I'd like to pass a predefined variable as the column number for an awk script. I've stripped out the unnecessary bits and below is an example of what I'd like to get done. Further below is a portion of what I've tried so far.
Reason: This is a semi-long script that currently works though I'd like to define the columns early in the script as this would make the script much easier to update as columns change.
I'd like for the "state" variable to be passed on to awk's column identifier, eg:
#/bin/bash
export state='$6'
cat ~/file | awk -v column="$state" 'state!="FAILED"'
Running the above code produces rows that do indeed have column 6 as "FAILED", so there must be something wrong. While awk '$6!="FAILED"' works as expected
Different things I've tried so far:
defining $state as 6 rather than '$6' and including the $ in the awk != command.
awk '{ENVIRON["state"]!="FAILED"}', with the same modifications as 1
This should work:
state=6
cat ~/file | awk -v column="$state" '$column != "FAILED"'
$var in awk will get the field specified by the value of variable var.
So, $NF will get the last field. Note that the awk variable here is column, not state.
For example:
% seq 1 20 | paste - - - -
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
17 18 19 20
% seq 1 20 | paste - - - - | awk -v column=3 '{print $column}'
3
7
11
15
19
Related
I have a dataframe like this
1 3 MAPQ=0;CT=3to5;SRMAPQ=60
2 34 MAPQ=60;CT=3to5;SRMAPQ=67
4 56 MAPQ=67;CT=3to5;SRMAPQ=50
5 7 MAPQ=44;CT=3to5;SRMAPQ=61
with using awk (or others)
I want to extract rows with only SRMAPQ over 60.
This means the output is
2 34 MAPQ=60;CT=3to5;SRMAPQ=67
5 7 MAPQ=44;CT=3to5;SRMAPQ=61
update: "SRMAPQ=60" can be anywhere in the line,
MAPQ=44;CT=3to5;SRMAPQ=61;DT=3to5
You don't have to extract the value out of SRMAPQ separately and do the comparison. If the format is fixed like above, just use = as the field separator and access the last field using $NF
awk -F= '$NF > 60' file
Or if SRMAPQ can occur anywhere in the line (as updated in the comments), use a generic approach
awk 'match($0, /SRMAPQ=([0-9]+)/){ l = length("SRMAPQ="); v = substr($0, RSTART+l, RLENGTH-l) } v > 60' file
I would use GNU AWK following way let file.txt content be
1 3 MAPQ=0;CT=3to5;SRMAPQ=60
2 34 MAPQ=60;CT=3to5;SRMAPQ=67;SOMETHING=2
4 56 MAPQ=67;CT=3to5;SRMAPQ=50
5 7 MAPQ=44;CT=3to5;SRMAPQ=61
then
awk 'BEGIN{FS="SRMAPQ="}$2>60' file.txt
output
2 34 MAPQ=60;CT=3to5;SRMAPQ=67;SOMETHING=2
5 7 MAPQ=44;CT=3to5;SRMAPQ=61
Note: added SOMETHING to test if it would work when SRMAPQ is not last. Explantion: I set FS to SRMAPQ= thus what is before that becomes first field ($1) and what is behind becomes second field ($2). In 2nd line this is 67;SOMETHING=2 with which GNU AWK copes by converting its' longmost prefix which constitute number in this case 67, other lines have just numbers. Disclaimer: this solution assumes that all but last field have trailing ;, if this does not hold true please test my solution fully before usage.
(tested in gawk 4.2.1)
I have a file as follows:
5 6
7 8
12 15
Using awk, how can I find the distance between the second column of one line with the first column of the next line. In this case, distance between 6 and 7 and 8 and 12 and print as follows, distance of first line set to zero:
5 6 0
7 8 1
12 15 4
awk '{print $0, (NR>1?$1-p:0); p=$2}' file
try:
awk 'NR==1{val=$2;print $0,"0";next} {print $0,$1-val;val=$2}' Input_file
Adding explanation now too successfully.
