awk combine 2 commands for csv file formatting - shell

I have a CSV file which has 4 columns. I want to first:
print the first 10 items of each column
only print the items in the third column
My method is to pipe the first awk command into another but i didnt get exactly what i wanted:
awk 'NR < 10' my_file.csv | awk '{ print $3 }'


The only missing thing was the -F.
awk -F "," 'NR < 10' my_file.csv | awk -F "," '{ print $3 }'


You don't need to run awk twice.
awk -F, 'NR<=10{print $3}'
This prints the third field for every line whose record number (line) is less than or equal to 10.
Note that < is different from <=. The former matches records one through nine, the latter matches records one through ten. If you need ten records, use the latter.
Note that this will walk through your entire file, so if you want to optimize your performance:
awk -F, '{print $3} NR>10{exit}'
This will print the third column. Then if the record number is greater than 10, it will exit. This does not step through your entire file.
Note also that awk's "CSV" matching is very simple; awk does not understand quoted fields, so the record:
red,"orange,yellow",green
has four fields, two of which have double quotes in them. YMMV depending on your input.

Related

Searching for a string between two characters

I need to find two numbers from lines which look like this
>Chr14:453901-458800
I have a large quantity of those lines mixed with lines that doesn't contain ":" so we can search for colon to find the line with numbers. Every line have different numbers.
I need to find both numbers after ":" which are separated by "-" then substract the first number from the second one and print result on the screen for each line
I'd like this to be done using awk
I managed to do something like this:
awk -e '$1 ~ /\:/ {print $0}' file.txt
but it's nowhere near the end result
For this example i showed above my result would be:
4899
Because it is the result of 458800 - 453901 = 4899
I can't figure it out on my own and would appreciate some help

With GNU awk. Separate the row into multiple columns using the : and - separators. In each row containing :, subtract the contents of column 2 from the contents of column 3 and print result.
awk -F '[:-]' '/:/{print $3-$2}' file
Output:
4899

Using awk
$ awk -F: '/:/ {split($2,a,"-"); print a[2] - a[1]}' input_file
4899

How to call awk from bash with last field as variable

Suppose I have a CSV input file called input.csv:
1,2,3,valueIWantToPrint
3,5,2,valueIWantToPrint
I currently need to print the last element of each row of that input with awk, which is easily accomplished with the field separator and NF variables:
awk -F"," '{print $NF}' input.csv
But now let's say that I want to make the field that I want to print a variable, because later perhaps the input format will change and it will be a different field.
Input file:
1,2,valueIWantToPrint,3
3,5,valueIWantToPrint,2
Script:
FIELD_TO_PRINT=3
awk -F"," -v fieldToPrint=FIELD_TO_PRINT '{print $fieldToPrint}' input.csv
Ok, that was easy. But now to make it as flexible as possible, I would like have the ability to set FIELD_TO_PRINT to the equivalent of NF so that I can print the last value regardless of the number of fields. What I'm after is something like this:
Input:
1,2,3,7,2,5,23,1,3,6,valueIWantToPrint
3,5,2,6,3,valueIWantToPrint
This script doesn't work, but illustrates what I am trying to accomplish:
FIELD_TO_PRINT=NF
awk -F"," -v fieldToPrint=FIELD_TO_PRINT '{print $fieldToPrint}' input.csv
Is there a convenient way to set a variable to mean "the last field in record?" This example is pretty trivial, but ultimately the FIELD_TO_PRINT variable will be put in a separate configuration file, and the awk script will be much larger and more complex. So having a good way to accomplish this will be very useful.

You can use this round-about way:
Negative n means NF:
n=-1
awk -F, -v n="$n" '{print (n<0 ? $NF : $n)}' f.csv
valueIWantToPrint
valueIWantToPrint
and when n > 0 print numbered field:
n=3
awk -F, -v n="$n" '{print (n<0 ? $NF : $n)}' f.csv
3
2

you can use this trick
$ awk -F, -v n=-1 '{print (n<0)?$(NF+n+1):$n}' file
valueIWantToPrint
valueIWantToPrint
assume negative indices start counting from NF backwards, so -2 will mean the penultimate field etc.

print 3 consecutive column after specific string from CSV

I need to print 2 columns after specific string (in my case it is 64). There can be multiple instances of 64 within same CSV row, however next instance will not occur within 3 columns of previous occurrence. Output of each instance should be in next line and unique. The problem is, the specific string does not fall in same column for all rows. All row is having kind of dynamic data and there is no header for CSV. Let say, below is input file (its just a sample, actual file is having approx 300 columns & 5 Million raws):
00:TEST,123453103279586,ABC,XYZ,123,456,65,906,06149,NIL TS21,1,64,906,06149,NIL TS22,1,64,916,06149,NIL BS20,1,64,926,06149,NIL BS30,1,64,906,06149,NIL CAML,1,ORIG,0,TERM,1,1,1,6422222222
00:TEST,123458131344169,ABC,XYZ,123,456,OCCF,1,1,1,64,857,19066,NIL TS21,1,64,857,19066,NIL TS22,1,64,857,19066,NIL BS20,1,64,857,19067,NIL BS30,1,64,857,19068,NIL PSS,1,E2 EPSDATA,GRANTED,NONE,1,N,N,256000,5
00:TEST,123458131016844,ABC,XYZ,123,456,HOLD,,1,64,938,36843,NIL TS21,1,64,938,36841,NIL TS22,1,64,938,36823,NIL BS20,1,64,938,36843,NIL BS30,1,64,938,36843,NIL CAML,1,ORIG,0,TERM,00,50000,N,N,N,N
00:TEST,123453102914690,ABC,XYZ,123,456,HOLD,,1,PBS,TS11,64,938,64126,NIL TS21,1,64,938,64126,NIL TS22,1,64,938,64126,NIL BS20,1,64,938,64226,NIL BS30,1,64,938,64326,NIL CAML,1,ORIG,0,TERM,1,1,1,6422222222,2222,R
Output required(only unique entries):
64,906,06149
64,857,19066
64,857,19067
64,857,19068
64,938,36843
64,938,36841
64,938,36823
64,938,36843
64,938,36843
64,938,64326
There is no performance related concerns. I have tried to search many threads but could not get anything near related. Please help.

