How to use grep or awk to process a specific column ( with keywords from text file ) - linux

I've tried many combinations of grep and awk commands to process text from file.
This is a list of customers of this type:
John,Mills,81,Crescent,New York,NY,john#mills.com,19/02/1954
I am trying to separate these records into two categories, MEN and FEMALES.
I have a list of some 5000 Female Names , all in plain text , all in one file.
How can I "grep" the first column ( since I am only matching first names) but still printing the entire customer record ?
I found it easy to "cut" the first column and grep --file=female.names.txt, but this way it's not going to print the entire record any longer.
I am aware of the awk option but in that case I don't know how to read the female names from file.
awk -F ',' ' { if($1==" ???Filename??? ") print $0} '
Many thanks !

You can do this with Awk:
awk -F, 'NR==FNR{a[$0]; next} ($1 in a)' female.names.txt file.csv
Would print the lines of your csv file that contain first names of any found in your file female.names.txt.
awk -F, 'NR==FNR{a[$0]; next} !($1 in a)' female.names.txt file.csv
Would output lines not found in female.names.txt.
This assumes the format of your female.names.txt file is something like:
Heather
Irene
Jane

Try this:
grep --file=<(sed 's/.*/^&,/' female.names.txt) datafile.csv
This changes all the names in the list of female names to the regular expression ^name, so it only matches at the beginning of the line and followed by a comma. Then it uses process substitution to use that as the file to match against the data file.

Another alternative is Perl, which can be useful if you're not super-familiar with awk.
#!/usr/bin/perl -anF,
use strict;
our %names;
BEGIN {
while (<ARGV>) {
chomp;
$names{$_} = 1;
}
}
print if $names{$F[0]};
To run (assume you named this file filter.pl):
perl filter.pl female.names.txt < records.txt

So, I've come up with the following:
Suppose, you have a file having the following lines in a file named test.txt:
abe 123 bdb 532
xyz 593 iau 591
Now you want to find the lines which include the first field having the first and last letters as vowels. If you did a simple grep you would get both of the lines but the following will give you the first line only which is the desired output:
egrep "^([0-z]{1,} ){0}[aeiou][0-z]+[aeiou]" test.txt
Then you want to the find the lines which include the third field having the first and last letters as vowels. Similary, if you did a simple grep you would get both of the lines but the following will give you the second line only which is the desired output:
egrep "^([0-z]{1,} ){2}[aeiou][0-z]+[aeiou]" test.txt
The value in the first curly braces {1,} specifies that the preceding character which ranges from 0 to z according to the ASCII table, can occur any number of times. After that, we have the field separator space in this case. Change the value within the second curly braces {0} or {2} to the desired field number-1. Then, use a regular expression to mention your criteria.

Related

How to replace some cells number of .csv file if specific lines found in Linux

Lets say I have the following file.csv file content
"US","BANANA","123","100","0.5","ok"
"US","APPLE","456","201","0.1", "no"
"US","PIE","789","109","0.8","yes"
"US","APPLE","245","201","0.4","no"
I want to search all lines that have APPLE and 201, and then replace the column 5 values to 0. So, my output would look like
"US","BANANA","123","100","0.5","ok"
"US","APPLE","456","201","0", "no"
"US","PIE","789","109","0.8","yes"
"US","APPLE","245","201","0","no"
I can do grep search
grep "APPLE" file.csv | grep 201
to find out the lines. But could not figure out how to modify column 5 values of these lines in the original file.
You can use awk for this:
awk -F, '$2=="\"APPLE\"" { for (i=1;i<=NF;i++) { if ($i=="\"201\"") { gsub($5,"\""substr($5,2,length($5)-1)*1.10"\"",$5) } } }1' file.csv
Set the field delimiter to , and then when the second field is equal to APPLE in quotes, loop through each field and check if it is equal to 201 in quotes. If it is, replace the 5th field with 0 in quotes using Awk's gsub function. Print each line, changed or otherwise with short-hand 1

How can I use grep and regular expression to display names with just 3 characters

I am new to grep and UNIX. I have a sample of data and want to display all the first names that only contain three characters e.g. Lee_example. but I having some difficulty doing that. I am currently using this code cat file.txt|grep -E "[A-Z][a-z]{2}" but it is displaying all the names that contain at least 3 characters and not only 3 characters
Sample data
name
number
Lee_example
1
Hector_exaple
2
You need to match the _ after the first name.
grep -E "[A-Z][a-z]{2}_"
With awk:
awk -F_ 'length($1)==3{print $1}'
-F_ tells awk to split the input lines by _. length($1) == 3 checks whether the first fields (the name) is 3 characters long and {print $1} prints the name in that case.

