filter out rows from data frame based on multiple string matching conditions in R

Question

I am reading external data using read.table() in R like:

student_record <- read.table("Address of data",fill = TRUE,col.names=c("student_id","name"))

Student id is a 20 character long string of the format say STU01000001010001001 and I want to keep rows where student id satisfy following conditions:

   ( 0 – 2 = STU) AND
(5 – 9  != 11111) AND
(10 – 11 != (00 or 10)) AND
(12 – 17  != 111111) AND
(18-19 = 04)

Here 0,2 and so on represent index of character in student id. How can I filter out records using such filter conditions?

I executed this after read.table() to filter:

stu_record <- student_record[grepl("^STU.{2}(?!11111).(?!(00|10)).(?!111111).04", student_record[,1], perl=T),]

but the output doesn't seems to come correct because everything gets filtered out and I get an empty frame

When I executed this:

stu_record <- student_record[grepl("^STU.{2}(?!11111).(?!(00|10)).(?!111111)04", student_record[,1], perl=T),]  

then I see records but they don't seems to be correct as I can see records like STU13120600500000002 which should not come as last two index should be 04

UPDATE: few rows that I see after executing above command are(The ids dont get filtered correctly as las two digits should be 04 but I see 01):

       student_id         Name    
  "STU01115000000000001"  "A"   
  "STU01115000000000001"  "B"   
  "STU01115000000000001"  "C"   
  "STU01115000000000001"  "D"   
  "STU01115000000000001"  "E"   
  "STU01115000000000001"  "F"   
  "STU01115000000000001"  "G"   
  "STU01115000000000001"  "H"   
  "STU01115000000000001"  "I"

while some of the ids which should have been there but got filtered out are:

      "STU01155000000000004"  "F"   
      "STU01135000000000004"  "G"   
      "STU01145000000000004"  "H"   
      "STU01125000000000004"  "I"

NOTE: There are certain index in string for which there is no condition like for index 3 and 4 there is no filtering condition so they can be anything.

Answer 1

This should work. I made up a test string.

string <- c("STU0100010", "STU0100010", "STU0300010", "STU0100090")

grepl("^STU(?!01).*(?!01|90)$", string, perl = T)
[1] FALSE FALSE  TRUE FALSE

The grep function looks for strings in the vector that start with STU, but are not followed by 01 (using a negative lookahead assertion) and there is not an 01 or 90 at the end (another negative lookahead and the end of string anchor).

Answer 2

Using df from @digEmAll

df[grepl("^STU.(?!01).{2}(?!(01|90))", df[,1], perl=T),]
#    student_id name
#1 STUx1000xx    A
#3 STU01008bb    C

Answer 3

You can use substr function :

# example data
df <- 
data.frame(
student_id=c('STUx1000xx','STU00110xx','STU01008bb','STU01090aa'),
name=c('A','B','C','D'),stringsAsFactors=F)

# > df
#   student_id name
# 1 STUx1000xx    A
# 2 STU00110xx    B
# 3 STU01008bb    C
# 4 STU01090aa    D

# create filter using substr function
condition <- substr(df$student_id,1,3) == 'STU' &
             substr(df$student_id,5,6) != '01' &
             substr(df$student_id,7,8) != '01' &
             substr(df$student_id,7,8) != '90' 

filtered <- df[condition,]

# > filtered
#   student_id name
# 1 STUx1000xx    A
# 3 STU01008bb    C

EDIT :

the new condition should be :

condition <- substr(df$student_id,1,3) == 'STU' &
             substr(df$student_id,6,10) != '11111' &
             substr(df$student_id,11,12) != '00' &
             substr(df$student_id,11,12) != '10' &
             substr(df$student_id,13,18) != '111111' &
             substr(df$student_id,19,20) == '04'