May 12, 2021 R language tutorial
2. Gets the structure of the data frame
3. A summary of the data in the data frame
A data frame is a table or two-dimensional pattern structure in which each column contains the value of a variable, and each row contains a set of values from each column.
The following are the characteristics of the data frame.
# Create the data frame.
emp.data <- data.frame(
emp_id = c (1:5),
emp_name = c("Rick","Dan","Michelle","Ryan","Gary"),
salary = c(623.3,515.2,611.0,729.0,843.25),
start_date = as.Date(c("2012-01-01", "2013-09-23", "2014-11-15", "2014-05-11",
"2015-03-27")),
stringsAsFactors = FALSE
)
# Print the data frame.
print(emp.data)
When we execute the code above, it produces the following results -
emp_id emp_name salary start_date 1 1 Rick 623.30 2012-01-01 2 2 Dan 515.20 2013-09-23 3 3 Michelle 611.00 2014-11-15 4 4 Ryan 729.00 2014-05-11 5 5 Gary 843.25 2015-03-27
The structure of the data frame can be seen by using the str() function.
# Create the data frame.
emp.data <- data.frame(
emp_id = c (1:5),
emp_name = c("Rick","Dan","Michelle","Ryan","Gary"),
salary = c(623.3,515.2,611.0,729.0,843.25),
start_date = as.Date(c("2012-01-01", "2013-09-23", "2014-11-15", "2014-05-11",
"2015-03-27")),
stringsAsFactors = FALSE
)
# Get the structure of the data frame.
str(emp.data)
When we execute the code above, it produces the following results -
'data.frame': 5 obs. of 4 variables: $ emp_id : int 1 2 3 4 5 $ emp_name : chr "Rick" "Dan" "Michelle" "Ryan" ... $ salary : num 623 515 611 729 843 $ start_date: Date, format: "2012-01-01" "2013-09-23" "2014-11-15" "2014-05-11" ...
You can get a statistical summary and nature of your data by applying the summaryy() function.
# Create the data frame.
emp.data <- data.frame(
emp_id = c (1:5),
emp_name = c("Rick","Dan","Michelle","Ryan","Gary"),
salary = c(623.3,515.2,611.0,729.0,843.25),
start_date = as.Date(c("2012-01-01", "2013-09-23", "2014-11-15", "2014-05-11",
"2015-03-27")),
stringsAsFactors = FALSE
)
# Print the summary.
print(summary(emp.data))
When we execute the code above, it produces the following results -
emp_id emp_name salary start_date Min. :1 Length:5 Min. :515.2 Min. :2012-01-01 1st Qu.:2 Class :character 1st Qu.:611.0 1st Qu.:2013-09-23 Median :3 Mode :character Median :623.3 Median :2014-05-11 Mean :3 Mean :664.4 Mean :2014-01-14 3rd Qu.:4 3rd Qu.:729.0 3rd Qu.:2014-11-15 Max. :5 Max. :843.2 Max. :2015-03-27
Use column names to extract specific columns from the data frame.
# Create the data frame.
emp.data <- data.frame(
emp_id = c (1:5),
emp_name = c("Rick","Dan","Michelle","Ryan","Gary"),
salary = c(623.3,515.2,611.0,729.0,843.25),
start_date = as.Date(c("2012-01-01","2013-09-23","2014-11-15","2014-05-11",
"2015-03-27")),
stringsAsFactors = FALSE
)
# Extract Specific columns.
result <- data.frame(emp.data$emp_name,emp.data$salary)
print(result)
When we execute the code above, it produces the following results -
emp.data.emp_name emp.data.salary 1 Rick 623.30 2 Dan 515.20 3 Michelle 611.00 4 Ryan 729.00 5 Gary 843.25
