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
Related
Consider the following data in a file file.txt:
$
$
$
FORCE 10 30 40
* 1 5 4
FORCE 11 20 22
* 2 3 0
FORCE 19 25 10
* 16 12 8
.
.
.
I want to print every 2nd element of the third column, starting from line 4, resulting in:
30
20
25
I have tried:
cat file.txt | sed 's/\|/ /' | awk 'NR%2==4 {print $3}'
However, this is not resulting in anything being printed and no errors generated either.
You might use awk checking that the row number > 3 and then check for an even row number with NR%2==0.
Note that you don't have to use cat
awk 'NR > 3 && NR%2==0 {
print $3
}' file.txt
Output
30
20
25
Using sed
$ sed -En '4~2s/([^ \t]*[ \t]+){2}([^ \t]*).*/\2/p' input_file
30
20
25
I have tried:
cat file.txt | sed 's/\|/ /' | awk 'NR%2==4 {print $3}'
However, this is not resulting in anything being printed and no errors
generated either.
You do not need cat whilst using GNU sed as it can read file on its' own, in this case it would be sed 's/\|/ /' file.txt.
You should consider if you need that part at all, your sample input does not have pipe character at all, so it would do nothing to it. You might also drop that part if lines holding values you want to print do not have that character.
Output is empty as NR%2==4 does never hold, remainder of division by x is always smaller than x (in particular case of %2 only 2 values are possible: 0 and 1)
This might work for you (GNU sed):
sed -nE '4~2s/^((\S+)\s*){3}.*/\2/p' file
Turn off implicit printing by setting the -n option and reduce back slashes in regexps by turning on -E.
From the fourth line and then every second line thereafter, capture the third column and print it.
N.B. The \2 represents the last inhabitant of that back reference which in conjunction with the {3} means the above.
Alternative:
sed -n '4,${s/^\(\(\S\+\)\s*\)\{3\}.*/\2/p;n}' file
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
I have multiple files with the same name (3pGtoA_freq.txt), but all located in different directories.
Each file looks like this:
pos 5pG>A
1 0.162421557770395
2 0.0989643268124281
3 0.0804131316857248
4 0.0616563298066399
5 0.0577551761714493
6 0.0582450832072617
7 0.0393129770992366
8 0.037037037037037
9 0.0301016419077404
10 0.0327510917030568
11 0.0301598837209302
12 0.0309050772626932
13 0.0262089331856774
14 0.0254612546125461
15 0.0226130653266332
16 0.0206971677559913
17 0.0181280059193489
18 0.0243993993993994
19 0.0181347150259067
20 0.0224429727740986
21 0.0175690211545357
22 0.0183916336098089
23 0.0196078431372549
24 0.0187983781791375
25 0.0173192771084337
I want to cut column 2 from each file and paste column by column in one file
I tried running:
for s in results_Sample_*_hg19/results_MapDamage_Sample_*/results_Sample_*_bwa_LongSeed_sorted_hg19_noPCR/3pGtoA_freq.txt; do awk '{print $2}' $s >> /home/users/istolarek/aDNA/3pGtoA_all; done
but it's not pasting the columns next to each other.
Also I wanted to name each column by the '*', which is the only string that changes in path.
Any help with that?
for i in $(find you_file_dir -name 3pGtoA_freq.txt);do awk '{print $2>>"NewFile"}' $i; done
I would do this by processing all files in parallel in awk:
awk 'BEGIN{printf "pos ";
for(i=1;i<ARGC;++i)
printf "%-19s",gensub("^results_Sample_","",1,gensub("_hg19.*","",1,ARGV[i]));
printf "\n";
while(getline<ARGV[1]){
printf "%-4s%-19s",$1,$2;
for(i=2;i<ARGC;++i){
getline<ARGV[i];
printf "%-19s",$2}
printf "\n"}}{exit}' \
results_Sample_*_hg19/results_MapDamage_Sample_*/results_Sample_*_bwa_LongSeed_sorted_hg19_noPCR/3pGtoA_freq.txt
If your awk doesn't have gensub (I'm using cygwin), you can remove the first four lines (printf-printf); headers won't be printed in that case.
When I wrote an answer for this question I used the following:
something | sed '$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 = $0 }'
This only assigns the contents of contents of the line to p if the first field is greater than 3.
I have file like below :
this is a sample file
this file will be used for testing
this is a sample file
this file will be used for testing
I want to count the words using AWK.
the expected output is
this 2
is 1
a 1
sample 1
file 2
will 1
be 1
used 1
for 1
the below AWK I have written but getting some errors
cat anyfile.txt|awk -F" "'{for(i=1;i<=NF;i++) a[$i]++} END {for(k in a) print k,a[k]}'
It works fine for me:
awk '{for(i=1;i<=NF;i++) a[$i]++} END {for(k in a) print k,a[k]}' testfile
used 1
this 2
be 1
a 1
for 1
testing 1
file 2
will 1
sample 1
is 1
PS you do not need to set -F" ", since its default any blank.
PS2, do not use cat with programs that can read data itself, like awk
You can add sort behind code to sort it.
awk '{for(i=1;i<=NF;i++) a[$i]++} END {for(k in a) print k,a[k]}' testfile | sort -k 2 -n
a 1
be 1
for 1
is 1
sample 1
testing 1
used 1
will 1
file 2
this 2
Instead of looping each line and saving the word in array ({for(i=1;i<=NF;i++) a[$i]++}) use gawk with multi-char RS (Record Separator) definition support option and save each field in array as following(It's a little bit fast):
gawk '{a[$0]++} END{for (k in a) print k,a[k]}' RS='[[:space:]]+' file
Output:
used 1
this 2
be 1
a 1
for 1
testing 1
file 2
will 1
sample 1
is 1
In above gawk command I defines space-character-class [[:space:]]+ (including one or more spaces or \new line character) as record separator.
Here is Perl code which provides similar sorted output to Jotne's awk solution:
perl -ne 'for (split /\s+/, $_){ $w{$_}++ }; END{ for $key (sort keys %w) { print "$key $w{$key}\n"}}' testfile
$_ is the current line, which is split based on whitespace /\s+/
Each word is then put into $_
The %w hash stores the number of occurrences of each word
After the entire file is processed, the END{} block is run
The keys of the %w hash are sorted alphabetically
Each word $key and number of occurrences $w{$key} is printed