Replace empty fields in tab separated file with "0" - awk

Say I have a data file with rows like a^I^I^I^I^I^I^Ib^Ic, which is separated by ^I (means a tab).
Now, I want to change the empty value of each column into 0, so the result should be like: a^I0^I0^I0^I0^I0^I0^Ib^Ic.
How can I achieve it with only one sed command?


This is easier to do using a tool with support for look-ahead:
perl -pe 's/\t(?=\t)/\t0/g' file
This puts a "0" in between any pair of tab characters. The look-ahead is useful as it matches the second tab without consuming it, so it can be used in the next match.
Here's a way you could do it using awk:
awk -F'\t' -v OFS='\t' '{ for (i = 1; i <= NF; ++i) sub(/^$/, 0, $i) } 1' file
This loops through all the fields, substituting all empty ones with a 0.


With GNU sed:
sed ':a;s/\t\(\t\|$\)/\t0\1/;ta' file
Replace all \t followed by \t or end of line with \t0.


another awk
$ awk -v RS='\t' -v ORS='\t' '$0==""{$0=0}1'
or with BEGIN block
$ awk 'BEGIN{RS=ORS="\t"} $0==""{$0=0}1'


$ awk '{while(gsub(/\t\t/,"\t0\t"));} 1' file
a 0 0 0 0 0 0 b c

Related

Can I delete a field in awk?

This is test.txt:
0x01,0xDF,0x93,0x65,0xF8
0x01,0xB0,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0xB2,0x00,0x76
If I run
awk -F, 'BEGIN{OFS=","}{$2="";print $0}' test.txt
the result is:
0x01,,0x93,0x65,0xF8
0x01,,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,,0x00,0x76
The $2 wasn't deleted, it just became empty.
I hope, when printing $0, that the result is:
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0x00,0x76

All the existing solutions are good though this is actually a tailor made job for cut:
cut -d, -f 1,3- file
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0x00,0x76
If you want to remove 3rd field then use:
cut -d, -f 1,2,4- file
To remove 4th field use:
cut -d, -f 1-3,5- file

I believe simplest would be to use sub function to replace first occurrence of continuous ,,(which are getting created after you made 2nd field NULL) with single ,. But this assumes that you don't have any commas in between field values.
awk 'BEGIN{FS=OFS=","}{$2="";sub(/,,/,",");print $0}' Input_file
2nd solution: OR you could use match function to catch regex from first comma to next comma's occurrence and get before and after line of matched string.
awk '
match($0,/,[^,]*,/){
print substr($0,1,RSTART-1)","substr($0,RSTART+RLENGTH)
}' Input_file

It's a bit heavy-handed, but this moves each field after field 2 down a place, and then changes NF so the unwanted field is not present:
$ awk -F, -v OFS=, '{ for (i = 2; i < NF; i++) $i = $(i+1); NF--; print }' test.txt
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01
0x01,0x00,0x76
$
Tested with both GNU Awk 4.1.3 and BSD Awk ("awk version 20070501" on macOS Mojave 10.14.6 — don't ask; it frustrates me too, but sometimes employers are not very good at forward thinking). Setting NF may or may not work on older versions of Awk — I was a little surprised it did work, but the surprise was a pleasant one, for a change.

If Awk is not an absolute requirement, and the input is indeed as trivial as in your example, sed might be a simpler solution.
sed 's/,[^,]*//' test.txt
This is especially elegant if you want to remove the second field. A more generic approach to remove, the nth field would require you to put in a regex which matches the first n - 1 followed by the nth, then replace that with just the the first n - 1.
So for n = 4 you'd have
sed 's/\([^,]*,[^,]*,[^,]*,\)[^,]*,/\1/' test.txt
or more generally, if your sed dialect understands braces for specifying repetitions
sed 's/\(\([^,]*,\)\{3\}\)[^,]*,/\1/' test.txt
Some sed dialects allow you to lose all those pesky backslashes with an option like -r or -E but again, this is not universally supported or portable.
In case it's not obvious, [^,] matches a single character which is not (newline or) comma; and \1 recalls the text from first parenthesized match (back reference; \2 recalls the second, etc).
Also, this is completely unsuitable for escaped or quoted fields (though I'm not saying it can't be done). Every comma acts as a field separator, no matter what.

