Add a new column with Line number to a pipe delimited file - macos

I am really struggling to add a new column with the line number in the terminal to a pipe delimited file
My current file looks like this:
ID|FirstName|LastName
12|John|Svernson
23|Mark|Wright
11|Chris|Watson
And I want the file to look like the following:
LN|ID|FirstName|LastName
1|12|John|Svernson
2|23|Mark|Wright
3|11|Chris|Watson
I have over 90k lines. I couldn't find a way to do this. Any help or guidance would be appreciated. Thanks

A fun alternative:
{ echo LN; seq $(( $(wc -l < file) - 1 )); } | paste -d'|' - file
although in reality I'd use
awk '{print (NR==1 ? "LN" : NR-1), $0}' OFS="|" file

could you please try following.
awk 'FNR==1{print "LN|"$0;next} {$1=++count "|" $1} 1' Input_file
To write output into Input_file itself append > temp_file && mv temp_file Input_file in above code too.

Use awk variable NR (record number):
awk -F'|' 'NR==1{print "LN" FS $0}NR!=1{print NR-1 FS $0}' file

I would use nl. You may run into issues if your line number goes over 6 digits, but you can add a -w to increase the number width if you need to:
{ read line; echo "LN|$line"; cat | nl -w 10 -ba -s \|; } < input |
sed 's/^ *//';

Related

delete all line after a specific date

I have a lot of *.csv files. I want to delete the content after a specific line. I will delete all lines after 20031231
How do I solve this problem with some lines of a shell script?
Test,20031231,000107,0.74843,0.74813
Test,20031231,000107,0.74838,0.74808
Test,20031231,000108,0.74841,0.74815
Test,20031231,000108,0.74835,0.74809
Test,20031231,000110,0.74842,0.74818
Test,20040101,000100,0.73342,0.744318
quick and dirty but without any other info about constraint
sed '1,/20031231/p;d' YourFile
If you want to use a shell script, the best is to use awk. This will do the trick:
awk 'BEGIN {FS=","} {if ($2 == "20031231") print $0}' input.csv > output.csv
This code will write to a different file only the lines that have 20031231.
ignores empty lines and unmatched data
awk file:
$ cat awk.awk
{
if($2<="20031231" && $0!=""){
print $0
}else{
next
}
}
execution:
$ awk -F',' -f awk.awk input
Test,20031231,000107,0.74843,0.74813
Test,20031231,000107,0.74838,0.74808
Test,20031231,000108,0.74841,0.74815
Test,20031231,000108,0.74835,0.74809
Test,20031231,000110,0.74842,0.74818
one liner:
$ awk -F',' '{if($2<="20031231" && $0!=""){print $0}else{next}}' input
Test,20031231,000107,0.74843,0.74813
Test,20031231,000107,0.74838,0.74808
Test,20031231,000108,0.74841,0.74815
Test,20031231,000108,0.74835,0.74809
Test,20031231,000110,0.74842,0.74818
with Miller (http://johnkerl.org/miller/doc/)
mlr --nidx --fs "," filter '$2>20031231' input
gives you
Test,20040101,000100,0.73342,0.744318
With awk please try:
awk -F, '$2<=20031231' input.csv

awk print something if column is empty

I am trying out one script in which a file [ file.txt ] has so many columns like
abc|pqr|lmn|123
pqr|xzy|321|azy
lee|cha| |325
xyz| |abc|123
I would like to get the column list in bash script using awk command if column is empty it should print blank else print the column value
I have tried the below possibilities but it is not working
cat file.txt | awk -F "|" {'print $2'} | sed -e 's/^$/blank/' // Using awk and sed
cat file.txt | awk -F "|" '!$2 {print "blank"} '
cat file.txt | awk -F "|" '{if ($2 =="" ) print "blank" } '
please let me know how can we do that using awk or any other bash tools.
Thanks
I think what you're looking for is
awk -F '|' '{print match($2, /[^ ]/) ? $2 : "blank"}' file.txt
match(str, regex) returns the position in str of the first match of regex, or 0 if there is no match. So in this case, it will return a non-zero value if there is some non-blank character in field 2. Note that in awk, the index of the first character in a string is 1, not 0.
Here, I'm assuming that you're interested only in a single column.
If you wanted to be able to specify the replacement string from a bash variable, the best solution would be to pass the bash variable into the awk program using the -v switch:
awk -F '|' -v blank="$replacement" \
'{print match($2, /[^ ]/) ? $2 : blank}' file.txt
This mechanism avoids problems with escaping metacharacters.
You can do it using this sed script:
sed -r 's/\| +\|/\|blank\|/g' File
abc|pqr|lmn|123
pqr|xzy|321|azy
lee|cha|blank|325
xyz|blank|abc|123
If you don't want the |:
sed -r 's/\| +\|/\|blank\|/g; s/\|/ /g' File
abc pqr lmn 123
pqr xzy 321 azy
lee cha blank 325
xyz blank abc 123
Else with awk:
awk '{gsub(/\| +\|/,"|blank|")}1' File
abc|pqr|lmn|123
pqr|xzy|321|azy
lee|cha|blank|325
xyz|blank|abc|123
You can use awk like this:
awk 'BEGIN{FS=OFS="|"} {for (i=1; i<=NF; i++) if ($i ~ /^ *$/) $i="blank"} 1' file
abc|pqr|lmn|123
pqr|xzy|321|azy
lee|cha|blank|325
xyz|blank|abc|123

