Ordering alphabetically after ordering once numerically [duplicate] - r

Question

Supose I have a data frame with 3 columns (name, y, sex) where name is character, y is a numeric value and sex is a factor.
sex<-c("M","M","F","M","F","M","M","M","F")
x<-c("MARK","TOM","SUSAN","LARRY","EMMA","LEONARD","TIM","MATT","VIOLET")
name<-as.character(x)
y<-rnorm(9,8,1)
score<-data.frame(x,y,sex)
score
name y sex
1 MARK 6.767086 M
2 TOM 7.613928 M
3 SUSAN 7.447405 F
4 LARRY 8.040069 M
5 EMMA 8.306875 F
6 LEONARD 8.697268 M
7 TIM 10.385221 M
8 MATT 7.497702 M
9 VIOLET 10.177969 F
If I wanted to order it by y I would use:
score[order(score$y),]
x y sex
1 MARK 6.767086 M
3 SUSAN 7.447405 F
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
5 EMMA 8.306875 F
6 LEONARD 8.697268 M
9 VIOLET 10.177969 F
7 TIM 10.385221 M
So far, so good... The names keep the correct score BUT how could I reorder it to have M and F levels not mixed. I need to order and at the same time keep factor levels separated.
Finally I would like to take a step further to involve character, the example doesn't help, but what if there were tied y values and I would have to order again within factor (e.g. TIM and TOM got 8.4 and I have to assign alphabetical order).
I was thinking about by function but it creates a list and doesn't help really. I think there must be some function like it to apply on data frames and get data frames as return.
TO MAKE CLEAR THE POINT:
sep<-split(score,score$sex)
sep$M<-sep$M[order(sep$M[,2]),]
sep$M
x y sex
1 MARK 6.767086 M
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
6 LEONARD 8.697268 M
7 TIM 10.385221 M
sep$F<-sep$F[order(sep$F[,2]),]
sep$F
x y sex
3 SUSAN 7.447405 F
5 EMMA 8.306875 F
9 VIOLET 10.177969 F
merged<-rbind(sep$M,sep$F)
merged
x y sex
1 MARK 6.767086 M
8 MATT 7.497702 M
2 TOM 7.613928 M
4 LARRY 8.040069 M
6 LEONARD 8.697268 M
7 TIM 10.385221 M
3 SUSAN 7.447405 F
5 EMMA 8.306875 F
9 VIOLET 10.177969 F
I know how to do that if I have 2 or 3 factors. But what if I had serious levels of factors, say 20, should I write a for loop?

Answer 1

order takes multiple arguments, and it does just what you want:
with(score, score[order(sex, y, x),])
## x y sex
## 3 SUSAN 6.636370 F
## 5 EMMA 6.873445 F
## 9 VIOLET 8.539329 F
## 6 LEONARD 6.082038 M
## 2 TOM 7.812380 M
## 8 MATT 8.248374 M
## 4 LARRY 8.424665 M
## 7 TIM 8.754023 M
## 1 MARK 8.956372 M

Answer 2

Here is a summary of all methods mentioned in other answers/comments (to serve future searchers). I've added a data.table way of sorting.
# Base R
do.call(rbind, by(score, score$sex, function(x) x[order(x$y),]))
with(score, score[order(sex, y, x),])
score[order(score$sex,score$x),]
# Using plyr
arrange(score, sex,y)
ddply(score, c('sex', 'y'))
# Using `data.table`
library("data.table")
score_dt <- setDT(score)
# setting a key works sorts the data.table
setkey(score_dt,sex,x)
print(score_dt)
Here is Another question that deals with the same

Answer 3

I think there must be some function like it to apply on data frames
and get data frames as return
Yes there is:
library(plyr)
ddply(score, c('y', 'sex'))

Answer 4

It sounds to me like you're trying to order by score within the males and females and return a combined data frame of sorted males and sorted females.
You are right that by(score, score$sex, function(x) x[order(x$y),]) returns a list of sorted data frames, one for male and one for female. You can use do.call with the rbind function to combine these data frames into a single final data frame:
do.call(rbind, by(score, score$sex, function(x) x[order(x$y),]))
# x y sex
# F.5 EMMA 7.526866 F
# F.9 VIOLET 8.182407 F
# F.3 SUSAN 9.677511 F
# M.4 LARRY 6.929395 M
# M.8 MATT 7.970015 M
# M.7 TIM 8.297137 M
# M.6 LEONARD 8.845588 M
# M.2 TOM 9.035948 M
# M.1 MARK 10.082314 M

Answer 5

I believe that the person asked how to sort it by the orders in the case of say 20.
I know how to do that if I have 2 or 3 factors. But what if I had
serious levels of factors, say 20, should I write a for loop?
I have one where there are 9 orders with various counts.
stage_name count
<ord> <int>
1 Closed Lost 957
2 Closed Won 1413
3 Evaluation 1773
4 Meeting Scheduled 4104
5 Nurture 1222
6 Opportunity Disqualified 805
7 Order Submitted 1673
8 Qualifying 5138
9 Quoted 4976
In this case you can see that it is displayed using alphabetical order of stage_name, but stage_name is actually an ordered factor that has a very different order.
This code orders the factor is a much different order:
# Make categoricals ----
check_stage$stage_name = ordered(check_stage$stage_name, levels=c(
'Opportunity Disqualified',
'Qualifying',
'Evaluation',
'Meeting Scheduled',
'Quoted',
'Order Submitted',
'Closed Won',
'Closed Lost',
'Nurture'))
Now we can just apply the factor as the method of ordering this is a dplyr function, but you might need forcats too. I have both libraries installed:
check_stage <- check_stage %>%
arrange(factor(stage_name))
This now gives the output in the factor order as desired:
check_stage
# A tibble: 9 × 2
stage_name count
<ord> <int>
1 Opportunity Disqualified 805
2 Qualifying 5138
3 Evaluation 1773
4 Meeting Scheduled 4104
5 Quoted 4976
6 Order Submitted 1673
7 Closed Won 1413
8 Closed Lost 957
9 Nurture 1222