With GNU sed you can add a number modifier to substitute nth match of non-comma characters followed by comma:
sed -E 's/[^,]*,//2' file

Using awk in a regex-free way, with the option to choose which line will be deleted:
awk '{ col = 2; n = split($0,arr,","); line = ""; for (i = 1; i <= n; i++) line = line ( i == col ? "" : ( line == "" ? "" : "," ) arr[i] ); print line }' test.txt
Step by step:
{
col = 2 # defines which column will be deleted
n = split($0,arr,",") # each line is split into an array
# n is the number of elements in the array
line = "" # this will be the new line
for (i = 1; i <= n; i++) # roaming through all elements in the array
line = line ( i == col ? "" : ( line == "" ? "" : "," ) arr[i] )
# appends a comma (except if line is still empty)
# and the current array element to the line (except when on the selected column)
print line # prints line
}

Another solution:
You can just pipe the output to another sed and squeeze the delimiters.
$ awk -F, 'BEGIN{OFS=","}{$2=""}1 ' edward.txt | sed 's/,,/,/g'
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0x00,0x76
$

Commenting on the first solution of #RavinderSingh13 using sub() function:
awk 'BEGIN{FS=OFS=","}{$2="";sub(/,,/,",");print $0}' Input_file
The gnu-awk manual: https://www.gnu.org/software/gawk/manual/html_node/Changing-Fields.html
It is important to note that making an assignment to an existing field changes the value of $0 but does not change the value of NF, even when you assign the empty string to a field." (4.4 Changing the Contents of a Field)
So, following the first solution of RavinderSingh13 but without using, in this case,sub() "The field is still there; it just has an empty value, delimited by the two colons":
awk 'BEGIN {FS=OFS=","} {$2="";print $0}' file
0x01,,0x93,0x65,0xF8
0x01,,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,,0x00,0x76

My solution:
awk -F, '
{
regex = "^"$1","$2
sub(regex, $1, $0);
print $0;
}'
or one line code:
awk -F, '{regex="^"$1","$2;sub(regex, $1, $0);print $0;}' test.txt
I found that OFS="," was not necessary

I would do it following way, let file.txt content be:
0x01,0xDF,0x93,0x65,0xF8
0x01,0xB0,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0xB2,0x00,0x76
then
awk 'BEGIN{FS=",";OFS=""}{for(i=2;i<=NF;i+=1){$i="," $i};$2="";print}' file.txt
output
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0x00,0x76
Explanation: I set OFS to nothing (empty string), then for 2nd and following column I add , at start. Finally I set what is now comma and value to nothing. Keep in mind this solution would need rework if you wish to remove 1st column.

Replace a letter with another from the last word from the last two lines of a text file

How could I possibly replace a character with another, selecting the last word from the last two lines of a text file in shell, using only a single command? In my case, replacing every occurrence of a with E from the last word only.
Like, from a text file containing this:
tree;apple;another
mango.banana.half
monkey.shelf.karma
to this:
tree;apple;another
mango.banana.hElf
monkey.shelf.kErmE
I tried using sed -n 'tail -2 'mytext.txt' -r 's/[a]+/E/*$//' but it doesn't work (my error: sed expression #1, char 10: unknown option to 's).

Could you please try following, tac + awk solution. Completely based on OP's samples only.
tac Input_file |
awk 'FNR<=2{if(/;/){FS=OFS=";"};if(/\./){FS=OFS="."};gsub(/a/,"E",$NF)} 1' |
tac
Output with shown samples is:
tree;apple;another
mango.banana.hElf
monkey.shelf.kErmE
NOTE: Change gsub to sub in case you want to substitute only very first occurrence of character a in last field.

This might work for you (GNU sed):
sed -E 'N;${:a;s/a([^a.]*)$/E\1/mg;ta};P;D' file
Open a two line window throughout the length of the file by using the N to append the next line to the previous and the P and D commands to print then delete the first of these. Thus at the end of the file, signified by the $ address the last two lines will be present in the pattern space.
Using the m multiline flag on the substitution command, as well as the g global flag and a loop between :a and ta, replace any a in the last word (delimited by .) by an E.
Thus the first pass of the substitution command will replace the a in half and the last a in karma. The next pass will match nothing in the penultimate line and replace the a in karmE. The third pass will match nothing and thus the ta command will fail and the last two lines will printed with the required changes.

