I have a text file 2 fields separated by :
i3583063:b3587412
i3583064:b3587412
i3583065:b3587412
i3583076:b3587421
i3583077:b3587421
i3583787:b3587954
i3584458:b3588416
i3584459:b3588416
i3584460:b3588416
i3584461:b3588416
i3584462:b3588416
i3584463:b3588416
i3584464:b3588416
i3584465:b3588416
Field 1 is always uniq but not field 2 it can be repeated. How can I identify first, 2nd 3rd etc. occurrence of field 2? Can I use count?
Thanks
I don't know if I've ever heard of a standard Unix count utility, but you can do this with Awk. Here's an Awk script that adds the count as a third column:
awk -F: 'BEGIN {OFS=":"} {$3=++count[$2]; print}' input.txt
It should generate the output:
i3583063:b3587412:1
i3583064:b3587412:2
i3583065:b3587412:3
i3583076:b3587421:1
i3583077:b3587421:2
i3583787:b3587954:1
i3584458:b3588416:1
i3584459:b3588416:2
i3584460:b3588416:3
i3584461:b3588416:4
i3584462:b3588416:5
i3584463:b3588416:6
i3584464:b3588416:7
i3584465:b3588416:8
The heart of the script {$3=++count[$2]; print} simply increments a counter indexed by the value of the second field, stores it in a new third field, and then outputs the line with this new field. Awk is a great little language and still well worth learning.
You can use the sort command with the -u parameter. This way redundant lines are removed.
sort -u filename.txt
If you want to count occurrences
sort -u filename.txt | wc -l
Related
FILE1.TXT
0020220101
or
01 20220101
Need to extra date part from file where text starts from 2
Options tried:
t_FILE_DT1='awk -F"2" '{PRINT $NF}' FILE1.TXT'
t_FILE_DT2='cut -d'2' -f2- FILE1.TXT'
echo "$t_FILE_DT1"
echo "$t_FILE_DT2"
1st output : 0101
2nd output : 0220101
Expected Output: 20220101
Im new to linux scripting. Could some one help guide where Im going wrong?
Use grep like so:
echo "0020220101\n01 20220101" | grep -P -o '\d{8}\b'
20220101
20220101
Here, GNU grep uses the following options:
-P : Use Perl regexes.
-o : Print the matches only (1 match per line), not the entire lines.
SEE ALSO:
grep manual
perlre - Perl regular expressions
Using any awk:
$ awk '{print substr($0,length()-7)}' file
20220101
20220101
The above was run on this input file:
$ cat file
0020220101
01 20220101
Regarding PRINT $NF in your question - PRINT != print. Get out of the habit of using all-caps unless you're writing Cobol. See correct-bash-and-shell-script-variable-capitalization for some reasons.
The 2 in your scripts is telling awka and cut to use the character 2 as the field separator so each will carve up the input into substrings everywhere a 2 occurs.
The 's in your question are single quotes used to make strings literal, you were intending to use backticks, `cmd`, but those are deprecated in favor of $(cmd) anyway.
I would instead of looking for "after" the 2 .. (not having to worry about whether there is a space involved as well) )
Think instead about extracting the last 8 characters, which you know for fact is your date ..
input="/path/to/txt/file/FILE1.TXT"
while IFS= read -r line
do
# read in the last 8 characters of $line .. You KNOW this is the date ..
# No need to worry about exact matching at that point, or spaces ..
myDate=${line: -8}
echo "$myDate"
done < "$input"
About the cut and awk commands that you tried:
Using awk -F"2" '{PRINT $NF}' file will set the field separator to 2, and $NF is the last field, so printing the value of the last field is 0101
Using cut -d'2' -f2- file uses a delimiter of 2 as well, and then print all fields starting at the second field, which is 0220101
If you want to match the 2 followed by 7 digits until the end of the string:
awk '
match ($0, /2[0-9]{7}$/) {
print substr($0, RSTART, RLENGTH)
}
' file
Output
20220101
The accepted answer shows how to extract the first eight digits, but that's not what you asked.
grep -o '2.*' file
will extract from the first occurrence of 2, and
grep -o '2[0-9]*' file
will extract all the digits after every occurrence of 2. If you specifically want eight digits, try
grep -Eo '2[0-9]{7}'
maybe also with a -w option if you want to only accept a match between two word boundaries. If you specifically want only digits after the first occurrence of 2, maybe try
sed -n 's/[^2]*\(2[0-9]*\).*/\1/p' file
Suppose i have | delimeted file,
Line1: 1|2|3|4
Line2: 5|6|7|8
Line3: 9|9|1|0
Now i need to read 3 field at second line which is 7 in above example how i can do that using Cut or Sed Command. I'm new to unix please help
A job for awk:
awk -F '|' 'NR==2{print $3}' file
or
awk -F '|' -v row=2 -v col=3 'NR==row{print $col}' file
Output:
7
This should work:
sed -n '2p' file |awk -F '|' '{print $3}'
This might work for you (GNU sed):
sed -rn '2s/^(([^|]*)\|?){3}.*/\2/p' file
Turn off automatic printing by setting the -n option, turn on easier regexp declaration by -r option. Use pattern matching and back references to replace the whole of the second line by the third field of the same line and print the result.
The address of the substitution command is limited to only the second line.
