When I wrote an answer for this question I used the following:
something | sed '$d' | awk '$1>3{print $0}'
e.g.
print only lines where the 1st field is bigger than 3 (awk)
but omit the last line sed '$d'.
This seems for me a bit of duplicate work, surely it is possible to do the above only with awk - without the sed?
I'm an awkdiot - so, can someone suggest a solution?
Here's one way you could do it:
$ printf "%s\n" {1..10} | awk 'NR>1&&p>3{print p}{p=$1}'
4
5
6
7
8
9
Basically, print the first field of the previous line, rather than the current one.
As Wintermute has rightly pointed out in the comments (thanks), in order to print the whole line, you can modify the code to this:
awk 'p { print p; p="" } $1 > 3 { p = $0 }'
This only assigns the contents of contents of the line to p if the first field is greater than 3.
Related
I am trying to split a file column using the substr awk command. So the input is as follows (it consists of 4 lines, one blank line):
#NS500645:122:HYGVMBGX2:4:21402:2606:16446:ACCTAGAAGG:R1
ACCTAGAAGGATATGCGCTTGCGCGTTAGAGATCACTAGAGCTAAGGAATTTGAGATTACAGTAAGCTATGATCC
/AAAAEEEEEEEEEEAAEEEAEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
I want to split the second line by the pattern "GATC" but keeping it on the right sub-string like:
ACCTAGAAGGATATGCGCTTGCGCGTTAGA GATCACTAGAGCTAAGGAATTTGAGATTACAGTAAGCTATGATCC
I want that the last line have the same length as the splitted one and regenerate the file like:
ACCTAGAAGGATATGCGCTTGCGCGTTAGA
/AAAAEEEEEEEEEEAAEEEAEEEEEEEEE
GATCACTAGAGCTAAGGAATTTGAGATTACAGTAAGCTAT
EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
GATCC
EEEEE
For split the last colum I am using this awk script:
cat prove | paste - - - - | awk 'BEGIN
{FS="\t"; OFS="\t"}\ {gsub("GATC","/tGATC", $2); {split ($2, a, "\t")};\ for
(i in a) print substr($4, length(a[i-1])+1,
length(a[i-1])+length(a[i]))}'
But the output is as follows:
/AAAAEEEEEEEEEEAAEEEAEEEEEEEEE
EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
Being the second and third line longer that expected.
I check the calculated length that are passed to the substr command and are correct:
1 30
31 70
41 45
Using these length the output should be:
/AAAAEEEEEEEEEEAAEEEAEEEEEEEEE
EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
EEEEE
But as I showed it is not the case.
Any suggestions?
I guess you're looking something like this, but your question formatting is really confusing
$ awk -v OFS='\t' 'NR==1 {next}
NR==2 {n=index($0,"GATC")}
/^[^+]/ {print substr($0,1,n-1),substr($0,n)}' file
ACCTAGAAGGATATGCGCTTGCGCGTTAGA GATCACTAGAGCTAAGGAATTTGAGATTACAGTAAGCTATGATCC
/AAAAEEEEEEEEEEAAEEEAEEEEEEEEE EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
I assumed your file is in this format
dummy header line to be ignored
ACCTAGAAGGATATGCGCTTGCGCGTTAGAGATCACTAGAGCTAAGGAATTTGAGATTACAGTAAGCTATGATCC
+
/AAAAEEEEEEEEEEAAEEEAEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
I have a big text file with 2 tab separated fields. as you see in the small example every 2 lines have a number in common. I want to summarize my text file in this way.
1- look for the lines that have the number in common and sum up the second column of those lines.
small example:
ENST00000054666.6 2
ENST00000054666.6_2 15
ENST00000054668.5 4
ENST00000054668.5_2 10
ENST00000054950.3 0
ENST00000054950.3_2 4
expected output:
ENST00000054666.6 17
ENST00000054668.5 14
ENST00000054950.3 4
as you see the difference is in both columns. in the 1st column there is only one repeat of each common and without "_2" and in the 2nd column the values is sum up of both lines (which have common number in input file).
I tried this code but does not return what I want:
awk -F '\t' '{ col2 = $2, $2=col2; print }' OFS='\t' input.txt > output.txt
do you know how to fix it?