If you want to use Sed, here's a solution:
tac input_file | sed -E '1,2{h;s/.*[^a-zA-Z]([a-zA-Z]+)/\1/;s/a/E/;x;s/(.*[^a-zA-Z]).*/\1/;G;s/\n//}' | tac
One tiny detail. In your question you say you want to replace a letter, but then you transform karma in kErme, so what is this? If you meant to write kErma, then the command above will work; if you meant to write kErmE, then you have to change it just a bit: the s/a/E/ should become s/a/E/g.

With tac+perl
$ tac ip.txt | perl -pe 's/\w+\W*$/$&=~tr|a|E|r/e if $.<=2' | tac
tree;apple;another
mango.banana.hElf
monkey.shelf.kErmE
\w+\W*$ match last word in the line, \W* allows any possible trailing non-word characters to be matched as well. Change \w and \W accordingly if numbers and underscores shouldn't be considered as word characters - for ex: [a-zA-Z]+[^a-zA-Z]*$
$&=~tr|a|E|r change all a to E only for the matched portion
e flag to enable use of Perl code in replacement section instead of string
To do it in one command, you can slurp the entire input as single string (assuming this'll fit available memory):
perl -0777 -pe 's/\w+\W*$(?=(\n.*)?\n\z)/$&=~tr|a|E|r/gme'

Using GNU awk forsplit() 4th arg since in the comments of another solution the field delimiter is every sequence of alphanumeric and numeric characters:
$ gawk '
BEGIN {
pc=2 # previous counter, ie how many are affected
}
{
for(i=pc;i>=1;i--) # buffer to p hash, a FIFO
if(i==pc && (i in p)) # when full, output
print p[i]
else if(i in p) # and keep filling
p[i+1]=p[i] # above could be done using mod also
p[1]=$0
}
END {
for(i=pc;i>=1;i--) {
n=split(p[i],t,/[^a-zA-Z0-9\r]+/,seps) # split on non alnum
gsub(/a/,"E",t[n]) # replace
for(j=1;j<=n;j++) {
p[i]=(j==1?"":p[i] seps[j-1]) t[j] # pack it up
}
print p[i] # output
}
}' file
Output:
tree;apple;another
mango.banana.hElf
monkey.shelf.kErmE

Would this help you ? on GNU awk
$ cat file
tree;apple;another
mango.banana.half
monkey.shelf.karma
$ tac file | awk 'NR<=2{s=gensub(/(.*)([.;])(.*)$/,"\\3",1);gsub(/a/,"E",s); print gensub(/(.*)([.;])(.*)$/,"\\1\\2",1) s;next}1' | tac
tree;apple;another
mango.banana.hElf
monkey.shelf.kErmE
Better Readable version :
$ tac file | awk 'NR<=2{
s=gensub(/(.*)([.;])(.*)$/,"\\3",1);
gsub(/a/,"E",s);
print gensub(/(.*)([.;])(.*)$/,"\\1\\2",1) s;
next
}1' | tac

With GNU awk you can set FS with the two separators, then gsub for the replacement in $3, the third field, if NR>1
awk -v FS=";|[.]" 'NR>1 {gsub("a", "E",$3)}1' OFS="." file
tree;apple;another
mango.banana.hElf
monkey.shelf.kErmE

With GNU awk for the 3rd arg to match() and gensub():
$ awk -v n=2 '
NR>n { print p[NR%n] }
{ p[NR%n] = $0 }
END {
for (i=0; i<n; i++) {
match(p[i],/(.*[^[:alnum:]])(.*)/,a)
print a[1] gensub(/a/,"E","g",a[2])
}
}
' file
tree;apple;another
mango.banana.hElf
monkey.shelf.kErmE
or with any awk:
awk -v n=2 '
NR>n { print p[NR%n] }
{ p[NR%n] = $0 }
END {
for (i=0; i<n; i++) {
match(p[i],/.*[^[:alnum:]]/)
lastWord = substr(p[i],1+RLENGTH)
gsub(/a/,"E",lastWord )
print substr(p[i],1,RLENGTH) lastWord
}
}
' file
If you want to do it for the last 50 lines of a file instead of the last 2 lines just change -v n=2 to -v n=50.
The above assumes there are at least n lines in your input.

