parsing out in bash

Question

my file looks like this

//
[297]((((21:0.125204,20:0.125204):0.00994299,(28:0.0790047,(7:0.0146105,5:0.0146105):0.0643943):0.0561423):0
segsites:
positions: 1 2 4 6 9 10 45 67 78 89 
01010101010101010101010101011111010101

I would need to parse it into three different files

//
[297]((((21:0.125204,20:0.125204):0.00994299,(28:0.0790047,(7:0.0146105,5:0.0146105):0.0643943):0.0561423):0

then

 segsites:
    positions: 1 2 4 6 9 10 45 67 78 89 

an

 01010101010101010101010101011111010101

the problem is that there are not a fixed number of lines starting with the brackets and so on. the final line could start with either a 0 or an 1 and that is a problem as I dont know how to tell grep to search for either 0 or 1. I would need to grep for the []. then the segsites and the line below which includes the pos and then the numeric line with 01

Answer 1

Well, in awk:

awk '/^[01]+$/ {print > "third-file"; next} 
 /(segsites|positions)/ {print > "second-file"; next} 
 {print > "first-file"}' input-file

Considering three cases:

• A line containing only 0s and 1s (/^[01]+$/)
• A line containing either segsites or positions: /(segsites|positions)/
• All other lines

In each case, we print to different file and proceed to the next line. Change "third-file", "second-file", "first-file" to appropriate names (keep the quotes).

Answer 2

Using grep with PCRE [-P] and Dotall modifier [(?s)]:

1st portion:

$ grep -Poz "(?s)^.*?(?=\nsegsites:)" file.txt 
//
[297]((((21:0.125204,20:0.125204):0.00994299,(28:0.0790047,(7:0.0146105,5:0.0146105):0.0643943):0.0561423):0

2nd portion:

$ grep -Poz "(?s)segsites.*?(?=\n[10]+$)" file.txt 
segsites:
positions: 1 2 4 6 9 10 45 67 78 89 

In case of third portion only grep will do (if you have just one line):

$ grep -E "^[01]+$" file.txt
01010101010101010101010101011111010101

If you are not sure (single/multiple lines):

$ grep -Poz "(?<=\n)(?:[10]|\n)*(?=\n$)" file.txt 
010101010101010101010101010111110101011
01010101010101010101010101011111010101010

$ grep -Poz "(?<=\n)(?:[10]|\n)*(?=\n$)" file.txt 
010101010101010101010101010111110101011

Explanations:

• grep -P represents PCRE (Perl Compatible Regular Expressions)

• grep -o will select only the matched portion

• grep -z will cause lines to be separated by ASCII NUL instead of new lines

• (?s) is called Dotall modifier, using this any character (including line breaks) can be matched by .

• (?=) is Positive lookahead, represents what follows our match

• (?<) is Positive lookbehind, represents what precedes our match