Confusion about awk command when dealing with if statement - awk

$ cat awk.txt
12 32 45
5 2 3
33 11 33
$ cat awk.txt | awk '{FS='\t'} $1==5 {print $0}'
5 2 3
$ cat awk.txt | awk '{FS='\t'} $1==33 {print $0}'
Nothing is returned when judging the first field is 33 or not. It's confusing.

By saying
awk '{FS='\t'} $1==5 {print}' file
You are defining the field separator incorrectly. To make it be a tab, you need to say "\t" (with double quotes). Further reading: awk not capturing first line / separator.
Also, you are setting it every line, so it does not affect the first one. You want to use:
awk 'BEGIN{FS='\t'} $1==5' file
Yes, but why did it work in one case but not in the other?
awk '{FS='\t'} $1==5' file # it works
awk '{FS='\t'} $1==33' file # it does not work
You're using single quotes around '\t', which means that you're actually concatenating 3 strings together: '{FS=', \t and '} $1==5' to produce your awk command. The shell interprets the \t as t, so your awk script is actually:
awk '{FS=t} $1==5'
The variable t is unset, so you're setting the field separator to the empty string "". This means that the line is split into as many fields as characters you have. You can see it doing awk 'BEGIN{FS='\t'} {print NF}' file, that will show how many fields each record has.
Then, $1 is just 3 and $2 contains the second 3.

first of all !. Could you explain better what you really want to do before you ask ?. look....!
more awk.txt
12 32 45
5 2 3
33 11 33
awk -F"[ \t]" '$1 == 5 { print $0}' awk.txt
5 2 3
awk -F"[ \t]" '$1 == 33 { print $0}' awk.txt
33 11 33
awk -F"[ \t]" '$1 == 12 { print $0}' awk.txt
12 32 45
http://www.staff.science.uu.nl/~oostr102/docs/nawk/nawk_23.html
Fcs

Related

Excise EOF character from AWK processing

Watch this:
echo one two three | awk 'BEGIN {RS=" "; FS=""} {print NF}'
3
3
6
That last count is suspicious. Let's have a look at the last field of each record:
echo one two three | awk 'BEGIN {RS=" "; FS=""} {print $NF}'
e
o
e
I will take a wild guess and assume AWK is picking up some sort of EOF character after the last field of the last record.
How do I tell AWK to ignore such extraneous character?
The input stream includes a trailing \n.
By default awk will treat that trailing \n as the record separator (as opposed to part of the data).
In your code you've redefined awk's record separator as a space, thus leaving the \n to be treated as part of the input data.
If your version of awk supports multiple record separators you could define a dual record separator of space and \n, eg:
$ echo one two three | awk 'BEGIN {RS="[ \n]"; FS=""} {print NF}'
3
3
5
Or, as mentioned in the comment, you could explicitly remove the \n, eg:
$ echo one two three | awk 'BEGIN {RS=" "; FS=""} {sub(/\n/,"");print NF}'
3
3
5
Alternatively you could strip the \n before calling awk, eg:
$ echo one two three | tr -d '\n' | awk 'BEGIN {RS=" "; FS=""} {print NF}'
3
3
5
Or you could even skip the generation of the \n in the first place, eg:
$ printf "one two three" | awk 'BEGIN {RS=" "; FS=""} {print NF}'
3
3
5

Sort a file preserving the header as first position with bash

When sorting a file, I am not preserving the header in its position:
file_1.tsv
Gene Number
a 3
u 7
b 9
sort -k1,1 file_1.tsv
Result:
a 3
b 9
Gene Number
u 7
So I am tryig this code:
sed '1d' file_1.tsv | sort -k1,1 > file_1_sorted.tsv
first='head -1 file_1.tsv'
sed '1 "$first"' file_1_sorted.tsv
What I did is to remove the header and sort the rest of the file, and then trying to add again the header. But I am not able to perform this last part, so I would like to know how can I copy the header of the original file and insert it as the first row of the new file without substituting its actuall first row.
You can do this as well :
{ head -1; sort; } < file_1.tsv
** Update **
For macos :
{ IFS= read -r header; printf '%s\n' "$header" ; sort; } < file_1.tsv
a simpler awk
$ awk 'NR==1{print; next} {print | "sort"}' file
$ head -1 file; tail -n +2 file | sort
Output:
Gene Number
a 3
b 9
u 7
Could you please try following.
awk '
FNR==1{
first=$0
next
}
{
val=(val?val ORS:"")$0
}
END{
print first
print val | "sort"
}
' Input_file
Logical explanation:
Check condition FNR==1 to see if its first line; then save its values to variable and move on to next line by next.
Then keep appending all lines values to another variable with new line till last line.
Now come to END block of this code which executes when Input_file is done being read, there print first line value and put sort command on rest of the lines value there.
This will work using any awk, sort, and cut in any shell on every UNIX box and will work whether the input is coming from a pipe (when you can't read it twice) or from a file (when you can) and doesn't involve awk spawning a subshell:
awk -v OFS='\t' '{print (NR>1), $0}' file | sort -k1,1n -k2,2 | cut -f2-
The above uses awk to stick a 0 at the front of the header line and a 1 in front of the rest so you can sort by that number then whatever other field(s) you want to sort on and then remove the added field again with a cut. Here it is in stages:
$ awk -v OFS='\t' '{print (NR>1), $0}' file
0 Gene Number
1 a 3
1 u 7
1 b 9
$ awk -v OFS='\t' '{print (NR>1), $0}' file | sort -k1,1n -k2,2
0 Gene Number
1 a 3
1 b 9
1 u 7
$ awk -v OFS='\t' '{print (NR>1), $0}' file | sort -k1,1n -k2,2 | cut -f2-
Gene Number
a 3
b 9
u 7

