How to use IFS in bash for an array of strings?

Question

Given an array of strings, I want IFS to parse each string by an underscore, join the first three separated values together using the same delimeter and output them to an empty list. For instance, if I have the array input=("L2Control_S3_L001_R1_001.fastq", "L2Control_S3_L001_R2_001.fastq") IFS would take the first string and separate it out by the underscore, so the output would be:

L2Control
S3
L001
R1
001.fastq

Afterward, it would take the first three separated values and join them together with an underscore: "L2Control_S3_L001". Lastly, this value would be appended onto a new array output=("L2Control_S3_L001") this process would continue until all values in the array are completed. I have tried the below implementation, but it seems to run infinitely.

#!/bin/bash

str=("L2Control_S3_L001_R1_001.fastq", "L2Control_S3_L001_R1_001.fastq")

IFS='_'
final=()

for (( c = 0; c = 2; c++ )); do
  read -ra  SEPA <<< "${str[$c]}"
  final+=("${SEPA[0]}_${SEPA[1]}_${SEPA[2]}")
done

Can someone help me with this, please?

Your for loop is not correct. The middle condition should be something like c < ${#str[@]} — siride
– siride, Commented Apr 2, 2022 at 18:34
(1) Don't use a , to separate array elements -- that comma is ending up as part of your data. (2) Note that you can set IFS local to a single read command. IFS=_ read -r -a arrayname won't change IFS for any command other than that read. — Charles Duffy
– Charles Duffy, Commented Apr 2, 2022 at 18:34
BTW, I don't see a reason to use read -a at all when you expect exactly three pieces. IFS=_ read -r first second third will put the first piece in "$first", the second in "$second", everything after the second in "$third" and there you are. — Charles Duffy
– Charles Duffy, Commented Apr 2, 2022 at 18:35
And why use a counter c at all? for a_str in "${str[@]}"; do will assign each piece to a_str in turn. — Charles Duffy
– Charles Duffy, Commented Apr 2, 2022 at 18:36
...but anyhow, none of the problems I see this code having have anything to do with IFS. If I'm missing something that is IFS-specific, could you explain more clearly? — Charles Duffy
– Charles Duffy, Commented Apr 2, 2022 at 18:36

glenn jackman · Accepted Answer · 2022-04-02 18:56:28Z

1

The ${array[*]} expansion joins elements using the first character of IFS. You can combine this with the ${var:offset:length} expansion.

output=()
for str in "${input[@]}"; do
    read -ra fields <<< "$str"
    output+=("${fields[*]:0:3}")
done

I find declare -p varname ... handy to inspect the contents of variables.

This can also be done with bash parameter expansion:

str="L2Control_S3_L001_R1_001.fastq"
IFS=_
suffix=${str#*"$IFS"*"$IFS"*"$IFS"}
first3=${str%"$IFS$suffix"}
declare -p str suffix first3

declare -- str="L2Control_S3_L001_R1_001.fastq"
declare -- suffix="R1_001.fastq"
declare -- first3="L2Control_S3_L001"

Can also do that in one line, but it's hairy:

first3="${str%"$IFS${str#*"$IFS"*"$IFS"*"$IFS"}"}"

edited Apr 2, 2022 at 18:56

answered Apr 2, 2022 at 18:49

glenn jackman

249k42 gold badges233 silver badges362 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Logios James Over a year ago

I tried to do your following example using: input=("L2Control_S3_L001_R1_001.fastq" "L2Control_S3_L001_R2_001.fastq") output=() for str in "${input[@]}"; do read -ra feilds <<< "$str" output+=("${feilds[*]:0:2}") done for str_a in "${output[@]}"; do echo "$str_a" done However it just printed out the samething that was in the input: "L2Control_S3_L001_R1_001.fastq" "L2Control_S3_L001_R2_001.fastq"

glenn jackman Over a year ago

When testing that, did you set IFS=_

markp-fuso · Accepted Answer · 2022-04-02 20:14:48Z

1

Setup (notice no comma needed to separate array entries):

str=("L2Control_S3_L001_R1_001.fastq" "L2Control_S3_L001_R1_001.fastq")

One idea using a while/read loop to parse the input strings into 4 parts based on a delimiter (IFS=_):

final=()

while IFS=_ read -r f1 f2 f3 ignore
do
    final+=("${f1}_${f2}_${f3}")
done < <(printf "%s\n" "${str[@]}")

typeset -p final

Where the variable ignore will be assigned fields #4-#n.

This generates:

declare -a final=([0]="L2Control_S3_L001" [1]="L2Control_S3_L001")

edited Apr 2, 2022 at 20:14

answered Apr 2, 2022 at 20:03

markp-fuso

38.5k5 gold badges24 silver badges48 bronze badges

Comments

KamilCuk · Accepted Answer · 2022-04-02 20:12:14Z

0

That's a lengthy description of taking the first 3 elements separated by _.

printf "%s\n" "${str[@]}" | cut -d_ -f-3

answered Apr 2, 2022 at 20:12

KamilCuk

146k8 gold badges84 silver badges154 bronze badges

Collectives™ on Stack Overflow

How to use IFS in bash for an array of strings?

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related