choices combination,order & tree

Question

I have following data that represents sequence of person's choice between four values (f1,f2,c1,c2) :

df=structure(list(combi = structure(c(24L, 8L, 3L, 19L, 4L, 23L, 
15L, 12L, 14L, 22L, 5L, 13L, 18L, 9L, 2L, 25L, 11L, 7L, 21L, 
10L, 6L, 17L, 20L, 16L), .Label = c("", "c1-c2-f1-f2", "c1-c2-f2-f1", 
"c1-f1-c2-f2", "c1-f1-f2-c2", "c1-f2-c2-f1", "c1-f2-f1-c2", "c2-c1-f1-f2", 
"c2-c1-f2-f1", "c2-f1-c1-f2", "c2-f1-f2-c1", "c2-f2-c1-f1", "c2-f2-f1-c1", 
"f1-c1-c2-f2", "f1-c1-f2-c2", "f1-c2-c1-f2", "f1-c2-f2-c1", "f1-f2-c1-c2", 
"f1-f2-c2-c1", "f2-c1-c2-f1", "f2-c1-f1-c2", "f2-c2-c1-f1", "f2-c2-f1-c1", 
"f2-f1-c1-c2", "f2-f1-c2-c1"), class = "factor"), nb = c(10L, 
0L, 2L, 4L, 1L, 5L, 1L, 2L, 1L, 3L, 1L, 0L, 3L, 5L, 0L, 18L, 
5L, 2L, 5L, 0L, 4L, 4L, 11L, 2L)), .Names = c("combi", "nb"), class = "data.frame", row.names = c(1L, 
3L, 5L, 7L, 9L, 11L, 13L, 15L, 17L, 19L, 21L, 23L, 25L, 27L, 
29L, 31L, 33L, 35L, 37L, 39L, 41L, 43L, 45L, 47L))

I'm wondering if there's tree representation (or else) that could quantifiy, for each step choices number, by taking in account sub chain that are commun. Example :

f2  (52) -f1 (28)  -c1-c2  (10)
                   -c2-c1  (18) 

f2(52) there is 52 times chains begining by f2. there is 28 times chain beginning by f2-f1.

Thanks a lot.

Answer 1

If you read the combi values in (using as.character) you can expand those values to character columns:

df2 <-  cbind(df, read.table(text=as.character(df$combi), sep="-",stringsAsFactors=FALSE)  )

Then you can tabulate at whatever level you want:

 xtabs(nb~V1, data=df2) # First level only
#V1
#c1 c2 f1 f2 
#10 12 15 52 

xtabs(nb~paste(V1,V2,sep="-"), data=df2) # first and second
#--
# paste(V1, V2, sep = "-")
#c1-c2 c1-f1 c1-f2 c2-c1 c2-f1 c2-f2 f1-c1 f1-c2 f1-f2 f2-c1 f2-c2 f2-f1 
#    2     2     6     5     5     2     2     6     7    16     8    28 

You can also deploy the addmargins function to compactly the display the two "most senior" position sub-totals:

 addmargins( xtabs(nb~V1+V2, data=df2))
 #=========
     V2
V1    c1 c2 f1 f2 Sum
  c1   0  2  2  6  10
  c2   5  0  5  2  12
  f1   2  6  0  7  15
  f2  16  8 28  0  52
  Sum 23 16 35 15  89

This could be "flattened" with ftable:

 ftable( addmargins( xtabs(nb~V1+V2, data=df2)), row.vars=1:2)
V1  V2     
c1  c1    0
    c2    2
    f1    2
    f2    6
    Sum  10
c2  c1    5
    c2    0
    f1    5
    f2    2
    Sum  12
f1  c1    2
    c2    6
    f1    0
    f2    7
    Sum  15
f2  c1   16
    c2    8
    f1   28
    f2    0
    Sum  52
Sum c1   23
    c2   16
    f1   35
    f2   15
    Sum  89

And the final tally would be:

xtabs(nb~paste(V1,V2,V3,V4,sep="-"), data=df2)
#-----
paste(V1, V2, V3, V4, sep = "-")
c1-c2-f1-f2 c1-c2-f2-f1 c1-f1-c2-f2 c1-f1-f2-c2 c1-f2-c2-f1 c1-f2-f1-c2 c2-c1-f1-f2 c2-c1-f2-f1 
          0           2           1           1           4           2           0           5 
c2-f1-c1-f2 c2-f1-f2-c1 c2-f2-c1-f1 c2-f2-f1-c1 f1-c1-c2-f2 f1-c1-f2-c2 f1-c2-c1-f2 f1-c2-f2-c1 
          0           5           2           0           1           1           2           4 
f1-f2-c1-c2 f1-f2-c2-c1 f2-c1-c2-f1 f2-c1-f1-c2 f2-c2-c1-f1 f2-c2-f1-c1 f2-f1-c1-c2 f2-f1-c2-c1 
          3           4          11           5           3           5          10          18 

