CODEWITHSUNDEEP
rdataframesubset
Stewart Wiseman
Stewart Wiseman

Reputation: 705

Using grep to help subset a data frame

I am having trouble subsetting my data. I want the data subsetted on column x, where the first 3 characters begin G45.

My data frame:

x <- c("G448", "G459", "G479", "G406")  
y <- c(1:4)
My.Data <- data.frame (x,y)

I have tried:

subset (My.Data, x=="G45*")

But I am unsure how to use wildcards. I have also tried grep() to find the indicies:

grep  ("G45*", My.Data$x)

but it returns all 4 rows, rather than just those beginning G45, probably also as I am unsure how to use wildcards.

Upvotes: 37

Views: 128620

Answers (2)

Ayan
Ayan

Reputation: 157

You may also use the stringr package

library(dplyr)
library(stringr)
My.Data %>% filter(str_detect(x, '^G45'))

You may not use '^' (starts with) in this case, to obtain the results you need

Upvotes: 1

A5C1D2H2I1M1N2O1R2T1
A5C1D2H2I1M1N2O1R2T1

Reputation: 193517

It's pretty straightforward using [ to extract:

grep will give you the position in which it matched your search pattern (unless you use value = TRUE). 

grep("^G45", My.Data$x)
# [1] 2

Since you're searching within the values of a single column, that actually corresponds to the row index. So, use that with [ (where you would use My.Data[rows, cols] to get specific rows and columns).

My.Data[grep("^G45", My.Data$x), ]
#      x y
# 2 G459 2

The help-page for subset shows how you can use grep and grepl with subset if you prefer using this function over [. Here's an example.

subset(My.Data, grepl("^G45", My.Data$x))
#      x y
# 2 G459 2

As of R 3.3, there's now also the startsWith function, which you can again use with subset (or with any of the other approaches above). According to the help page for the function, it's considerably faster than using substring or grepl.

subset(My.Data, startsWith(as.character(x), "G45"))
#      x y
# 2 G459 2

Upvotes: 69

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