A data frame is a table or a two-dimensional array-like structure in which each column contains values of one variable and each row contains one set of values from each column.
Following are the characteristics of a 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 above code, it produces the following result −
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 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 above code, it produces the following result −
'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" ...
The statistical summary and nature of the data can be obtained by applying summary() 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 above code, it produces the following result −
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
Extract specific column from a data frame using 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 ) # Extract Specific columns. result <- data.frame(emp.data$emp_name,emp.data$salary) print(result)
When we execute the above code, it produces the following result −
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 all 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 above code, it produces the following result −
emp_id emp_name salary start_date 1 1 Rick 623.3 2012-01-01 2 2 Dan 515.2 2013-09-23
Extract 3rd and 5th row with 2nd and 4th column
# 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 above code, it produces the following result −
emp_name start_date 3 Michelle 2014-11-15 5 Gary 2015-03-27
A data frame can be expanded by adding columns and rows.
Just add the column vector using a 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 above code, it produces the following result −
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 add more rows permanently to an existing data frame, we need to bring in the new rows in the same structure as the existing data frame and use the rbind() function.
In the example below we create a data frame with new rows and merge it with the 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 above code, it produces the following result −
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