Get the last occurrence of a pattern before another pattern

Question

In a file like this one :

...
Pattern2:TheWrongBar
foo 
Pattern2:TheRightBar
foo 
First Pattern
foo
...

I need to find the last occurrence of Pattern2 that is before First Pattern which would be in this case Pattern2:TheRightBar

My first idea is to get all the remaining file before First pattern with :

sed -e '/First Pattern/,$d' myfile | tac | grep -m1 "Pattern I need to get"

Isn't there a way to optimise this code?

Answer 1

With awk:

awk '/Pattern2/ {line=$0; next}; /First Pattern/ {print line; exit}' file.txt

• /Pattern2/ {line=$0; next}: If the pattern Pattern2 is matched, saving the line in variable line, and going to the next line

• /First Pattern/ {print line; exit}: if First Pattern is found, printing the variable line, and exiting

Example:

% cat file.txt                                                                 
...
Pattern2:TheWrongBar
foo 
Pattern2:TheRightBar
foo 
First Pattern
foo
...

% awk '/Pattern2/ {line=$0; next}; /First Pattern/ {print line; exit}' file.txt
Pattern2:TheRightBar

Answer 2

You could run

sed '/PATTERN2/h;/PATTERN1/!d;x;/PATTERN2/!d;q' infile

How it works:

sed '/PATTERN2/h         # if line matches PATTERN2 save it to hold buffer 
/PATTERN1/!d             # if it doesn't match PATTERN1 delete it
x                        # exchange buffers
/PATTERN2/!d             # if current pattern space doesn't match delete it
q' infile                # quit (auto-printing the current pattern space)

This would only exit if there's at least one line matching PATTERN2 before some line matching PATTERN1 so with an input like

1
2
PATTERN1
PATTERN2--1st
3
PATTERN2--2nd
PATTERN1
...

it will print

PATTERN2--2nd

If you wanted instead to exit on first match of PATTERN1 regardless, you would run

sed -n '/PATTERN2/h;/PATTERN1/!d;x;/PATTERN2/p;q' infile

which prints nothing with the above input (this one does exactly what your solution does).

Answer 3

Turns out the most efficient way in my case was:

grep -m1 "First Pattern" my_file -B 10000000 | tac | grep -m1 "Pattern2"

Obviously the -B option cannot be used in some examples but grep is so much faster than awk or sed that I went with that solution. If the value for option -B gets higher, the search is way less efficient.

Answer 4

Finds the line count of "First Pattern", then uses head to display lines above it, pipes through tac and greps it.

head --lines=+"$(grep -nm1 "First Pattern" file | cut -d\: -f1)" file | tac | grep -m1 "Pattern2" 

Eg.

head --lines=+6 file | tac | grep -m1 "Pattern2" 

---

This is more reliable then using -m 1000000 in grep. Since speed is important to OP, I checked the run time and it also appears to be faster then all other current answers (on my system)

wc -l file
25910209 file

time awk '/Pattern2/ {line=$0; next}; /First Pattern/ {print line; exit}' file
Pattern2:TheRightBar

real  0m2.881s
user  0m2.844s
sys 0m0.036s

time sed '/Pattern2/h;/First Pattern/!d;x;/Pattern2/!d;q' file
Pattern2:TheRightBar

real  0m5.218s
user  0m5.192s
sys 0m0.024s

time (grep -m1 "First Pattern" file -B 10000000 | tac | grep -m1 "Pattern2")

real  0m0.624s
user  0m0.552s
sys 0m0.124s

time (head --lines=+"$(grep -nm1 "First Pattern" file | cut -d\: -f1)" file | tac | grep -m1 "Pattern2")
Pattern2:TheRightBar

real  0m0.586s
user  0m0.528s
sys 0m0.160s