5.2 Data Frames
source: http://www.r-exercises.com/2016/01/04/data-frame-exercises/
5.2.1 Create the following data frame, afterwards invert Sex for all individuals.
Name <- c("Alex", "Lilly", "Mark", "Oliver", "Martha", "Lucas", "Caroline")
Age <- c(25, 31, 23, 52, 76, 49, 26)
Height <- c(177, 163, 190, 179, 163, 183, 164)
Weight <- c(57, 69, 83, 75, 70, 83, 53)
Sex <- as.factor(c("F", "F", "M", "M", "F", "M", "F"))
df <- data.frame (row.names = Name, Age, Height, Weight, Sex)
levels(df$Sex) # The order of levels: "F" "M". So we can reassign them differently to invert Sex for all.## [1] "F" "M"levels(df$Sex) <- c("M","F")
df## Age Height Weight Sex
## Alex 25 177 57 M
## Lilly 31 163 69 M
## Mark 23 190 83 F
## Oliver 52 179 75 F
## Martha 76 163 70 M
## Lucas 49 183 83 F
## Caroline 26 164 53 M5.2.2 Create this data frame (make sure you import the variable Working as character and not factor).
Working <- c("Yes","No","No","Yes","Yes","No","Yes")
df2 <- data.frame(row.names = Name,Working) #Name has been already defined in exercise 1.
df2## Working
## Alex Yes
## Lilly No
## Mark No
## Oliver Yes
## Martha Yes
## Lucas No
## Caroline YesAdd this data frame column-wise to the previous one.
a) How many rows and columns does the new data frame have?
b) What class of data is in each column?
df <- cbind(df,df2)
dim(df) #The output is [1] nrow ncolumn## [1] 7 5str(df)## 'data.frame': 7 obs. of 5 variables:
## $ Age : num 25 31 23 52 76 49 26
## $ Height : num 177 163 190 179 163 183 164
## $ Weight : num 57 69 83 75 70 83 53
## $ Sex : Factor w/ 2 levels "M","F": 1 1 2 2 1 2 1
## $ Working: Factor w/ 2 levels "No","Yes": 2 1 1 2 2 1 25.2.3 Check what class of data is the (built-in data set) state.center and convert it to data frame.
class(state.center) #state.center is a list of 2 numeric vectors.## [1] "list"data.state <- as.data.frame(state.center) # Now it is a data frame5.2.4 Create a simple data frame from 3 vectors. Order the entire data frame by the first column.
a <- (rnorm(10))
b <- letters[4:13]
c <- c("yes","no","no","no","no","yes","no","yes","yes","no")
df3 <- data.frame(a,b,c)
df3[with (df3, order(a)),] ## a b c
## 5 -1.4874427 h no
## 6 -0.9261379 i yes
## 8 -0.8676770 k yes
## 10 -0.4776941 m no
## 2 -0.3444281 e no
## 9 -0.1275440 l yes
## 4 0.4226900 g no
## 1 0.4440771 d yes
## 3 0.4682689 f no
## 7 0.9045589 j no5.2.5 Create a data frame from a matrix of your choice, change the row names so every row says id_i (where i is the row number) and change the column names to variable_i (where i is the column number). I.e., for column 1 it will say variable_1, and for row 2 will say id_2 and so on.
matrix.data <- matrix(1:40, nrow = 10, ncol = 4)
df <- as.data.frame(matrix.data)
colnames(df) <- paste("variable_", 1:ncol(df))
rownames(df) <- paste("id_", 1:nrow(df))
df## variable_ 1 variable_ 2 variable_ 3 variable_ 4
## id_ 1 1 11 21 31
## id_ 2 2 12 22 32
## id_ 3 3 13 23 33
## id_ 4 4 14 24 34
## id_ 5 5 15 25 35
## id_ 6 6 16 26 36
## id_ 7 7 17 27 37
## id_ 8 8 18 28 38
## id_ 9 9 19 29 39
## id_ 10 10 20 30 405.2.6 For this exercise, we’ll use the (built-in) dataset VADeaths.
- Make sure the object is a data frame, if not change it to a data frame.
- Create a new variable, named Total, which is the sum of each row.