Solution 1st: Following awk may help you on same.
awk '{sub(/_.*/,"",$1)} {a[$1]+=$NF} END{for(i in a){print i,a[i]}}' Input_file
Solution 2nd: In case your Input_file is sorted by 1st field then following may help you.
awk '{sub(/_.*/,"",$1)} prev!=$1 && prev{print prev,val;val=""} {val+=$NF;prev=$1} END{if(val){print prev,val}}' Input_file
Use > output.txt at the end of the above codes in case you need the output in a output file too.
If order is not a concern, below may also help :
awk -v FS="\t|_" '{count[$1]+=$NF}
END{for(i in count){printf "%s\t%s%s",i,count[i],ORS;}}' file
ENST00000054668.5 14
ENST00000054950.3 4
ENST00000054666.6 17
Edit :
If the order of the output does matter, below approach using a flag helps :
$ awk -v FS="\t|_" '{count[$1]+=$NF;++i;
if(i==2){printf "%s\t%s%s",$1,count[$1],ORS;i=0}}' file
ENST00000054666.6 17
ENST00000054668.5 14
ENST00000054950.3 4
How to print all lines if certain condition matches.
Example:
echo "$ip"
this is a sample line
another line
one more
last one
If this file has more than 3 lines then print the whole variable.
I am tried:
echo $ip| awk 'NR==4'
last one
echo $ip|awk 'NR>3{print}'
last one
echo $ip|awk 'NR==12{} {print}'
this is a sample line
another line
one more
last one
echo $ip| awk 'END{x=NR} x>4{print}'
Need to achieve this:
If this file has more than 3 lines then print the whole file. I can do this using wc and bash but need a one liner.
The right way to do this (no echo, no pipe, no loops, etc.):
$ awk -v ip="$ip" 'BEGIN{if (gsub(RS,"&",ip)>2) print ip}'
this is a sample line
another line
one more
last one
You can use Awk as follows,
echo "$ip" | awk '{a[$0]; next}END{ if (NR>3) { for(i in a) print i }}'
one more
another line
this is a sample line
last one
you can also make the value 3 configurable from an awk variable,
echo "$ip" | awk -v count=3 '{a[$0]; next}END{ if (NR>count) { for(i in a) print i }}'
The idea is to store the contents of the each line in {a[$0]; next} as each line is processed, by the time the END clause is reached, the NR variable will have the line count of the string/file you have. Print the lines if the condition matches i.e. number of lines greater than 3 or whatever configurable value using.
And always remember to double-quote the variables in bash to avoid undergoing word-splitting done by the shell.
Using James Brown's useful comment below to preserve the order of lines, do
echo "$ip" | awk -v count=3 '{a[NR]=$0; next}END{if(NR>3)for(i=1;i<=NR;i++)print a[i]}'
this is a sample line
another line
one more
last one
Another in awk. First test files:
$ cat 3
1
2
3
$ cat 4
1
2
3
4
Code:
$ awk 'NR<4{b=b (NR==1?"":ORS)$0;next} b{print b;b=""}1' 3 # look ma, no lines
[this line left intentionally blank. no wait!]
$ awk 'NR<4{b=b (NR==1?"":ORS)$0;next} b{print b;b=""}1' 4
1
2
3
4
Explained:
NR<4 { # for tghe first 3 records
b=b (NR==1?"":ORS) $0 # buffer them to b with ORS delimiter
next # proceed to next record
}
b { # if buffer has records, ie. NR>=4
print b # output buffer
b="" # and reset it
}1 # print all records after that
awk newbie here! I am asking for help to solve a simple specific task.
Here is file.txt
1
2
3
5
6
7
8
9
As you can see a single number (the number 4) is missing. I would like to print on the console the number 4 that is missing. My idea was to compare the current line number with the entry and whenever they don't match I would print the line number and exit. I tried
cat file.txt | awk '{ if ($NR != $1) {print $NR; exit 1} }'
But it prints only a newline.
I am trying to learn awk via this small exercice. I am therefore mainly interested in solutions using awk. I also welcome an explanation for why my code does not do what I would expect.
Try this -
awk '{ if (NR != $1) {print NR; exit 1} }' file.txt
4
since you have a solution already, here is another approach, comparing with previous values.
awk '$1!=p+1{print p+1} {p=$1}' file
you positional comparison won't work if you have more than one missing value.