Is there a way to remove only the followed duplicates?

I have a CSV input with these columns:
1,zzzz,xxxx,
1,xxxx,xyxy,
2,xxxx,xxxx,
3,yyyy,xxxx,
3,xxxx,yyyy,
3,xxxx,zzzz,
1,ffff,xxxx,
1,aaaa,xxxx,
And I need to discard lines where the first field matches that of the preceding line:
1,zzzz,xxxx,
2,xxxx,xxxx,
3,yyyy,xxxx,
1,ffff,xxxx,
I tried sort | uniq alone but didn't work because all lines are different with exception of first field (number).
Use awk instead of uniq:
awk -F, '$1 != last { last=$1; print }'
-F, sets the field separator to comma. $1 is the contents of the first field, so this prints the line whenever the first field changes.
Got the wanted output with uniq --check-chars=N; the uniq will check only a specified number of characters in the lines, and since the input isn't sorted this will allow the characters to appear later on the list.

CSV grep but keep the header

I have a CSV file that look like this:
A,B,C
1,2,3
4,4,4
1,2,6
3,6,9
Is there an easy way to grep all the rows in which the B column is 2, and keep the header? For example, I want the output be like
A,B,C
1,2,3
1,2,6
I am working under linux
Using awk:
awk -F, 'NR==1 || $2==2' file
NR==1 -> if first line,
$2==2 -> if second column is equal to 2. Lines are printed if either of the above is true.
To choose the column using the header column name:
awk -F, -v col="B" 'NR==1{for(i=1;i<=NF;i++)if($i==col)break;print;next}$i==2' file
Replace B with the appropriate name of the column which you want to check against.
You can use addresses in sed:
sed -n '1p;/^[^,]*,2/p'
It means:
1p Print the first line.
/ Start a match.
^ Match the beginnning of a line.
[^,] Match anything but a comma
* zero or more times.
, Match a comma.
2 Match a 2.
/p End of match, if it matches, print.
If the header can contain the value you are looking for, you should be more careful:
sed -n '1p;1!{/^[^,]*,2/p}'
1!{ ... } just means "Do the following for lines other then the first one".
For column number n>2, you can add a quantifier:
sed -n '1p;1!{/^\([^,]*,\)\{M\}2/p}'
where M=n-1. The quantifier just means repetition, so the non-comma-0-or-more-times-comma thing is repeated M times.
For true CSV files where a value can contain a comma, switch to Perl and Text::CSV.
$ awk -F, 'NR==1 { for (i=1;i<=NF;i++) h[$i] = i; print; next } $h["B"] == 2' file
A,B,C
1,2,3
1,2,6
By the way, sed is an excellent tool for simple substitutions on a single line, for anything else, just use awk - the code will be clearer and MUCH easier to enhance in future if necessary.

Replacing a column of data in text files with Linux command

I have several text files whose lines are tab-delimited.
The second column contains incorrect data.
How do I change everything in the second column to a specific text string?
awk ' { $2="<STRING>"; print } ' <FILENAME>
cat INFILE | perl -ne '$ln=$_;#x=split(/","/); #a=split(/","/, $ln,8);#b=splice(#a,0,7); $l=join("\",\"", #b); $r=join("\",\"", splice(#x,8)); print "$l\",\"10\",\"$r"'
This is an example that changes the 10th column to "10". I prefer this as I don't have to count the matching parenthesis like in the sed technique.
A simple and cheap hack:
cat INFILE | sed 's/\(.*\)\t\(.*\)\t\(.*\)/\1\tREPLACEMENT\t\3/' > OUTFILE
testing it:
echo -e 'one\ttwo\tthree\none\ttwo\tthree' | sed 's/\(.*\)\t\(.*\)\t\(.*\)/\1\tREPLACEMENT\t\3/'
takes in
one two three
one two three
and produces
one REPLACEMENT three
one REPLACEMENT three

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