Reshape data using dcast?

Question

I don't know if using dcast() is the right way, but I want to reshape the following data.frame:

df <- data.frame(x=c("p1","p1","p2"),y=c("a","b","a"),z=c(14,14,16))
df
   x y  z
1 p1 a 14
2 p1 b 14
3 p2 a 16

so that it looks like this one:

df2 <- data.frame(x=c("p1","p2"),a=c(1,1),b=c(1,0),z=c(14,16))
   x a b  z
1 p1 1 1 14
2 p2 1 0 16

The variable y in df should be broken so that its elements are new variables, each dummy coded. All other variables (in this case just z ) are equal for each person (p1,p2 etc.). The only variable where a specific person p has different values is y .
The reason I want this is because I need to merge this dataset with other ones by variable x . Thing is, it needs to be one row per person (p1,p2 etc).

Answer 1

The following works, but seems cumbersome.

df2 <- df
df2$y <- as.numeric(y)
df$y2 <- as.numeric(df$y)

df2 <- dcast(df, x+z~y, value.var="y2")

df2
   x  z a  b
1 p1 14 1  2
2 p2 16 1 NA

Answer 2

This is almost a duplicate of a previous question , and the same basic answer I used there works again. No need for any external packages either.

aggregate(model.matrix(~ y - 1, data=df),df[c("x","z")],max)

   x  z ya yb
1 p1 14  1  1
2 p2 16  1  0

---

To explain this, as it is a bit odd looking, the model.matrix call at its most basic returns a binary indicator variable for each unique value for each row of your data.frame, like so:

  ya yb
1  1  0
2  0  1
3  1  0

If you aggregate that intermediate result by your two id variables ( x and z ), you are then essentially acting on the initial data.frame of:

   x  z ya yb
1 p1 14  1  0
2 p1 14  0  1
3 p2 16  1  0

So if you take the max value of ya and yb within each combination of x and z , you basically do:

   x  z ya      yb
1 p1 14  1*max*  0
2 p1 14  0       1*max*

--collapse--

   x  z ya      yb
1 p1 14  1       1

...and repeat that for each unique x / z combination to give the final result:

   x  z ya yb
1 p1 14  1  1
2 p2 16  1  0

---

Things get a bit crazy to generalise this to more columns, but it can be done, courtesy of this question eg:

df <- data.frame(x=c("p1","p1","p2"),y=c("a","b","a"),z=c("14","15","16"))
intm <- model.matrix(~ y + z - 1, data=df,
                 contrasts.arg = sapply(df[2:3], contrasts, contrasts=FALSE))
aggregate(intm,df[c("x")],max)

   x ya yb z14 z15 z16
1 p1  1  1   1   1   0
2 p2  1  0   0   0   1

Answer 3

I'm not sure much of this you have to do but if you need a way to automate it, I wrote this little function that might help:

First run dcast:

new = dcast(df, x+z~y, value.var="y")

Load into your R environment:

 # args to be passed: 
 # df is your dataframe 
 # cols is a list of format c("colname1", "colname2", ... , "colnameN")
    binarizeCols = function(df, cols){
      for(i in cols){
        column = which(colnames(df) == i)
        truthRow = is.na(df[,column])
        for(j in 1:length(truthRow)){
          if(truthRow[j] == FALSE){
            df[j,column] = 1
          }else{
             df[j,column] = 0
           }
        }
      }
      return(df)
    }

then run:

new = binarizeCols(new, c("a", "b"))

and you get:

     x  z  a  b
   1 p1 14 1  1 
   2 p2 16 1  0

not as fast as using _apply() but there's no hardcoding, you can enter any colnames you want (maybe you want to skip one in the middle?) and you dont create a new instance of your df. note: I use "=" instead of "<-" because I thought it was being phased out but they can be replaced if need be.

Answer 4

df <- data.frame(x=c("p1","p1","p2","p3"),
                 y=c("a","b","a","c"),
                 z=c(14,14,16,17))  # wanted larger test case.
new <- dcast(df, x+z~y, value.var="y")
new[3:5] <- sapply(lapply(new[3:5], '%in%', unique(df$y) ), as.numeric)
new
   x  z a b c
1 p1 14 1 1 0
2 p2 16 1 0 0
3 p3 17 0 0 1

First check for containment in a vector that summarizes the possible values to create columns of logical values. Then 'dummify' by taking as.numeric of those logical values.