Match blank lines before a word awk

Question

I have a long file (showing only a piece):

145 27262253 2093226 747883433 76303046 2.74331
146 27992017 2188217 747883433 76303046 2.8678
147 30385435 2433407 747883433 76303046 3.18913
148 31218703 2514902 747883433 76303046 3.29594
149 33852828 2660530 747883433 76303046 3.48679
150 36161756 2836045 747883433 76303046 3.71682

Alignments      747883433       76303046
Bases   111613795461    11392665612

1 40000373 2754292 838333186 82982133 3.31914
2 35955786 2451917 838333186 82982133 2.95475
3 33056935 2241392 838333186 82982133 2.70105
4 32241895 2172229 838333186 82982133 2.61771
145 29490370 2184347 838333186 82982133 2.63231
146 30252912 2282821 838333186 82982133 2.75098
147 32862262 2544600 838333186 82982133 3.06644
148 33769718 2631164 838333186 82982133 3.17076
149 36673113 2787718 838333186 82982133 3.35942
150 39222287 2975755 838333186 82982133 3.58602

Alignments      838333186       82982133
Bases   125129342261    12391027833

1 35736929 2509527 741319423 80995147 3.09837
2 32185143 2238927 741319423 80995147 2.76427
3 29595482 2043259 741319423 80995147 2.52269
4 28861157 1978254 741319423 80995147 2.44244

I want to match the blank line before Alignments word and the word itself. Expecting:

Alignments      747883433       76303046   

Alignments      838333186       82982133

Is it possible? I have many others blank lines and Alignments words. My try: | awk '{if($1 ~ /^[[:space:]]*Alignments/) {print $0}}'. However, I get:

Alignments      747883433       76303046
Alignments      838333186       82982133

You are after a multiline grep/awk. There are several questions that ask the same question. — Sparhawk, Aug 30 '19 at 05:39

score 2 · Answer 1 · answered Aug 30 '19 at 05:33

2

Why don't we use grep? :

grep -A1 "^$" file | grep -B1 'Alignments' | grep -v -- "^--$"

answered Aug 30 '19 at 05:33

Siva

9,077

score 2 · Accepted Answer · answered Aug 30 '19 at 06:23

2

$ awk '/^$|^Alignments/' input.txt | uniq

Alignments      747883433       76303046

Alignments      838333186       82982133

the uniq makes sure there will be no more than one blank line before, after, or between any Alignments lines.

grep could be used instead. or sed -n. or perl -n. e.g.

$ grep -E '^$|Alignments' input.txt | uniq

answered Aug 30 '19 at 06:23

cas

78,579

How can achieve the same behavior of uniq without using the pipe? I have created a variable in awk so I wouldn't like to lose it. I was thinking something like: awk '/(^$){1}|^Alignments/' input.txt but it matches more than just one blank lines before. – giannkas Aug 30 '19 at 12:27
1

generic answer is "do more stuff in the awk script before piping to uniq"... what exactly are you trying to do? what variable have you created and what's it for? why do you think you would lose it? – cas Aug 30 '19 at 12:47
Thank you @cas for your help. Finally, I realized that I needed to delete the blank line after Bases and 1 because is not part of the original file that I was editing. I achieved it by using this: | awk 'm=($1 == "==>") {save_location=$2} !m&&$1=="1",$1=="Bases" {print > save_location}' which is a piece of a longer instruction. However, your answer responds what I was asking and it's neat! – giannkas Aug 31 '19 at 14:44

score 2 · Answer 3 · answered Aug 30 '19 at 06:31

2

Using GNU awk:

awk -v RS='\nAlignments[ 0-9]*' '{print RT}' file

The record separtor RS is set to the expected match and is printed for every record using RT (record terminator).

answered Aug 30 '19 at 06:31

oliv

2,636

score 1 · Answer 4 · answered Aug 30 '19 at 05:22

1

You can use the below Awk for your problem. Match a empty line and see if the line starting with Alignments immediately starts after that

awk '!NF { line = NR; next } (NR = line + 1 ) && /^Alignments/{ printf "\n%s\n",$0; }' file

answered Aug 30 '19 at 05:22

Inian

12,807

2

we are hardcoring the newline in output... it would print the same if we don't have an empty line... – Siva Aug 30 '19 at 05:29
2

@msp9011 : No only if the previous line is empty! – Inian Aug 30 '19 at 05:46
1

just check by removing empty line in input file... – Siva Aug 30 '19 at 05:52

score 1 · Answer 5 · answered Aug 30 '19 at 06:09

1

Sed excels in such tasks. First we stick the next line to the current provided the current is empty. Then interrogate and print upon meeting the criterion set.

$ sed -ne '
    /./!N
    /^\nAlignments/p
' file.txt

answered Aug 30 '19 at 06:09

Rakesh Sharma

836

score 0 · Answer 6 · answered Sep 01 '19 at 16:29

0

Tried with Below command and it worked fine

awk '{a[++i]=$0}/Alignments/{for(x=NR-1;x<=NR;x++)print a[x]}' filename| sed -n '/^$/,+1p'

output

Alignments      747883433       76303046

Alignments      838333186       82982133

answered Sep 01 '19 at 16:29

Praveen Kumar BS

5,211

Match blank lines before a word awk

6 Answers6