We can use a pipe of two commands... first to put the 64's leading on a line and a second to print first three columns if we see a leading 64.
sed 's/,64[,\n]/\n64,/g' | awk -F, '/^64/ { print $1 FS $2 FS $3 }'
There are ways of doing this with a single awk command, but this felt quick and easy to me.
Though the sample data from the question contains redundant lines, karakfa (see below) reminds me that the question speaks of a "unique data" requirement. This version uses the keys of an associative array to keep track of duplicate records.
sed 's/,64[,\n]/\n64,/g' | awk -F, 'BEGIN { split("",a) } /^64/ && !((x=$1 FS $2 FS $3) in a) { a[x]=1; print x }'

gawk:
awk -F, '{for(i=0;++i<=NF;){if($i=="64")a=4;if(--a>0)s=s?s","$i:$i;if(a==1){print s;s=""}}}' file

Sed for fun
sed -n -e 's/$/,n,n,n/' -e ':a' -e 'G;s/[[:blank:],]\(64,.*\)\(\n\)$/\2\1/;s/.*\(\n\)\(64\([[:blank:],][^[:blank:],]\{1,\}\)\{2\}\)\([[:blank:],][^[:blank:],]\{1,\}\)\{3\}\([[:blank:],].*\)\{0,1\}$/\1\2\1\5/;s/^.*\n\(.*\n\)/\1/;/^64.*\n/P;s///;ta' YourFile | sort -u
assuming column are separated by blank space or comma
need a sort -u for uniq (possible in sed but a new "simple" action of the same kind to add in this case)

awk to the rescue!
$ awk -F, '{for(i=1;i<=NF;i++)
if($i==64)
{k=$i FS $(++i) FS $(++i);
if (!a[k]++)
print k
}
}' file
64,906,06149
64,916,06149
64,926,06149
64,857,19066
64,857,19067
64,857,19068
64,938,36843
64,938,36841
64,938,36823
64,938,64126
64,938,64226
64,938,64326
ps. your sample output doesn't match the given input.

Get the contents of one column given another column

I have a tab separated file with 3 columns. I'd like to get the contents of the first column, but only for the rows where the 3rd column is equal to 8. How do I extract these values? If I just wanted to extract the values in the first column, I would do the following:
cat file1 | tr "\t" "~" | cut -d"~" -f1 >> file_with_column_3
I'm thinking something like:
cat file1 | tr "\t" "~" | if cut -d"~" -f3==8; then cut -d"~" -f1 ; fi>> file_with_column_3
But that doesn't quite seem to work.

Given that your file is tab delimited, it seems like this problem would be well suited for awk.
Something simple like below should work for you, though without any sample data I can't say for sure (try to always include this on questions on SO)
awk -F'\t' '$3==8 {print $1}' inputfile > outputfile
The -F'\t' sets the input delimiter as tab.
$3==8 compares if the 3rd column based on that delimiter is 8.
If so, the {print $1} is executed, which prints the first column.
Otherwise, nothing is done and awk proceeds to the next line.
If your file had a header you wanted to preserve, you could just modify this like the following, which tells awk to print if the current record number is 1.
awk -F'\t' 'NR==1 {print;} $3==8 {print $1}' inputfile > outputfile

awk can handle this better:
awk -F '\t' '$3 == 8 { print $1 }' file1

You can do it with bash only too:
cat x | while read y; do split=(${y}); [ ${split[2]} == '8' ] && echo $split[0]; done
The input is read in variable y, then split into an array. The IFS (input field separator) defaults to <space><tab<>newline>, so it splits on tabs too. The third field of the array is then compared to '8'. If it equals, it prints the first field of the array. Remember that fields in arrays start counting at zero.

Filtering /etc/passwd in Unix

I'd like to filter the content of /etc/passwd, only showing the lines for which the value in the third column is greater than 999.
Is there an easy way to do this with a one liner? I'd like to do it without writing a boring for-loop.

This is a simple way to do it:
awk -F: '$3 > 999' /etc/passwd
This uses awk with a field separator of : and instructs it to print the line if the third field is greater than 999. If you want to only print the first field (username) or construct some new lines based on the fields, this is a starting point:
awk -F: '{if ($3 > 999) print "user", $1, "uid", $3}' /etc/passwd

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