grep the last occurrence of a pattern before another pattern

Question

I have a huge file contains two types of patterns say pattern1 and pattern2, pattern1 may appear many times before pattern2 appears. I want to grep the last occurrence of each pattern1 before each pattern2.

Input file:

some text
pattern1=1
some lines
pattern1=2
some lines
pattern1=3
some lines
pattern2
some lines
pattern1=4
some lines
pattern1=5
some lines
pattern1=6
some lines
pattern1=7
some lines
pattern2

Desired output:

pattern1=3
pattern1=7

I tried with grep when I know the numbers of lines between pattern2 and the previous pattern1:

grep -B400 "pattern2" | grep "pattern1"

but I need a unique command that can be run over any file regardless of the number of lines between the two patterns.

Answer 1

$ awk '/pattern1/{x=$0} /pattern2/{print x}' input
pattern1=3
pattern1=7

Saves the pattern1 matches (the whole line) to the variable x and prints that when pattern2 happens. Will print a blank line if there's a pattern2 before any pattern1, which would take more logic to detect if that's not desirable. Will drop any trailing pattern1 that are not followed by a pattern2 before the end of the input.

Answer 2

@thrig's answer is good, but I made some modifications to handle some extra test cases. The following script:

• Will not print an empty line if pattern2 appears before the first appearance of pattern1.
• Will not print duplicate lines if pattern2 appears multiple times after pattern1.

With the modified input file:

pattern2
some text
pattern1=1
some lines
pattern1=2
some lines
pattern1=3
some lines
pattern2
pattern2
some lines
pattern1=4
some lines
pattern1=5
pattern2
some lines
pattern1=6
some lines
pattern1=7
some lines
pattern2

The following script seems to do what you describe in the text:

$ awk '/pattern1/{x=$0} length(x) && /pattern2/{print x;x=""}' file
pattern1=3
pattern1=5
pattern1=7

Answer 3

Three grep calls:

1. Extract only lines that match ^pattern1= or ^pattern2$ from the original input file

   grep -e '^pattern1=' -e '^pattern2$' file
   

2. Get the lines that match ^pattern2$, and the lines immediately before these (using the non-standard -B option):

   grep -B1 '^pattern2$'
   

3. Get all lines matching ^pattern1= from these:

   grep '^pattern1='
   

All together:

grep -e '^pattern1=' -e '^pattern2$' file |
grep -B1 '^pattern2$' |
grep '^pattern1='

This handles the same edge cases as user000001's answer, namely that it does not output duplicate lines if there are many pattern2 lins with no pattern1 line between them, and it would not produce empty lines for pattern2 lines at the start of the file.

---

Using sed:

sed -e '/^pattern1=/ { h; d; }' \
    -e '/^pattern2$/ x' \
    -e '/^pattern1=/ !d' file

1. If the current line is a pattern1 line, it saves it into the hold space and discards it.

2. If the current line a pattern2 line, it swaps in the hold space.

3. If the current line now is not a pattern1 line, it is discarded.

4. (implicitly) print the current line. The current line, by the preceding commands, must be a pattern1 line swapped in from the hold space due to finding a pattern2 line. The hold space would, therefore, by necessity, hold a pattern2 line, ensuring that the pattern1 line would not be outputted multiple times.

Answer 4

❯ printf 'g/pattern2/?pattern1?p\n' | ed -s in.txt
pattern1=3
pattern1=7

Explanation:

g/pattern2/ → For every match of the pattern2, do

?pattern1? → Search backwards for pattern1

p → print

Limitations:

1. Search backwards wraps back to end of file and selects the last match of pattern1
2. Prints the same line multiple times if there are multiple pattern2

E.g.

pattern1=1
pattern2
pattern2

This will print pattern1 twice.

Answer 5

egrep "^pattern1|^pattern2" <file> | grep -B 1 "^pattern2" | grep "^pattern1"

The first egrep will get only the lines that contain either pattern (stripping all other unknown lines from the output). The second grep will get the pattern2 and whatever line is before it. This will be used to remove lines where there is no pattern1 before the pattern2. The third grep will just return the remaining pattern1 lines.

Answer 6

 awk '{a[++i]=$0}/pattern2/{for(x=NR-2;x<=NR;x++)print a[x]}' file1.txt| awk '/pattern1/'

output

pattern1=3
pattern1=7

As per Input pattern1 is separated by one line so used above awk command to extract the required output