bash awk first 1st column and 3rd column with everything after

I am working on the following bash script:
# contents of dbfake file
1 100% file 1
2 99% file name 2
3 100% file name 3
#!/bin/bash
# cat out data
cat dbfake |
# select lines containing 100%
grep 100% |
# print the first and third columns
awk '{print $1, $3}' |
# echo out id and file name and log
xargs -rI % sh -c '{ echo %; echo "%" >> "fake.log"; }'
exit 0
This script works ok, but how do I print everything in column $3 and then all columns after?
You can use cut instead of awk in this case:
cut -f1,3- -d ' '
awk '{ $2 = ""; print }' # remove col 2
If you don't mind a little whitespace:
awk '{ $2="" }1'
But UUOC and grep:
< dbfake awk '/100%/ { $2="" }1' | ...
If you'd like to trim that whitespace:
< dbfake awk '/100%/ { $2=""; sub(FS "+", FS) }1' | ...
For fun, here's another way using GNU sed:
< dbfake sed -r '/100%/s/^(\S+)\s+\S+(.*)/\1\2/' | ...
All you need is:
awk 'sub(/.*100% /,"")' dbfake | tee "fake.log"
Others responded in various ways, but I want to point that using xargs to multiplex output is rather bad idea.
Instead, why don't you:
awk '$2=="100%" { sub("100%[[:space:]]*",""); print; print >>"fake.log"}' dbfake
That's all. You don't need grep, you don't need multiple pipes, and definitely you don't need to fork shell for every line you're outputting.
You could do awk ...; print}' | tee fake.log, but there is not much point in forking tee, if awk can handle it as well.

Explode to Array

I put together this shell script to do two things:
Change the delimiters in a data file ('::' to ',' in this case)
Select the columns and I want and append them to a new file
It works but I want a better way to do this. I specifically want to find an alternative method for exploding each line into an array. Using command line arguments doesn't seem like the way to go. ANY COMMENTS ARE WELCOME.
# Takes :: separated file as 1st parameters
SOURCE=$1
# create csv target file
TARGET=${SOURCE/dat/csv}
touch $TARGET
echo #userId,itemId > $TARGET
IFS=","
while read LINE
do
# Replaces all matches of :: with a ,
CSV_LINE=${LINE//::/,}
set -- $CSV_LINE
echo "$1,$2" >> $TARGET
done < $SOURCE
Instead of set, you can use an array:
arr=($CSV_LINE)
echo "${arr[0]},${arr[1]}"
The following would print columns 1 and 2 from infile.dat. Replace with
a comma-separated list of the numbered columns you do want.
awk 'BEGIN { IFS='::'; OFS=","; } { print $1, $2 }' infile.dat > infile.csv
Perl probably has a 1 liner to do it.
Awk can probably do it easily too.
My first reaction is a combination of awk and sed:
Sed to convert the delimiters
Awk to process specific columns
cat inputfile | sed -e 's/::/,/g' | awk -F, '{print $1, $2}'
# Or to avoid a UUOC award (and prolong the life of your keyboard by 3 characters
sed -e 's/::/,/g' inputfile | awk -F, '{print $1, $2}'
awk is indeed the right tool for the job here, it's a simple one-liner.
$ cat test.in
a::b::c
d::e::f
g::h::i
$ awk -F:: -v OFS=, '{$1=$1;print;print $2,$3 >> "altfile"}' test.in
a,b,c
d,e,f
g,h,i
$ cat altfile
b,c
e,f
h,i
$

get Nth line in file after parsing another file

I have one of my large file as
foo:43:sdfasd:daasf
bar:51:werrwr:asdfa
qux:34:werdfs:asdfa
foo:234:dfasdf:dasf
qux:345:dsfasd:erwe
...............
here 1st column foo, bar and qux etc. are file names. and 2nd column 43,51, 34 etc. are line numbers. I want to print Nth line(specified by 2nd column) for each file(specified in 1st column).
How can I automate above in unix shell.
Actually above file is generated while compiling and I want to print warning line in code.
-Thanks,
while IFS=: read name line rest
do
head -n $line $name | tail -1
done < input.txt
while IFS=: read file line message; do
echo "$file:$line - $message:"
sed -n "${line}p" "$file"
done <yourfilehere
awk 'NR==4 {print}' yourfilename
or
cat yourfilename | awk 'NR==4 {print}'
The above one will work for 4th line in your file.You can change the number as per your requirement.
Just in awk, but probably worse performance than answers by #kev or #MarkReed.
However it does process each file just once. Requires GNU awk
gawk -F: '
BEGIN {OFS=FS}
{
files[$1] = 1
lines[$1] = lines[$1] " " $2
msgs[$1, $2] = $3
}
END {
for (file in files) {
split(lines[file], l, " ")
n = asort(l)
count = 0
for (i=1; i<=n; i++) {
while (++count <= l[i])
getline line < file
print file, l[i], msgs[file, l[i]]
print line
}
close(file)
}
}
'
This might work for you:
sed 's/^\([^,]*\),\([^,]*\).*/sed -n "\2p" \1/' file |
sort -k4,4 |
sed ':a;$!N;s/^\(.*\)\(".*\)\n.*"\(.*\)\2/\1;\3\2/;ta;P;D' |
sh
sed -nr '3{s/^([^:]*):([^:]*):.*$/\1 \2/;p}' namesNnumbers.txt
qux 34
-n no output by default,
-r regular expressions (simplifies using the parens)
in line 3 do {...;p} (print in the end)
s ubstitute foobarbaz with foo bar
So to work with the values:
fnUln=$(sed -nr '3{s/^([^:]*):([^:]*):.*$/\1 \2/;p}' namesNnumbers.txt)
fn=$(echo ${fnUln/ */})
ln=$(echo ${fnUln/* /})
sed -n "${ln}p" "$fn"

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