Exact Match with Grepl R

Exact match with grepl R

I think this is just:

Relationships[Relationships$Results==Names,]

If you end up doing ^Word1$ you're just doing a straight subset.
If you have multiple names, then instead use:

Relationships[Relationships$Results %in% Names,]

Using Grepl to get an exact match, whilst in a for loop

Try this:

list_df<-list()
for (i in 1:length(my_vector))
{
  df<-my_df%>%
  filter(grepl(paste0("\\b",my_vector[i],"\\b"),My_vehicles))
  list_df[[i]]=df
}

How to use grep()/gsub() to find exact match

Use word boundary \b which matches a between a word and non-word character,

string = c("apple", "apples", "applez")
grep("\\bapple\\b", string)
[1] 1

OR

Use anchors. ^ Asserts that we are at the start. $ Asserts that we are at the end.

grep("^apple$", string)
[1] 1

You could store the regex inside a variable and then use it like below.

pat <- "\\bapple\\b"
grep(pat, string)
[1] 1
pat <- "^apple$"
grep(pat, string)
[1] 1

Update:

paste("^",pat,"$", sep="")
[1] "^apple$"
string
[1] "apple"   "apple:s" "applez" 
pat
[1] "apple"
grep(paste("^",pat,"$", sep=""), string)
[1] 1

R grep and exact matches

You want to use word boundaries \b around your word patterns. A word boundary does not consume any characters. It asserts that on one side there is a word character, and on the other side there is not. You may also want to consider using the inline (?i) modifier for case-insensitive matching.

grep('(?i)(?=.*\\bplant\\b)(?=.*\\bcoal\\b)', df$Sources, perl=T, value=T)

Working Demo

Filter column using grepl to keep particular string match

Use ^ to match the beginning of a string, as in

df2 <- df1[,grepl("^H|^TCGA", colnames(df1))]

We can also use dplyr with starts_with():

library(dplyr)

df1 %>%
    select(starts_with(c('H', 'TCGA'))

Pattern match with grepl() function in R

The result is correct.

grepl is looking for the pattern of xx-xx-xx, where x is a digit, and that does appear in the first query. If you want to query starting from the beginning of the string, you can use the ^ symbol.

For example, if you were to run grepl("^[0-9]{2}-[0-9]{2}-[0-9]{2}", "2010-04-09"), you'd get FALSE, but grepl("^[0-9]{4}-[0-9]{2}-[0-9]{2}", "2010-04-09") would return TRUE.

PS: On the opposite end, $ indicates the end of the string.

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