To see it all in a column:

as.matrix( xtabs(nb~paste(V1,V2,V3,V4,sep="-"), data=df2) )
#----------------
            [,1]
c1-c2-f1-f2    0
c1-c2-f2-f1    2
c1-f1-c2-f2    1
c1-f1-f2-c2    1
c1-f2-c2-f1    4
c1-f2-f1-c2    2
c2-c1-f1-f2    0
c2-c1-f2-f1    5
c2-f1-c1-f2    0
c2-f1-f2-c1    5
c2-f2-c1-f1    2
c2-f2-f1-c1    0
f1-c1-c2-f2    1
f1-c1-f2-c2    1
f1-c2-c1-f2    2
f1-c2-f2-c1    4
f1-f2-c1-c2    3
f1-f2-c2-c1    4
f2-c1-c2-f1   11
f2-c1-f1-c2    5
f2-c2-c1-f1    3
f2-c2-f1-c1    5
f2-f1-c1-c2   10
f2-f1-c2-c1   18

I suppose a "final answer with all the subtotals might be:

 ftable( addmargins( xtabs(nb~V1+V2+paste(V3,V4,sep="-"), data=df2)), row.vars=1:3)

However, that has so many zero entries that I hesitate to recommend. You could strip out zero rows:

my.ftable <- ftable( addmargins( xtabs(nb~V1+V2+paste(V3,V4,sep="-"), data=df2)), row.vars=1:3)
my.df.table <- as.data.frame(my.ftable)
names(my.df.table)[3] <- "3rd_4th"
my.df.table[ my.df.table$Freq > 0,  ]
#---------
     V1  V2 3rd_4th Freq
14   f2  f1   c1-c2   10
15  Sum  f1   c1-c2   10
18   f1  f2   c1-c2    3
20  Sum  f2   c1-c2    3
23   f1 Sum   c1-c2    3
24   f2 Sum   c1-c2   10
25  Sum Sum   c1-c2   13
34   f2  c2   c1-f1    3
35  Sum  c2   c1-f1    3
42   c2  f2   c1-f1    2
45  Sum  f2   c1-f1    2
47   c2 Sum   c1-f1    2
49   f2 Sum   c1-f1    3
50  Sum Sum   c1-f1    5
# and many more rows
#...  until
321  c1 Sum     Sum   10
322  c2 Sum     Sum   12
323  f1 Sum     Sum   15
324  f2 Sum     Sum   52
325 Sum Sum     Sum   89

Answer 2

The data.tree package specialises in tree representation. It is based on splitting variables in a hierarchal order, for example world -> continent -> country -> city. In your case, you've mentioned every order for c1, c2, f1 and f2. Likely you'd need to do four tree plots e.g. c1 --> either c2, f1 or f2, each leading to the two unused values, and then plot them.

A basic example starting with c1, and then splitting off, and not including specific values:

library(data.tree)
c1 <- Node$new("c1")      # 1st level chain, "c1"
c2 <- c1$AddChild("c2")   # new 2nd level chain, "c2", off c1
f1 <- c2$AddChild("f1-f2")   # new level off c2
f2 <- c2$AddChild("f2-f1")   # new level off c2
f1 <- c1$AddChild("f1")   # new 2nd level chain, "f1", off c1
c2 <- f1$AddChild("c2-f2")   # new level off f1
f2 <- f1$AddChild("f2-c2")   # new level off f1
f2 <- c1$AddChild("f2")   # new 2nd level chain, "f2", off c1
c2 <- f2$AddChild("c2-f1")   # new level off f2
f1 <- f2$AddChild("f1-c2")   # new level off f2

print(c1)
       levelName
1  c1           
2   ¦--c2       
3   ¦   ¦--f1-f2
4   ¦   °--f2-f1
5   ¦--f1       
6   ¦   ¦--c2-f2
7   ¦   °--f2-c2
8   °--f2       
9       ¦--c2-f1
10      °--f1-c2

plot(c1)