- Change the order of the columns so total is the first variable.
class(VADeaths) #Which shows that it is a matrix## [1] "matrix"df <- data.frame(VADeaths)
df$Total <- rowSums(df)
df <- df[,c(5,1,2,3,4)]
df## Total Rural.Male Rural.Female Urban.Male Urban.Female
## 50-54 44.2 11.7 8.7 15.4 8.4
## 55-59 67.7 18.1 11.7 24.3 13.6
## 60-64 103.5 26.9 20.3 37.0 19.3
## 65-69 161.6 41.0 30.9 54.6 35.1
## 70-74 241.4 66.0 54.3 71.1 50.05.2.7 For this exercise we’ll use the (built-in) dataset state.x77.
- Make sure the object is a data frame, if not change it to a data frame.
class(state.x77)## [1] "matrix"df2 <- data.frame(state.x77)- Find out how many states have an income of less than 4300.
nrow(df2[df2$Income < 4300,]) # We are finding the number of rows in the table which is filtered as Income < 4300## [1] 20#length(df2$Income[df2$Income < 4300]) # So there are several ways.
#count(df2$Income < 4300) # You can see a frequency table with this statement- Find out which is the state with the highest income.
row.names(df2[which.max(df2$Income),]) # When I need to find a max value, "which.max" statement comes to my mind in a moment but you can find the solution with different ways.## [1] "Alaska"5.2.8 With the dataset swiss, create a data frame of only the rows 1, 2, 3, 10, 11, 12 and 13, and only the variables Examination, Education and Infant.Mortality.
df3 <- data.frame(swiss[c(1,2,3,10,11,12,13),c("Examination", "Education", "Infant.Mortality")])- The infant mortality of Sarine is wrong, it should be a NA, change it.
df3$Infant.Mortality[4] <- NA- Create a row that will be the total sum of the columns, name it Total.
Total <- colSums(df3, na.rm = TRUE) #We have a NA value in 3rd column. That's why it's safe to ignore NA values.
df4 <- rbind(df3,Total)
rownames(df4) = c(rownames(df3),"Total")
df4## Examination Education Infant.Mortality
## Courtelary 15 12 22.2
## Delemont 6 9 22.2
## Franches-Mnt 5 5 20.2
## Sarine 16 13 NA
## Veveyse 14 6 24.5
## Aigle 21 12 16.5
## Aubonne 14 7 19.1
## Total 91 64 124.7- Create a new variable that will be the proportion of Examination (Examination / Total)
Prop <- df4$Examination / df4["Total","Examination"]
Prop## [1] 0.16483516 0.06593407 0.05494505 0.17582418 0.15384615 0.23076923
## [7] 0.15384615 1.00000000df4 <- cbind(df4,Prop)
df4## Examination Education Infant.Mortality Prop
## Courtelary 15 12 22.2 0.16483516
## Delemont 6 9 22.2 0.06593407
## Franches-Mnt 5 5 20.2 0.05494505
## Sarine 16 13 NA 0.17582418
## Veveyse 14 6 24.5 0.15384615
## Aigle 21 12 16.5 0.23076923
## Aubonne 14 7 19.1 0.15384615
## Total 91 64 124.7 1.000000005.2.9 Create a data frame with the datasets state.abb, state.area, state.division, state.name, state.region. The row names should be the names of the states.
df <- data.frame(state.abb, state.area, state.division, state.region, row.names = state.name)- Rename the column names so only the first 3 letters after the full stop appear (e.g. States.abb will be abb).
colnames(df) <- substr(colnames(df), 7, 9)5.2.10 Add the previous data frame column-wise to state.x77
new.df <- cbind(state.x77,df)- Remove the variable div.
new.df$div <- NULL- Also remove the variables Life Exp, HS Grad, Frost, abb, and are.
#subsetting is easier when you need to remove more columns
new.df <- subset(new.df, ,-c(4, 6, 7, 9, 10))- Add a variable to the data frame which should categorize the level of illiteracy: [0,1) is low, [1,2) is some, [2, inf) is high.
# For such categorization, ifelse statement is very very useful.
new.df$Illiteracy.Levels <- ifelse(new.df$Illiteracy >= 0 & new.df$Illiteracy < 1, "Low",
ifelse(new.df$Illiteracy >= 1 & new.df$Illiteracy < 2, "Some",
"High"))- Find out which state from the west, with low illiteracy, has the highest income, and what that income is.
x <- subset(new.df, reg == "West" & Illiteracy.Levels == "Low")
row.names(x[which.max(x$Income),]) ## [1] "Nevada"max(x$Income)## [1] 5149