Extract Single Column as DataFrame in R
In this article, we will see how a single column can be extracted as Dataframe using R programming language.
Database in use:
Method 1: Extract Data from Data Frame using column name
In this method, a specific column can be extracted from the dataframe using its name using $.
Syntax:
dataframe$column_name
The column returned will be in a form of a vector, but it can be converted to a dataframe explicitly using data.frame() function of R language.
Example:
R
# Create the data frame. gfg.data <- data.frame ( GFG_Id = c (1:7), GFG_Name = c ( "Damon" , "Joe" , "Jenna" , "Ryan" , "Bonnie" , "Stefan" , "William" ), GFG_Sal = c (6200,5152,6110,7290,8485,7654, 2341), Start_date = as.Date ( c ( "2021-01-01" , "2016-06-27" , "2014-1-09" , "2017-02-14" , "2009-03-26" , "2019-06-27" , "2020-09-27" )), stringsAsFactors = FALSE ) # Print the data frame. print (gfg.data) # Extract Specific columns # Convert it explicitly to a dataframe specific_col <- data.frame (gfg.data$ GFG_Name, gfg.data$ GFG_Sal) print (specific_col) |
Output
Method 2: Extract Data from Data Frame using square bracket
The column can also be extracted as a dataframe using square brackets with an index of the column in the dataframe given to it as input.
Example:
R
# Create the data frame. gfg.data <- data.frame ( GFG_Id = c (1:7), GFG_Name = c ( "Damon" , "Joe" , "Jenna" , "Ryan" , "Bonnie" , "Stefan" , "William" ), GFG_Sal = c (6200,5152,6110,7290,8485,7654, 2341), Start_date = as.Date ( c ( "2021-01-01" , "2016-06-27" , "2014-1-09" , "2017-02-14" , "2009-03-26" , "2019-06-27" , "2020-09-27" )), stringsAsFactors = FALSE ) # Print the data frame. print (gfg.data) # Extract Single Variable as Data Frame # Using Square Brackets gfg.data[1] gfg.data[2] |
Output
Method 3: Extract Single Column as DataFrame
We extract columns as a vector from an R data frame, but sometimes we might need a column as a data frame. Thus, we can use as.data.frame to extract columns that we want to extract as a data frame with single square brackets. The purpose behind this could be merging the column with another data frame.
Example:
R
# Create the data frame. gfg.data <- data.frame ( GFG_Id = c (1:7), GFG_Name = c ( "Damon" , "Joe" , "Jenna" , "Ryan" , "Bonnie" , "Stefan" , "William" ), GFG_Sal = c (6200,5152,6110,7290,8485,7654, 2341), Start_date = as.Date ( c ( "2021-01-01" , "2016-06-27" , "2014-1-09" , "2017-02-14" , "2009-03-26" , "2019-06-27" , "2020-09-27" )), stringsAsFactors = FALSE ) # Print the data frame. print (gfg.data) # Extracting 2nd column as a separate # dataframe x2 <- as.data.frame (gfg.data[,2]) print (x2) # Extracting 3rd column as a separate # dataframe x3 <- as.data.frame (gfg.data[,3]) print (x3) |
Output
Method 4: Extract Single Variable as Data Frame Using drop Argument
If the drop argument is provided with the value FALSE the columns are not converted to vector objects.
Example:
R
# Create the data frame. gfg.data <- data.frame ( GFG_Id = c (1:7), GFG_Name = c ( "Damon" , "Joe" , "Jenna" , "Ryan" , "Bonnie" , "Stefan" , "William" ), GFG_Sal = c (6200,5152,6110,7290,8485,7654, 2341), Start_date = as.Date ( c ( "2021-01-01" , "2016-06-27" , "2014-1-09" , "2017-02-14" , "2009-03-26" , "2019-06-27" , "2020-09-27" )), stringsAsFactors = FALSE ) # Print the data frame. print (gfg.data) # Extract Single Variable as Data Frame # Using drop Argument gfg.data[ , 1, drop = FALSE ] gfg.data[ , 2, drop = FALSE ] |
Output