I have a file with thousands of rows. I want to print the rows which do not contain a period.
awk '{print$2}' file.txt | head
I have used this to print the column I am interested in, column 2 (The file only has two columns).
I have removed the head and then did
awk '{print$2}' file.txt | grep -v "." | head
But I only get blank lines not any actual values which is expected, I think it has included the spaces between the rows but I am not sure.
Is there an alternative command?
As suggested by Jim, I did-
awk '{print$2}' file.txt | grep -v "\." | head
However the number of lines is greater than before, is this expected? Also, my output is a list of numbers but with spaces in between them (Vertical), is this normal?
file.txt example below-
120.4 3
270.3 7.9
400.8 3.9
200.2 4
100.2 8.7
300.2 3.4
102.3 6
49.0 2.3
38.0 1.2
So the expected (and correct) output would be 3 lines, as there is 3 values in column 2 without the period:
$ awk '{print$2}' file.txt | grep -v "\." | head
3
4
6
However, when running the code as above, I instead get 5, which is also counting the spaces between the rows I think:
$ awk '{print$2}' file.txt | grep -v "\." | head
3
4
6
You seldom need to use grep if you're already using awk
This would print the second column on each line where that second column doesn't contain a dot:
awk '$2 !~ /\./ {print $2}'
But you also wanted to skip empty lines, or perhaps ones where the second column is not empty. So just test for that, too:
awk '$2 != "" && $2 !~ /\./ {print $2}'
(A more amusing version would be awk '$2 ~ /./ && $2 !~ /\./ {print $2}' )
As you said, grep -v "." gives you only blank lines. That's because the dot means "any character", and with -v, the only lines printed are those that don't contain, well, any characters.
grep is interpreting the dot as a regex metacharacter (the dot will match any single character). Try escaping it with a backslash:
awk '{print$2}' file.txt | grep -v "\." | head
If I understand well, you can try this sed
sed ':A;N;${s/.*/&\n/};/\n$/!bA;s/\n/ /g;s/\([^ ]*\.[^ ]* \)//g' file.txt
output
3
4
6
Related
I have made a shell script for getting the list of mac address using awk and arp-scan command. I want to strip the mac address to only last 4 digits i.e (i want to print only the letters yy)
ac:1e:04:0e:yy:yy
ax:8d:5c:27:yy:yy
ax:ee:fb:55:yy:yy
dx:37:42:c9:yy:yy
cx:bf:9c:a4:yy:yy
Try cut -d: -f5-
(Options meaning: delimiter : and fields 5 and up.)
EDIT: Or in awk, as you requested:
awk -F: '{ print $5 ":" $6 }'
here are a few
line=cx:bf:9c:a4:yy:yy
echo ${line:(-5)}
line=cx:bf:9c:a4:yy:yy
echo $line | cut -d":" -f5-
I imagine you want to strip the trailing spaces, but it isn't clear whether you want yy:yy or yyyy.
Anyhow, there are multiple ways to it but you already are running AWK and have the MAC in $2.
In the first case it would be:
awk '{match($2,/([^:]{2}:[^:]{2}) *$/,m); print m[0]}'
yy:yy
In the second (no colon :):
awk 'match($2,/([^:]{2}):([^:]{2}) *$/,m); print m[1]m[2]}'
yyyy
In case you don't have match available in your AWK, you'd need to resort to gensub.
awk '{print gensub(/.*([^:]{2}:[^:]{2}) *$/,"\\1","g",$2)}'
yy:yy
or:
awk '{print gensub(/.*([^:]{2}):([^:]{2}) *$/,"\\1\\2","g",$0)}'
yyyy
Edit:
I now realized the trailing spaces were added by anubhava in his edit; they were not present in the original question! You can then simply keep the last n characters:
awk '{print substr($2,13,5)}'
yy:yy
or:
awk '{print substr($2,13,2)substr($2,16,2)}'
yyyy
Taking into account that the mac address always is 6 octets, you probably could just do something like this to get the last 2 octets:
awk '{print substr($0,13)}' input.txt
While testing on the fly by using arp -an I notice that the output was not always printing the mac addresses in some cases it was returning something like:
(169.254.113.54) at (incomplete) on en4 [ethernet]
Therefore probably is better to filter the input to guarantee a mac address, this can be done by applying this regex:
^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$
Applying the regex in awk and only printing the 2 last octecs:
arp -an | awk '{if ($4 ~ /^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$/) print substr($4,13)}'
This will filter the column $4 and verify that is a valid MAC address, then it uses substr to just return the last "letters"
You could also split by : and print the output in multiple ways, for example:
awk '{if ($4 ~ /^([0-9A-Fa-f]{2}[:-]){5}([0-9A-Fa-f]{2})$/) split($4,a,":"); print a[5] ":" a[6]}
Notice the exp ~ /regexp/
This is true if the expression exp (taken as a string) is matched by regexp.
The following example matches, or selects, all input records with the upper-case letter `J' somewhere in the first field:
$ awk '$1 ~ /J/' inventory-shipped
-| Jan 13 25 15 115
-| Jun 31 42 75 492
-| Jul 24 34 67 436
-| Jan 21 36 64 620
So does this:
awk '{ if ($1 ~ /J/) print }' inventory-shipped
I have an gz file which contains values in $12 and $33, where they contains strings (ex $12: 33-A and $33: 33A), I am trying to create an awk command that reads the values and counts the number of times "-" is in $12 but not in $13.