awk: print each column of a file into separate files

I have a file with 100 columns of data. I want to print the first column and i-th column in 99 separate files, I am trying to use
for i in {2..99}; do awk '{print $1" " $i }' input.txt > data${i}; done
But I am getting errors
awk: illegal field $(), name "i"
input record number 1, file input.txt
source line number 1
How to correctly use $i inside the {print }?
Following single awk may help you too here:
awk -v start=2 -v end=99 '{for(i=start;i<=end;i++){print $1,$i > "file"i;close("file"i)}}' Input_file
An all awk solution. First test data:
$ cat foo
11 12 13
21 22 23
Then the awk:
$ awk '{for(i=2;i<=NF;i++) print $1,$i > ("data" i)}' foo
and results:
$ ls data*
data2 data3
$ cat data2
11 12
21 22
The for iterates from 2 to the last field. If there are more fields that you desire to process, change the NF to the number you'd like. If, for some reason, a hundred open files would be a problem in your system, you'd need to put the print into a block and add a close call:
$ awk '{for(i=2;i<=NF;i++){f=("data" i); print $1,$i >> f; close(f)}}' foo
If you want to do what you try to accomplish :
for i in {2..99}; do
awk -v x=$i '{print $1" " $x }' input.txt > data${i}
done
Note
the -v switch of awk to pass variables
$x is the nth column defined in your variable x
Note2 : this is not the fastest solution, one awk call is fastest, but I just try to correct your logic. Ideally, take time to understand awk, it's never a wasted time

Print every second consequtive field in two columns - awk

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

How to print last two columns using awk

All I want is the last two columns printed.
You can make use of variable NF which is set to the total number of fields in the input record:
awk '{print $(NF-1),"\t",$NF}' file
this assumes that you have at least 2 fields.
awk '{print $NF-1, $NF}' inputfile
Note: this works only if at least two columns exist. On records with one column you will get a spurious "-1 column1"
#jim mcnamara: try using parentheses for around NF, i. e. $(NF-1) and $(NF) instead of $NF-1 and $NF (works on Mac OS X 10.6.8 for FreeBSD awkand gawk).
echo '
1 2
2 3
one
one two three
' | gawk '{if (NF >= 2) print $(NF-1), $(NF);}'
# output:
# 1 2
# 2 3
# two three
using gawk exhibits the problem:
gawk '{ print $NF-1, $NF}' filename
1 2
2 3
-1 one
-1 three
# cat filename
1 2
2 3
one
one two three
I just put gawk on Solaris 10 M4000:
So, gawk is the cuplrit on the $NF-1 vs. $(NF-1) issue. Next question what does POSIX say?
per:
http://www.opengroup.org/onlinepubs/009695399/utilities/awk.html
There is no direction one way or the other. Not good. gawk implies subtraction, other awks imply field number or subtraction. hmm.
Please try this out to take into account all possible scenarios:
awk '{print $(NF-1)"\t"$NF}' file
or
awk 'BEGIN{OFS="\t"}' file
or
awk '{print $(NF-1), $NF} {print $(NF-1), $NF}' file
try with this
$ cat /tmp/topfs.txt
/dev/sda2 xfs 32G 10G 22G 32% /
awk print last column
$ cat /tmp/topfs.txt | awk '{print $NF}'
awk print before last column
$ cat /tmp/topfs.txt | awk '{print $(NF-1)}'
32%
awk - print last two columns
$ cat /tmp/topfs.txt | awk '{print $(NF-1), $NF}'
32% /