I would like to compare if two columns (mid) in a file is identical to each other. I am not sure of how to do it...Since the original file that I am working one is rather huge (in Gb)
file 1 (column1 and column 4 - to check if they are identical)
mid A1 A2 mid A3 A4 A5 A6
18 we gf 18 32 23 45 89
19 ew fg 19 33 24 46 90
21 ew fg 21 35 26 48 92
Thanks
M
if you just need to find the different row, awk will do,
awk '$1!=$4{print $1,$4}' data
You can check using diff and awk for advance difference.
diff <(awk '{print $1}' data) <(awk '{print $4}' data)
The status code ($?)of this command will tell if they are same (zero) or different (non-zero).
You can use that in base expression like this too,
if diff <(awk '{print $1}' data) <(awk '{print $4}' data) >& /dev/null;
then
echo same;
else
echo different;
fi
Something like this:
awk '{ if ($1 == $4) { print "same"; } else { print "different"; } }' < foo.txt
Completing a litle bit the Shiplu Mokaddim answer, if you have another delimiter, for example in a csv file, you can use:
awk -F; '$1!=$4{print $1,$4}' data.csv | sed -r 's/ /;/g'
In this sample, the delimiter is a ";". The sed command in the end is to replace again the delimiter to the original one. Be sure that you don´t have another space in you answer, i.e. and date time.
Question: Compare two columns value in the same file.
Answer:
cut -d, -f1 a.txt > b.txt ; cut -d, -f3 a.txt > c.txt ; cmp b.txt c.txt && echo "Column values are same"; rm -rf b.txt c.txt
Related
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
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 ($1 ~ /J/) print }' inventory-shipped
I want to sum all the numbers in a file (columns and lines) given by the first parameter, but my program shows sum=sum+$i instead of the numeric sum:
sum=0;
file=$1
for i in $file
do
sum=sum+$i;
done;
echo "The sum is: " $sum
Input file:
$cat file.txt
10 20 10
40
50
Expected output :
The sum is: 21
Maybe if there is an awk method to solve this?
Try this -
$cat file1.txt
10 20 10
40
50
$awk '{for(i=1;i<=NF;i++) {sum+=$i}} END {print sum}' file1.txt
130
OR
$xargs < file1.txt| tr ' ' + | bc
130
cat file.txt | xargs | sed -e 's/\ /+/g' | bc
You can also use a simple read and an array to sum the value relying on word splitting to separate the values into an array via the default IFS (Internal Field Separator), e.g.
#!/bin/bash
declare -i sum=0
fn="${1:-/dev/stdin}" ## read from file as 1st argument (default stdin)
while read -r line; do ## read each line
a=( $line ) ## separate values into array
for i in ${a[#]}; do ## for each value in array
((sum += i)) ## add to sum
done
done <"$fn"
echo "sum: $sum"
Example Input File
$ cat dat/numfile.txt
10 20 10
40
50
Example Use/Output
$ bash sumnumfile.sh dat/numfile.txt
sum: 130
Another for some awks (at least mawk and gawk):
$ awk -v RS="[^0-9]" '{s+=$1}END{print s}' file
130
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
I have two sets of text files. First set is in AA folder. Second set is in BB folder. The content of ff.txt file from first set(AA folder) is shown below.
Name number marks
john 1 60
maria 2 54
samuel 3 62
ben 4 63
I would like to print the second column(number) from this file if marks>60. The output will be 3,4. Next, read the ff.txt file in BB folder and delete the lines containing numbers 3,4. How can I do this with bash?
files in BB folder looks like this. second column is the number.
marks 1 11.824 24.015 41.220 1.00 13.65
marks 1 13.058 24.521 40.718 1.00 11.82
marks 3 12.120 13.472 46.317 1.00 10.62
marks 4 10.343 24.731 47.771 1.00 8.18
awk 'FNR == NR && $3 > 60 {array[$2] = 1; next} {if ($2 in array) next; print}' AA/ff.txt BB/filename
This works, but is not efficient (is that matter?)
gawk 'BEGIN {getline} $3>60{print $2}' AA/ff.txt | while read number; do gawk -v number=$number '$2 != number' BB/ff.txt > /tmp/ff.txt; mv /tmp/ff.txt BB/ff.txt; done
Of course, the second awk can be replaced with sed -i
For multi files:
ls -1 AA/*.txt | while read file
do
bn=`basename $file`
gawk 'BEGIN {getline} $3>60{print $2}' AA/$bn | while read number
do
gawk -v number=$number '$2 != number' BB/$bn > /tmp/$bn
mv /tmp/$bn BB/$bn
done
done
I didn't test it, so if there is a problem, please comment.