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How to check if a character column only contains alphabets in R data frame?
To check if a character column only contains alphabets in R data frame, we can follow the below steps −
- First of all, create a data frame with a character column.
- Then, use grepl function to check if all the values in the column contains only alphabets.
Example 1
Let's create a data frame as shown below −
x<-sample(c("India","UK","USA","Japan","China2"),20,replace=TRUE)
df1<-data.frame(x)
df1On executing, the above script generates the below output(this output will vary on your system due to randomization) −
x 1 India 2 USA 3 China2 4 USA 5 Japan 6 UK 7 China2 8 Japan 9 USA 10 Japan 11 Japan 12 Japan 13 Japan 14 India 15 Japan 16 UK 17 India 18 UK 19 India 20 USA
Check whether data frame column contains only alphabets
Using grepl function to check whether column x contains values only having alphabets −
x<-sample(c("India","UK","USA","Japan","China2"),20,replace=TRUE)
df1<-data.frame(x)
grepl("^[A-Za-z]+$",df1$x)Output
[1] TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [13] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
Example2
Let’s create a data frame as shown below −
y<-sample(c("1","23","14","11","1F","3"),20,replace=TRUE)
df2<-data.frame(y)
df2On executing, the above script generates the below output(this output will vary on your system due to randomization) −
y 1 11 2 1F 3 1 4 11 5 23 6 1F 7 11 8 23 9 14 10 14 11 1F 12 14 13 1F 14 11 15 1 16 23 17 1F 18 1 19 3 20 23
Check whether data frame column contains only alphabets
Using grepl function to check whether column y contains values only having alphabets −
y<-sample(c("1","23","14","11","1F","3"),20,replace=TRUE)
df2<-data.frame(y)
grepl("^[A-Za-z]+$",df2$y)Output
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [13] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
Updated on: 13-Aug-2021
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