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It appears that while grep has an invert argument, grepl does not.
I would like to subset for using 2 filters
data$ID[grepl("xyx", data$ID) & data$age>60]
How can I subset for age>60 and ID not containing "xyx"? What I did is
data$ID[abs(grepl("xyx", data.frame$ID)-1) & data$age>60]
which apparently works, but looks awful and unintuitive. Is there a nicer solution/argument?
712
Both functions allow you to see whether a certain pattern exists in a character string, but they return different results: grepl() returns TRUE when a pattern exists in a character string. grep() returns a vector of indices of the character strings that contain the pattern.
The grep command can search for a string in groups of files. When it finds a pattern that matches in more than one file, it prints the name of the file, followed by a colon, then the line matching the pattern.
The grep() function in R is used to check for matches of characters or sequences of characters in a given string.
The grepl in R is a built-in function that searches for matches of a string or string vector. The grepl() method takes a pattern and returns TRUE if a string contains the pattern, otherwise FALSE.
grepl returns a logical vector. You can use the ! operator if you want the opposite result.
data$ID[!grepl("xyx", data$ID) & data$age>60]
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