You can let sed repeat changing an a into E only for the last word with a label.
tac mytext.txt| sed -r ':a; 1,2s/a(\w*)$/E\1/; ta' | tac

Exact string match in awk

I have a file test.txt with the next lines
1997 100 500 2010TJ
2010TJXML 16 20 59
I'm using the next awk line to get information only about string 2010TJ
awk -v var="2010TJ" '$0 ~ var {print $0}' test.txt
But the code print the two lines. I want to know how to get the line containing the exact string
1997 100 500 2010TJ
the string can be placed in any column of the file.

Several options:
Use a gawk word boundary (not POSIX awk...):
$ gawk '/\<2010TJ\>/' file
An actual space or tab or what is separating the columns:
$ awk '/^2010TJ /' file
Or compare the field directly to the string:
$ awk '$1=="2010TJ"' file
You can loop over the fields to test each field if you wish:
$ awk '{for (i=1;i<=NF;i++) if ($i=="2010TJ") {print; next}}' file
Or, given your example of setting a variable, those same using a variable:
$ gawk -v s=2010TJ '$0~"\\<" s "\\>"'
$ awk -v s=2010TJ '$0~"^" s " "'
$ awk -v s=2010TJ '$1==s'
Note the first is a little different than the second and third. The first is the standalone string 2010TJ anywhere in $0; the second and third is a string that starts with that string.

Try this (for testing only column 1) :
awk '$1 == "2010TJ" {print $0}' test.txt
or grep like (all columns) :
gawk '/\<2010TJ\>/ {print $0}' test.txt
Note
\< \> is word boundarys

another awk with word boundary
awk '/\y2010TJ\y/' file
note \y matches either beginning or end of a word.

Grep part of string after symbol and shuffle columns

I would like to take the number after the - sign and put is as column 2 in my matrix. I know how to grep the string but not how to print it after the text string.
in:
1-967764 GGCTGGTCCGATGGTAGTGGGTTATCAGAACT
3-425354 GCATTGGTGGTTCAGTGGTAGAATTCTCGCC
4-376323 GGCTGGTCCGATGGTAGTGGGTTATCAGAAC
5-221398 GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT
6-180339 TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT
out:
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339

awk -F'[[:space:]-]+' '{print $3,$2}' file

Seems like a simple substitution should do the job:
sed -E 's/[0-9]+-([0-9]+)[[:space:]]*(.*)/\2 \1/' file
Capture the parts you're interested in and use them in the replacement.
Alternatively, using awk:
awk 'sub(/^[0-9]+-/, "") { print $2, $1 }' file
Remove the leading digits and - from the start of the line. When this is successful, sub returns true, so the action is performed, printing the second field, followed by the first.

Using regex ( +|-) as field separator:
$ awk -F"( +|-)" '{print $3,$2}' file
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339

here is another awk
$ awk 'split($1,a,"-") {print $2,a[2]}' file

awk '{sub(/.-/,"");print $2,$1}' file
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339

How to find if substring is in a variable in awk

i am using awk and need to find if a variable , in this case $24 contains the word 3:2- if so to print the line (for sed command)- the variable may include more letters or spaces or \n.......
for ex.
$24 == "3:2" {print "s/(inter = ).*/\\1\"" "3:2_pulldown" "\"/" >> NR }
in my above line- it never find such a string although it exists.
can you help me with the command please??

If you're looking for "3:2" within $24, then you want $24 ~ /3:2/ or index($24, "3:2") > 0
Why are you using awk to generate a sed script?
Update
To pass a variable from the shell to awk, use the -v option:
val="3:2" # or however you determine this value
awk -v v="$val" '$24 ~ v {print}'

awk '$24~/3:2/' file_name
this will serach for "3:2" in field 24