R Data Transformations - r

Question

I have a need to look at the data in a data frame in a different way. Here is the problem..
I have a data frame as follows
Person Item BuyOrSell
1 a B
1 b S
1 a S
2 d B
3 a S
3 e S
One of the requirements I have is to see the data as follows. Show the sum of all transactions made by the Person on individual items broken by the transaction type (B or S)
Person aB aS bB bS dB dS eB eS
1 1 1 0 1 0 0 0 0
2 0 0 0 0 1 0 0 0
3 1 0 0 0 0 0 0 1
So i created a new column and appended the values of both the Item and BuyOrSell.
df$newcol<-paste(Item,"-",BuyOrSell,sep="")
table(Person,newcol)
and was able to achieve the above results.
The last transformation requirement which was a tough nut to crack was as follows....
aB aS bB bS dB dS eB eS
aB 1 1 0 1 0 0 0 0
aS 1 2 0 1 0 0 0 1
bB 0 0 0 0 0 0 0 0
bS 1 1 0 0 0 0 0 0
dB 0 0 0 0 1 0 0 0
dS 0 0 0 0 0 0 0 0
eB 0 0 0 0 0 0 0 0
eS 0 1 0 0 0 0 0 1
where the above table had to be filled in with the number of people who made a particular transaction also made a transaction on another item.
I tried table(newcol,newcol) but it generated counts only for aB-aB,aS-aS,bB-bB,..... and 0s for all other combinations.
Any ideas on what package or command will let me crack this nut ?

Answer 1

Isn't the final result just:
# Following Ricardo's solution for casting, but using `acast` instead
A <- acast(Person~Item+BuyOrSell,data=df,fun.aggregate=length,drop=FALSE)
# A' * A
> t(A) %*% A
# a_B a_S b_B b_S d_B d_S e_B e_S
# a_B 1 1 0 1 0 0 0 0
# a_S 1 2 0 1 0 0 0 1
# b_B 0 0 0 0 0 0 0 0
# b_S 1 1 0 1 0 0 0 0
# d_B 0 0 0 0 1 0 0 0
# d_S 0 0 0 0 0 0 0 0
# e_B 0 0 0 0 0 0 0 0
# e_S 0 1 0 0 0 0 0 1

Answer 2

I think there is a better way, but here's a method using the package reshape2.
require(reshape2)
#reshapes data so each item and buy/sell event interaction occurs once
df2 <- dcast(Person~Item+BuyOrSell,data=df,fun.aggregate=length,drop=FALSE)
df2
# Person a_B a_S b_B b_S d_B d_S e_B e_S
# 1 1 1 1 0 1 0 0 0 0
# 2 2 0 0 0 0 1 0 0 0
# 3 3 0 1 0 0 0 0 0 1
#reshapes data so every row is an interaction by person
df3 <- melt(df2,id.vars="Person")
head(df3)
# Person variable value
# 1 1 a_B 1
# 2 2 a_B 0
# 3 3 a_B 0
# 4 1 a_S 1
# 5 2 a_S 0
# 6 3 a_S 1
#removes empty rows where no action occurred
#removes value column
df4 <- with(df3,
data.frame(Person=rep.int(Person,value),variable=rep.int(variable,value))
#performs a self-merge: now each row is
#every combination of two actions that one person has done
df5 <- merge(df4,df4,by="Person")
head(df5)
# Person variable.x variable.y
# 1 1 a_B a_B
# 2 1 a_B a_S
# 3 1 a_B b_S
# 4 1 a_S a_B
# 5 1 a_S a_S
# 6 1 a_S b_S
#tabulates variable interactions
with(df5,table(variable.x,variable.y))

Answer 3

Blue Magister,your solution works perfectly and i analyzed each an every step that you performed.
The output of df4 was follows:
Person variable
1 1 a_B
2 1 a_S
3 3 a_S
4 1 b_S
5 2 d_B
6 3 e_S
The output of with(df5,table(variable.x,variable.y)) was
variable.y
variable.x a_B a_S b_B b_S d_B d_S e_B e_S
a_B 1 1 0 1 0 0 0 0
a_S 1 2 0 1 0 0 0 1
b_B 0 0 0 0 0 0 0 0
b_S 1 1 0 1 0 0 0 0
d_B 0 0 0 0 1 0 0 0
d_S 0 0 0 0 0 0 0 0
e_B 0 0 0 0 0 0 0 0
e_S 0 1 0 0 0 0 0 1
which is exactly what i want.
When i look at the output of d4 it was almost similar to the my newcol solution ( using paste )
> df
Person newcol
1 1 a-B
2 1 b-S
3 1 a-S
4 2 d-B
5 3 a-S
6 3 e-S
The only difference here is the ordering of the rows when compared to your df4.
So, i ended up running this command
dfx <- merge(df,df,by="Person")
with(dfx,table(newcol.x,newcol.y))
and it generated the following...
newcol.y
newcol.x a-B a-S b-S d-B e-S
a-B 1 1 1 0 0
a-S 1 2 1 0 1
b-S 1 1 1 0 0
d-B 0 0 0 1 0
e-S 0 1 0 0 1
The above output ignored few rows and columns. What am i doing different from you ?