Checking for NR==1(when first line of Input_file) is there, then create a variable named val tp second field of the Input_file and then print the current line with "0" then do next(which will skip all further statements). Then printing the current line along with $1-val's value and then assigning the value of variable of val to $2 of the current line then.
Short awk approach:
awk 'NR==1{ $3=0 }NR>1{ $3=$1-p }{ p=$2 }1' file
The output:
5 6 0
7 8 1
12 15 4
p=$2 - capture the 2nd field value (p - considered as previous line value)
awk newbie here! I am asking for help to solve a simple specific task.
Here is file.txt
1
2
3
5
6
7
8
9
As you can see a single number (the number 4) is missing. I would like to print on the console the number 4 that is missing. My idea was to compare the current line number with the entry and whenever they don't match I would print the line number and exit. I tried
cat file.txt | awk '{ if ($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.
I have large tab delimited file with 1000 columns. I want to rearrange so that certain columns have to be moved to the end.
Could anyone help using awk
Example input:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Move columns 5,6,7,8 to the end.
Output:
1 2 3 4 9 10 11 12 13 14 15 16 17 18 19 20 5 6 7 8
This prints columns 1 to a, then b to the last, and then columns a+1 to b-1:
$ awk -v a=4 -v b=9 '{for (i=1;i<=NF;i+=i==a?b-a:1) {printf "%s\t",$i};for (i=a+1;i<b;i++) {printf "%s\t",$i};print""}' file
1 2 3 4 9 10 11 12 13 14 15 16
17 18 19 20 5 6 7 8
The columns are moved in this way for every line in the input file, however many lines there are.
How it works
-v a=4 -v b=9
This defines the variables a and b which determine the limits on which columns will be moved.
for (i=1;i<=NF;i+=i==a?b-a:1) {printf "%s\t",$i}
This prints all columns except the ones from a+1 to b-1.
In this loop, i is incremented by one except when i==a in which case it is incremented by b-a so as to skip over the columns to be moved. This is done with awk's ternary statement:
i += i==a ? b-a : 1
+= simply means "add to." i==a ? b-a : 1 is the ternary statement. The value that it returns depends on whether i==a is true or false. If it is true, the value before the colon is returned. If it is false, the value after the colon is returned.
for (i=a+1;i<b;i++) {printf "%s\t",$i}
This prints columns a+1 to b-1.
print""
This prints a newline character to end the line.
Alternative solution that avoids printf
This approach assembles the output into the variable out and then prints with a plain print command, avoiding printf and the need for percent signs:
awk -v a=4 -v b=9 '{out="";for (i=1;i<=NF;i+=i==a?b-a:1) out=out $i"\t";for (i=a+1;i<b;i++) out=out $i "\t";print out}' file
One way to rearrange 2 columns ($5 become $20 and $20 become $5) the rest stay unchanged :
$ awk '{x=$5; $5=$20; $20=x; print}' file.txt
for 4 columns :
$ awk '{
x=$5; $5=$20; $9=x;
y=$9; $9=$10; $10=y;
print
}' file.txt
My approach:
awk 'BEGIN{ f[5];f[6];f[7];f[8] } \
{ for(i=1;i<=NF;i++) if(!(i in f)) printf "%s\t", $i; \
for(c in f) printf "%s\t", $c; printf "\n"} ' file
It's splitted in 3 parts:
The BEGIN{} part determines which field should be moved to the end. The indexes of the array f are moved. In the example it's 5, 6, 7 and 8.
Cycle trough every field (doesn't matter if there are 1000 fields or more) and check if they are in the array. If not print them.
Now we need the skipped fields. Cycle trough the f array and print those values.
Another way in awk
Switch last A-B with last N fields
awk -vA=4 -vB=8 '{x=B-A;for(i=A;i<=B;i++){y=$i;$i=$(t=(NF-x--));$t=y}}1' file
Put N rows from end into positon A
awk -vA=3 -vB=8 '{split($0,a," ");x=A++;while(x++<B)$x=a[NF-(B-x)];while(B++<NF)$B=a[A++]}1' file