Maybe this will help:
seq $(tail -1 file)|diff - file|grep -Po '.*(?=d)'
4
Since I am learning awk as well
awk 'BEGIN{i=0}{i++;if(i!=$1){print i;i=$1}}' file
4
`awk` explanation read each number from `$1` into array `i` and increment that number list line by line with `i++`, if the number is not sequential, then print it.
cat file
1
2
3
5
6
7
8
9
11
12
13
15
awk 'BEGIN{i=0}{i++;if(i!=$1){print i;i=$1}}' file
4
10
14
Assume the following file
#zvview.exe
#begin Present/3
77191.0000 189.320100 0 0 3 0111110 16 1
-8.072430+6-8.072430+6 77190 0 1 37111110 16 2
37 2 111110 16 3
8.115068+6 0.000000+0 8.500000+6 6.390560-2 9.000000+6 6.803440-1111110 16 4
9.500000+6 1.685009+0 1.000000+7 2.582780+0 1.050000+7 3.260540+0111110 16 5
37 2 111110 16 18
What I would like to do, is print in two columns, the fields after line 6. This can be done using NR. The tricky part is the following : Every second field, should go in one column as well as adding an E before the sign, so that the output file will look like this
8.115068E+6 0.000000E+0
8.500000E+6 6.390560E-2
9.000000E+6 6.803440E-1
9.500000E+6 1.685009E+0
1.000000E+7 2.582780E+0
1.050000E+7 3.260540E+0
From the output file you see that I want to keep in $6 only length($6)=10 characters.
How is it possible to do it in awk?
can do all in awk but perhaps easier with the unix toolset
$ sed -n '6,7p' file | cut -c2-66 | tr ' ' '\n' | pr -2ats' '
8.115068+6 0.000000+0
8.500000+6 6.390560-2
9.000000+6 6.803440-1
9.500000+6 1.685009+0
1.000000+7 2.582780+0
1.050000+7 3.260540+0
Here is a awk only solution or comparison
$ awk 'NR>=6 && NR<=7{$6=substr($6,1,10);
for(i=1;i<=6;i+=2) {f[++c]=$i;s[c]=$(i+1)}}
END{for(i=1;i<=c;i++) print f[i],s[i]}' file
8.115068+6 0.000000+0
8.500000+6 6.390560-2
9.000000+6 6.803440-1
9.500000+6 1.685009+0
1.000000+7 2.582780+0
1.050000+7 3.260540+0
Perhaps shorter version,
$ awk 'NR>=6 && NR<=7{$6=substr($6,1,10);
for(i=1;i<=6;i+=2) print $i FS $(i+1)}' file
8.115068+6 0.000000+0
8.500000+6 6.390560-2
9.000000+6 6.803440-1
9.500000+6 1.685009+0
1.000000+7 2.582780+0
1.050000+7 3.260540+0
to convert format to standard scientific notation, you can pipe the result to
sed or embed something similar in awk script (using gsub).
... | sed 's/[+-]/E&/g'
8.115068E+6 0.000000E+0
8.500000E+6 6.390560E-2
9.000000E+6 6.803440E-1
9.500000E+6 1.685009E+0
1.000000E+7 2.582780E+0
1.050000E+7 3.260540E+0
With GNU awk for FIELDWIDTHS:
$ cat tst.awk
BEGIN { FIELDWIDTHS="9 2 9 2 9 2 9 2 9 2 9 2" }
NR>5 && NR<8 {
for (i=1;i<NF;i+=4) {
print $i "E" $(i+1), $(i+2) "E" $(i+3)
}
}
$ awk -f tst.awk file
8.115068E+6 0.000000E+0
8.500000E+6 6.390560E-2
9.000000E+6 6.803440E-1
9.500000E+6 1.685009E+0
1.000000E+7 2.582780E+0
1.050000E+7 3.260540E+0
If you really want to get rid of the leading blanks then there's various ways to do it (simplest being gsub(/ /,"",$<field number>) on the relevant fields) but I left them in because the above allows your output to line up properly if/when your numbers start with a -, like they do on line 4 of your sample input.
If you don't have GNU awk, get it as you're missing a LOT of extremely useful functionality.
I tried to combine #karafka 's answer using substr, so the following does the trick!
awk 'NR>=6 && NR<=7{$6=substr($6,1,10);for(i=1;i<=6;i+=2) print substr($i,1,8) "E" substr($i,9) FS substr($(i+1),1,8) "E" substr($(i+1),9)}' file
and the output is
8.115068E+6 0.000000E+0
8.500000E+6 6.390560E-2
9.000000E+6 6.803440E-1
9.500000E+6 1.685009E+0
1.000000E+7 2.582780E+0
1.050000E+7 3.260540E+0