Melt and cast the shape of your data.frame – Exercises

June 22, 2018
By

(This article was first published on R-exercises, and kindly contributed to R-bloggers)

 

Datasets often arrive to us in a form that is different from what we need for our modelling or visualisations functions who in turn don’t necessary require the same format.

Reshaping data.frames is a step that all analysts need but many struggle with. Practicing this meta-skill will in the long-run result in more time to focus on the actual analysis.

The solutions to this set will rely on data.table, mostly melt() and dcast() which are originally from the reshape2 package. However, you can also get practice out if it using your favourite base-R, tidyverse or any other method and then compare the results.

Solutions are available here.

 

Exercise 1

Take the following data.frame from this form

df <- data.frame(id = 1:2, q1 = c("A", "B"), q2 = c("C", "A"), stringsAsFactors = FALSE)
df
  id q1 q2
1  1  A  C
2  2  B  A 

to this

  id question value
1  1       q1     A
2  2       q1     B
3  1       q2     C
4  2       q2     A

 

Exercise 2

Do the opposite; return the data.frame back to it’s original form.

Exercise 3

Set up the data.frame in terms of questions. Such as the following:

  question id_1 id_2
1       q1    A    B
2       q2    C    A

 

Exercise 4

The data entry behind this data.frame went a little bit wrong. Get all the C and B entries into their corresponding columns

df2 <- data.frame(
  A = c("A1", "A12", "A31", "A4"), 
  B = c("B4", "C7", "C3", "B9"), 
  C = c("C3", "B16", "B3", "C4")
)

 

Exercise 5

Get this data.frame

df3 <- data.frame(
  Join_ID = rep(1:3, each = 2), 
  Type    = rep(c("a", "b"), 3), 
  v2      = c(8, 9, 7, 6, 5, 4)*10
)

 

To look like this:

  Join_ID a_v2 b_v2
1       1   80   90
2       2   70   60
3       3   50   40

 

Exercise 6

Revisiting a dataset used in an earlier exercise set on data exploration;
load the AER package and run the command data("Fertility") which loads the dataset Fertility to your workspace.
Melt it into the following format, with one row per child.

head(ferl)
  morekids age afam hispanic other work mother_id order gender
1       no  27   no       no    no    0         1     1   male
2       no  30   no       no    no   30         2     1 female
3       no  27   no       no    no    0         3     1   male
4       no  35  yes       no    no    0         4     1   male
5       no  30   no       no    no   22         5     1 female
6       no  26   no       no    no   40         6     1   male

 

Exercise 7

Take this

d1 = data.frame(
  ID=c(1,1,1,2,2,4,1,2), 
  medication=c(1,2,3,1,2,7,2,8)
)
d1
  ID medication
1  1          1
2  1          2
3  1          3
4  2          1
5  2          2
6  4          7
7  1          2
8  2          8

to this form:

  
   ID medications
1:  1  1, 2, 3, 2
2:  2     1, 2, 8
3:  4           7

 

Note the solution doesn’t use melt() nor dcast(), so you might look at other options.

Exercise 8

Get this

dfs <- data.frame(
  Name = c(rep("name1",3),rep("name2",2)),
  MedName = c("atenolol 25mg","aspirin 81mg","sildenafil 100mg", "atenolol 50mg","enalapril 20mg")
)
dfs
   Name          MedName
1 name1    atenolol 25mg
2 name1     aspirin 81mg
3 name1 sildenafil 100mg
4 name2    atenolol 50mg
5 name2   enalapril 20mg

 

Into the following format:

    Name  medication_1   medication_2     medication_3
1: name1 atenolol 25mg   aspirin 81mg sildenafil 100mg
2: name2 atenolol 50mg enalapril 20mg             

 

Exercise 9

Get the following data.frame organized in standard form

df7 <- data.table(
  v1 = c("name1, name2", "name3", "name4, name5"),
  v2 = c("1, 2", "3", "4, 5"), 
  v3 = c(1, 2, 3)
)
df7
             v1   v2 v3
1: name1, name2 1, 2  1
2:        name3    3  2
3: name4, name5 4, 5  3

 

Expected output:

 
      v1 v2 v3
1: name1  1  1
2: name2  2  1
3: name3  3  2
4: name4  4  3
5: name5  5  3

 

The solution doesn’t use melt() nor dcast() and can be suprisingly hard.

Exercise 10

Convert this:

 
df <- data.frame(
 Method = c("10.fold.CV Lasso", "10.fold.CV.1SE", "BIC", "Modified.BIC"),
 n = c(30, 30, 50, 50, 50, 50, 100, 100),
 lambda = c(1, 3, 1, 2, 2, 0, 1, 2), 
df = c(21, 17, 29, 26, 25, 32, 34, 32) ) 
> df
            Method   n lambda df
1 10.fold.CV Lasso  30      1 21
2   10.fold.CV.1SE  30      3 17
3              BIC  50      1 29
4     Modified.BIC  50      2 26
5 10.fold.CV Lasso  50      2 25
6   10.fold.CV.1SE  50      0 32
7              BIC 100      1 34
8     Modified.BIC 100      2 32

Into

 
            Method lambda_30 lambda_50 lambda_100 df_30 df_50 df_100
1 10.fold.CV Lasso         1         2               21    25       
2   10.fold.CV.1SE         3         0               17    32       
3              BIC                   1          1          29     34
4     Modified.BIC                   2          2          26     32  

 

(Image by Joe Alterio)

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