I have: gzcat test.gz | awk '{if ($12!=$33 && $12~/ -/ && $33!~/ -/) wc -l; else null} | wc -l'
But that command doesn't seem to work and get me the outcome I would like.
no need to check equality separately since it's implied, and no need to use wc, awk is capable of counting
... | awk '$12~/-/ && $33!~/-/{count++} END{print count+0}'
ps. your script is not a valid awk script. Also is the field 33 or 13?
I have a text file /tmp/some.txt with below values
JOHN YES 6 6 2345762
SHAUN NO 6 6 2345748
I want to create a csv file with below format (i.e based on rows. NOT based on columns).
JOHN,YES,6,6,2345762
SHAUN,NO,6,6,2345748
i tried below code
for i in `wc -l /tmp/some.txt | awk '{print $1}'`
do
awk 'NR==$i' /tmp/some.txt | awk '{print $1","$2","$3","$4","$5}' >> /tmp/some.csv
done
here wc -l /tmp/some.txt | awk '{print $1}' will get the value as 2 (i.e 2 rows in text file).
and for each row awk 'NR==$i' /tmp/some.txt | awk '{print $1","$2","$3","$4","$5}' will print the 5 fields into some.csvfile which is separated by comma.
when i execute each command separately it will work. but when i make it as a shell script i'm getting empty some.csv file.
#Kart: Could you please try following.
awk '{$1=$1;} 1' OFS=, Input_file > output.csv
I hope this helps you.
I suggest:
sed 's/[[:space:]]\+/,/g' /tmp/some.txt
You almost got it. awk already process the file row by row, so you don't need to iterate with the for loop.
So you just need to run:
awk '{print $1","$2","$3","$4","$5}' /tmp/some.txt >> /tmp/some.csv
With tr, squeezing (-s), and then transliterating space/tab ([:blank:]):
tr -s '[:blank:]' ',' <file.txt
With sed, substituting one or more space/tab with ,:
sed 's/[[:blank:]]\+/,/g' file.txt
With awk, replacing one ore more space/tab with , using gsub() function:
awk 'gsub("[[:blank:]]+", ",", $0)' file.txt
Example
% cat foo.txt
JOHN YES 6 6 2345762
SHAUN NO 6 6 2345748
% tr -s '[:blank:]' ',' <foo.txt
JOHN,YES,6,6,2345762
SHAUN,NO,6,6,2345748
% sed 's/[[:blank:]]\+/,/g' foo.txt
JOHN,YES,6,6,2345762
SHAUN,NO,6,6,2345748
% awk 'gsub("[[:blank:]]+", ",", $0)' foo.txt
JOHN,YES,6,6,2345762
SHAUN,NO,6,6,2345748
Let's say I have the following line in my file:
HELLO,1410250216446000,1410250216470330,1410250216470367,329,PE,B,T,GALU,[ , , T, I],3.38,3,A,A, , , , ,0, ,0,0, ,-Infinity,-Infinity,-Infinity, ,,0
if I use
grep -a -w HELLO my_file | head -10 | awk -F '[\t,]' '{print NF}' | less
output is 32.
But I don't want to count the commas within []. I mean [ , , T, I] must be counted as a single word. So that the output of my query is 29.
What will be one line command for doing this in Linux?
Remove content inside brackets using sed. Then continue counting
grep -a -w HELLO my_file|sed "s/\[.*\]//g" | head -10 | awk -F '[\t,]' '{print NF}' | less
output
29
I want to find the last appearance of a string in a text file with linux commands. For example
1 a 1
2 a 2
3 a 3
1 b 1
2 b 2
3 b 3
1 c 1
2 c 2
3 c 3
In such a text file, i want to find the line number of the last appearance of b which is 6.
I can find the first appearance with
awk '/ b / {print NR;exit}' textFile.txt
but I have no idea how to do it for the last occurrence.
cat -n textfile.txt | grep " b " | tail -1 | cut -f 1
cat -n prints the file to STDOUT prepending line numbers.
grep greps out all lines containing "b" (you can use egrep for more advanced patterns or fgrep for faster grep of fixed strings)
tail -1 prints last line of those lines containing "b"
cut -f 1 prints first column, which is line # from cat -n
Or you can use Perl if you wish (It's very similar to what you'd do in awk, but frankly, I personally don't ever use awk if I have Perl handy - Perl supports 100% of what awk can do, by design, as 1-liners - YMMV):
perl -ne '{$n=$. if / b /} END {print "$n\n"}' textfile.txt
This can work:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
We check every second file being "b" and we record the number of line. It is appended, so by the time we finish reading the file, it will be the last one.
Test:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
6
Update based on sudo_O advise:
$ awk '{if ($2=="b") a=NR} END{print a}' your_file
to avoid having some abc in 2nd field.
It is also valid this one (shorter, I keep the one above because it is the one I thought :D):
$ awk '$2=="b" {a=NR} END{print a}' your_file
Another approach if $2 is always grouped (may be more efficient then waiting until the end):
awk 'NR==1||$2=="b",$2=="b"{next} {print NR-1; exit}' file
or
awk '$2=="b"{f=1} f==1 && $2!="b" {print NR-1; exit}' file