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