The regexp groups the non-delimited characters followed by a delimiter a specific number of times. The second group, only retains the non-delimited characters for the specific number. Each grouping is replaced by the next and so the last grouping is reported, the .* consumes the remainder of the line and so only the third field (contents of second group) is printed.
N.B. the delimiter would be present following the final column and is therefore optional \|?
I need to find a duplicate entry in 2 different columns and keep only one of the duplicate and all unique entries. For me if A123 is in the first column and it show up later in the second column it's a duplicate. I also know for sure that A123 will always be paired to B123 by either being A123,B123 or B123,A123. I only need to keep one and it doesn't matter which one it is.
Ex: My input file would contain:
A123,B123
A234,B234
C123,D123
B123,A123
B234,A234
I'd like the output to be:
A123,B123
A234,B234
C123,D123
The best I can do is to extract the unique entries with :
awk -F',' 'NR==FNR{x[$1]++;next}; !x[$2]' file1 file1
or get only the duplicates with
awk -F',' 'NR==FNR{x[$1]++;next}; x[$2]' file1 file1
Any help would be greatly appreciated.
This can be shorter!
First print if the element is not yet present in the array. Then add the first field to the array. Only one run over the inputfile is necessary:
awk -F, '!x[$2];{x[$1]++}' file1
This awk one-liner works for your example:
awk -F, '!($2 in a){a[$1]=$0}END{for(x in a)print a[x]}' file
The conventional, idiomatic awk solution:
$ awk -F, '!seen[$1>$2 ? $1 : $2]++' file
A123,B123
A234,B234
C123,D123
By convention we always use seen (rather than x or anything else) as the array name when it represents a set where you want to check if it's index has been seen before, and using a ternary expression to produce the largest of the possible key values as the index ensures the order they appear in the input doesn't matter.
The above doesn't care about your unique situation where every $2 is paired to a specific $1 - it simply prints unique individual occurrences across a pair of fields. If you wanted it to work on the pair of fields combined (and assuming you have more fields so just using $0 as the index wouldn't work) that'd be:
awk -F, '!seen[$1>$2 ? $1 FS $2 : $2 FS $1]++' file
I need to calculate median value for the below input file. It is working fine for odd occurrences but not for even occurrences. Below is the input file and the script used. Could you please check what is wrong with this command and correct the same.
Input file:
col1,col2
AR,2.52
AR,3.57
AR,1.29
AR,6.66
AR,3.05
AR,5.52
Desired Output:
AR,3.31
Unix command:
cat test.txt | sort -t"," -k2n,2 | awk '{arr[NR]=$1} END { if (NR%2==1) print arr[(NR+1)/2]; else print (arr[NR/2]+arr[NR/2+1])/2}'
Don't forget that your input file has an additional line, containing the header. You need to take an additional step in your awk script to skip the first line.
Also, due to the fact you're using the default field separator, $1 will contain the whole line, so your code arr[NR/2]+arr[NR/2+1])/2 is never going to work. I would suggest that you changed it so that awk splits the input on a comma, then use the second field $2.
sort -t, -k2n,2 file | awk -F, 'NR>1{a[++i]=$2}END{if(i%2==1)print a[(i+1)/2];else print (a[i/2]+a[i/2+1])/2}'
I also removed your useless use of cat. Most tools, including sort and awk, are capable of reading in files directly, so you don't need to use cat with them.
Testing it out:
$ cat file
col1,col2
AR,2.52
AR,3.57
AR,1.29
AR,6.66
AR,3.05
AR,5.52
$ sort -t, -k2n,2 file | awk -F, 'NR>1{a[++i]=$2}END{if(i%2==1)print a[(i+1)/2];else print (a[i/2]+a[i/2+1])/2}'
3.31
It shouldn't be too difficult to modify the script slightly to change the output to whatever you want.
I have a file that is '|' delimited. One of the fields within the file is a time stamp. The field is in the following format: MM-dd-yyyy HH:mm:ss I'd like to be able to print to a file unique dates. I can use the cut command (cut -f1 -d'|' _file_name_ |sort|uniq) to extract unique dates. However, with the time portion of the field, I'm seeing hundreds of results. After I run the cut command, I'd like to take the substring of the first eleven characters to display unique dates. I tried using an awk command such as:
awk ' { print substr($1,1-11) }' | cut -f1 -d'|' _file_name_ |sort|uniq > _output_file_
I'm having no luck. Am I going about this the wrong way? Is there a more simple way of extracting the data I need. Any help would be appreciated.
cut -c1-11 will display characters 1-11 of each input line.
if the date is the first (space separated) field in the file, then the list of unique dates is just:
cut -f1 -d' ' filename | sort -u
Update: in addition to #shellter's correct answer, I'll just present an alternative to demonstrate other awk facilities:
awk '{split($10, a); date[a[1]]++} END {for (d in date) print d}' filename
You're all most there. This is based on the idea that the date time stamp is in field 1.
Edit : changed field to 10, also used -u option to sort instead of sep process with uniq
You don't need the cut, awk will do that for you.
awk -F"|" ' { print substr($10,1,11) }' _file_name_ |sort -u > _output_file_
I hope this helps.
P.S. as you appear to be a new user, if you get an answer that helps you please remember to mark it as accepted, or give it a + (or -) as a useful answer