 3 1
# 10 1 3
#
# $X8
# dep2 bet3
# 1 1 3
# 2 2 2
# 3 4 5
# 4 5 1
# 5 1 3
# 6 5 2
# 7 2 4
# 8 1 1
# 9 4 1
# 10 4 3
#
# $X9
# dep3 bet3
# 1 3 3
# 2 2 2
# 3 4 5
# 4 3 1
# 5 3 3
# 6 2 2
# 7 1 4
# 8 4 1
# 9 5 1
# 10 5 3