Extract the first two rows, and then extract all the columns
# Create the data frame.
emp.data <- data.frame(
emp_id = c (1:5),
emp_name = c("Rick","Dan","Michelle","Ryan","Gary"),
salary = c(623.3,515.2,611.0,729.0,843.25),
start_date = as.Date(c("2012-01-01", "2013-09-23", "2014-11-15", "2014-05-11",
"2015-03-27")),
stringsAsFactors = FALSE
)
# Extract first two rows.
result <- emp.data[1:2,]
print(result)
When we execute the code above, it produces the following results -
emp_id emp_name salary start_date 1 1 Rick 623.3 2012-01-01 2 2 Dan 515.2 2013-09-23
Use columns 2 and 4 to extract rows 3 and 5
# Create the data frame.
emp.data <- data.frame(
emp_id = c (1:5),
emp_name = c("Rick","Dan","Michelle","Ryan","Gary"),
salary = c(623.3,515.2,611.0,729.0,843.25),
start_date = as.Date(c("2012-01-01", "2013-09-23", "2014-11-15", "2014-05-11",
"2015-03-27")),
stringsAsFactors = FALSE
)
# Extract 3rd and 5th row with 2nd and 4th column.
result <- emp.data[c(3,5),c(2,4)]
print(result)
When we execute the code above, it produces the following results -
emp_name start_date 3 Michelle 2014-11-15 5 Gary 2015-03-27
You can extend data frames by adding columns and rows.
Simply add the column vector with the new column name.
# Create the data frame.
emp.data <- data.frame(
emp_id = c (1:5),
emp_name = c("Rick","Dan","Michelle","Ryan","Gary"),
salary = c(623.3,515.2,611.0,729.0,843.25),
start_date = as.Date(c("2012-01-01", "2013-09-23", "2014-11-15", "2014-05-11",
"2015-03-27")),
stringsAsFactors = FALSE
)
# Add the "dept" coulmn.
emp.data$dept <- c("IT","Operations","IT","HR","Finance")
v <- emp.data
print(v)
When we execute the code above, it produces the following results -
emp_id emp_name salary start_date dept 1 1 Rick 623.30 2012-01-01 IT 2 2 Dan 515.20 2013-09-23 Operations 3 3 Michelle 611.00 2014-11-15 IT 4 4 Ryan 729.00 2014-05-11 HR 5 5 Gary 843.25 2015-03-27 Finance
To permanently add more rows to an existing data frame, we need to introduce new rows with the same structure as the existing
data frame and use the rbind()
function.
In the following example, we create a data frame that contains a new row and merge it with an existing data frame to create the final data frame.
# Create the first data frame.
emp.data <- data.frame(
emp_id = c (1:5),
emp_name = c("Rick","Dan","Michelle","Ryan","Gary"),
salary = c(623.3,515.2,611.0,729.0,843.25),
start_date = as.Date(c("2012-01-01", "2013-09-23", "2014-11-15", "2014-05-11",
"2015-03-27")),
dept = c("IT","Operations","IT","HR","Finance"),
stringsAsFactors = FALSE
)
# Create the second data frame
emp.newdata <- data.frame(
emp_id = c (6:8),
emp_name = c("Rasmi","Pranab","Tusar"),
salary = c(578.0,722.5,632.8),
start_date = as.Date(c("2013-05-21","2013-07-30","2014-06-17")),
dept = c("IT","Operations","Fianance"),
stringsAsFactors = FALSE
)
# Bind the two data frames.
emp.finaldata <- rbind(emp.data,emp.newdata)
print(emp.finaldata)
When we execute the code above, it produces the following results -
emp_id emp_name salary start_date dept 1 1 Rick 623.30 2012-01-01 IT 2 2 Dan 515.20 2013-09-23 Operations 3 3 Michelle 611.00 2014-11-15 IT 4 4 Ryan 729.00 2014-05-11 HR 5 5 Gary 843.25 2015-03-27 Finance 6 6 Rasmi 578.00 2013-05-21 IT 7 7 Pranab 722.50 2013-07-30 Operations 8 8 Tusar 632.80 2014-06-17 Fianance