Print Previous Line If Condition Is Met

Print previous line if condition is met

This can be a way:

$ awk '$1=="BB" && $2>1 {print f} {f=$1}' file
AAAAAAAAAAAAA

Explanation

• $1=="BB" && $2>1 {print f} if the 1st field is exactly BB and 2nd field is bigger than 1, then print f, a stored value.
• {f=$1} store the current line in f, so that it is accessible when reading the next line.

Keep current and previous line only if current line fulfills a given condition

Yet another awk solution:

awk '/^>/ { header = $0; next } length > 50 && length <= 800 { print header ORS $0 }'

Print previous line along with matched line

You are on right path, You forgot to print the current line simply do.

awk '
/pattern/{
  if(prev)  { print prev ORS $0 }
  if(FNR==1){ print             }
  next
}
{
  prev=$0
}
' Input_file

I have also taken care of 1st first line here, in case pattern comes in very first line then it will simply print that line, since there is no previous line.

Explanation: Adding detailed explanation for above.

awk '                                 ##Starting awk program from here.
/pattern/{                            ##Checking for specific pattern here.
  if(prev)  { print prev ORS $0 }     ##Checking if prev is NOT NULL then print prev newline current line.
  if(FNR==1){ print             }     ##Checking if its first line then simply print the line.
  next                                ##next will skip all further statements from here.
}
{
  prev=$0                             ##Setting prev to current line here.
}
' Input_file                          ##Mentioning Input_file name here.

print a previous line based on a condition in python

Use tee to run a pair of iterators across inf. This only stores five lines in memory at any given time:

from itertools import tee

with open("SFU.txt") as inf:
    # set up iterators
    cfg,res = tee(inf)
    # advance cfg by four lines
    for i in range(4):
        next(cfg)

    for c,r in zip(cfg, res):
        if "configuration" in c:
            print(r)

and, as expected, results in

This is planet Mars

Edit: if you want to edit the -4th line, I suggest

def edited(r):
    # make your changes to r
    return new_r

with open("SFU.txt") as inf, open("edited.txt", "w") as outf:
    # set up iterators
    cfg, res = tee(inf)
    for i in range(4):
        next(cfg)

    # iterate through in tandem
    for c, r in zip(cfg, res):
        if "configuration" in c:
            r = edited(r)
        outf.write(r)

    # reached end - write out remaining queued values
    for r in res:
        outf.write(r)

Python to print previous line if an condition met in current line

When iterating over the lines, can't you just keep the previous line in each iteration?

# Holds the previous line
prev_line = None
for line in open(<something>, 'r'):
    if is_ora_line(line) and prev_line is not None:
        do_something_with_database_line(prev_line)
    # Remember now the current line as the previous line
    prev_line = line

Match a line and return a previous line before the match containing a pattern

This sed command should extract it:

sed -n '
    /^Id: / {                 # If the line starts with "Id: "
        s///                  # Remove the "Id: "
        h                     # Store what is left in the hold space
    }
    /^  email: '"$email"'/ {  # If the line starts with "  email: " plus the email
        x                     # Swap pattern and hold space
        p                     # Print pattern space
        q                     # Stop processing
    }
' infile

where $email is the shell variable containing the escaped version of test_1@somedomain.com:

raw='test_1@somedomain.com'
email=$(sed 's|[]/.*^$\[]|\\&|g' <<< "$raw")

This escapes the sed special characters .*/^$[]\.

Or, more compact:

sed -n '/^Id: /{s///;h};/^  email: '"$email"'/{x;p;q}' infile

macOS sed requires an extra ; before each closing }.

And yes, it's probably easier with awk /p>

move short lines to previous line

Here is one way with awk. It delays printing until it saw the next line.

Setting FS="" works with gawk 4.2.1 and above and mawk 1.3.4 20200120.

Fails with "original" awk version 20121220. (Thanks @Ed Morton for the reminder!). Other versions not tested.

% awk -F '' 'NR==1{ line=$0; next }
             NF<=60{ prev=$0; printf("%s%s",line,prev); line=""; next }
             NF>60{ print line } { line=$0 }
             END{ print line }' file
1 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw2 owenuof wuoef wue fiwuf wiuenf wie
3 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw
4 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw5 owenuof wuoef wue fiwuf wiuenf wie6 owenuof wuoef wue fiwuf wiuenf wie
7 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw
8 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw

With "original" awk using length()

% oawk 'NR==1{ line=$0; next }
        length($0)<=60{ prev=$0; printf("%s%s",line,prev); line=""; next }
        length($0)>60{ print line } { line=$0 }
        END{ print line }' file  
1 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw2 owenuof wuoef wue fiwuf wiuenf wie
3 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw
4 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw5 owenuof wuoef wue fiwuf wiuenf wie6 owenuof wuoef wue fiwuf wiuenf wie
7 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw
8 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw

Data

% cat file
1 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw
2 owenuof wuoef wue fiwuf wiuenf wie
3 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw
4 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw
5 owenuof wuoef wue fiwuf wiuenf wie
6 owenuof wuoef wue fiwuf wiuenf wie
7 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw
8 ewf nwje fwkjef wkejf wkej fwkejf wkejfwkjef woief nowienfw

calculate current line column less previous line column using awk

You can try this awk

$ awk 'NR==1{ print $0 } NR>1{ print $0,$2 - pre } { pre=$2 }' file
a 9
b 2 -7
c 5 3
d 3 -2
e 7 4

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