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 $1} prints the name in that case.
Related
I have a text file with two columns.
Product Cost
Abc....def 10
Abc.def 20
ajsk,,lll 04
I want to search for product starts from "Abc" and ends with "def" then for those entries I want to add Cost.
I have used :
grep "^Abc|def$" myfile
but it is not working
Use awk. cat myfile | awk '{print $1}' | grep query
If you can use awk, try this:
text.txt
--------
Product Cost
Abc....def 10
Abc.def 20
ajsk,,lll 04
With only awk:
awk '$1 ~ /^Abc.*def$/ { SUM += $2 } END { print SUM } ' test.txt
Result: 30
With grep and awk:
grep "^Abc.*def.*\d*$" test.txt | awk '{SUM += $2} END {print SUM}'
Result: 30
Explanation:
awk reads each line and matches the first column with a regular expression (regex)
The first column has to start with Abc, followed by anything (zero or more times), and ends with def
If such match is found, add 2nd column to SUM variable
After reading all lines print the variable
Grep extracts each line that starts with Abc, followed by anything, followed by def, followed by anything, followed by a number (zero or more times) to end. Those lines are fed/piped to awk. Awk just increments SUM for each line it receives. After reading all lines received, it prints the SUM variable.
Thanks edited. Do you want the command like this?
grep "^Abc.*def *.*$"
If you don't want to use cat, and also show the line numbers:
awk '{print $1}' filename | grep -n keyword
If applicable, you may consider caret ^: grep -E '^foo|^bar' it will match text at the beginning of the string. Column one is always located at the beginning of the string.
Regular expression > POSIX basic and extended
^ Matches the starting position within the string. In line-based tools, it matches the starting position of any line.
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.
I have a text file with lines like this:
Sequences (1:4) Aligned. Score: 4
Sequences (100:3011) Aligned. Score: 77
Sequences (12:345) Aligned. Score: 100
...
I want to be able to extract the values into a new tab delimited text file:
1 4 4
100 3011 77
12 345 100
(like this but with tabs instead of spaces)
Can anyone suggest anything? Some combination of sed or cut maybe?
You can use Perl:
cat data.txt | perl -pe 's/.*?(\d+):(\d+).*?(\d+)/$1\t$2\t$3/'
Or, to save to file:
cat data.txt | perl -pe 's/.*?(\d+):(\d+).*?(\d+)/$1\t$2\t$3/' > data2.txt
Little explanation:
Regex here is in the form:
s/RULES_HOW_TO_MATCH/HOW_TO_REPLACE/
How to match = .*?(\d+):(\d+).*?(\d+)
How to replace = $1\t$2\t$3
In our case, we used the following tokens to declare how we want to match the string:
.*? - match any character ('.') as many times as possible ('*') as long as this character is not matching the next token in regex (which is \d in our case).
\d+:\d+ - match at least one digit followed by colon and another number
.*? - same as above
\d+ - match at least one digit
Additionally, if some token in regex is in parentheses, it means "save it so I can reference it later". First parenthese will be known as '$1', second as '$2' etc. In our case:
.*?(\d+):(\d+).*?(\d+)
$1 $2 $3
Finally, we're taking $1, $2, $3 and printing them out separated by tab (\t):
$1\t$2\t$3
You could use sed:
sed 's/[^0-9]*\([0-9]*\)/\1\t/g' infile
Here's a BSD sed compatible version:
sed 's/[^0-9]*\([0-9]*\)/\1'$'\t''/g' infile
The above solutions leave a trailing tab in the output, append s/\t$// or s/'$'\t''$// respectively to remove it.
If you know there will always be 3 numbers per line, you could go with grep:
<infile grep -o '[0-9]\+' | paste - - -
Output in all cases:
1 4 4
100 3011 77
12 345 100
My solution using sed:
sed 's/\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]\)*/\1 \2 \3/g' file.txt
How can I cut off the first n and the last n columns from a tab delimited file?
I tried this to cut first n column. But I have no idea to combine first and last n column
cut -f 1-10 -d "<CTR>v <TAB>" filename
Cut can take several ranges in -f:
Columns up to 4 and from 7 onwards:
cut -f -4,7-
or for fields 1,2,5,6 and from 10 onwards:
cut -f 1,2,5,6,10-
etc
The first part of your question is easy. As already pointed out, cut accepts omission of either the starting or the ending index of a column range, interpreting this as meaning either “from the start to column n (inclusive)” or “from column n (inclusive) to the end,” respectively:
$ printf 'this:is:a:test' | cut -d: -f-2
this:is
$ printf 'this:is:a:test' | cut -d: -f3-
a:test
It also supports combining ranges. If you want, e.g., the first 3 and the last 2 columns in a row of 7 columns:
$ printf 'foo:bar:baz:qux:quz:quux:quuz' | cut -d: -f-3,6-
foo:bar:baz:quux:quuz
However, the second part of your question can be a bit trickier depending on what kind of input you’re expecting. If by “last n columns” you mean “last n columns (regardless of their indices in the overall row)” (i.e. because you don’t necessarily know how many columns you’re going to find in advance) then sadly this is not possible to accomplish using cut alone. In order to effectively use cut to pull out “the last n columns” in each line, the total number of columns present in each line must be known beforehand, and each line must be consistent in the number of columns it contains.
If you do not know how many “columns” may be present in each line (e.g. because you’re working with input that is not strictly tabular), then you’ll have to use something like awk instead. E.g., to use awk to pull out the last 2 “columns” (awk calls them fields, the number of which can vary per line) from each line of input:
$ printf '/a\n/a/b\n/a/b/c\n/a/b/c/d\n' | awk -F/ '{print $(NF-1) FS $(NF)}'
/a
a/b
b/c
c/d
You can cut using following ,
-d: delimiter ,-f for fields
\t used for tab separated fields
cut -d$'\t' -f 1-3,7-
To use AWK to cut off the first and last fields:
awk '{$1 = ""; $NF = ""; print}' inputfile
Unfortunately, that leaves the field separators, so
aaa bbb ccc
becomes
[space]bbb[space]
To do this using kurumi's answer which won't leave extra spaces, but in a way that's specific to your requirements:
awk '{delim = ""; for (i=2;i<=NF-1;i++) {printf delim "%s", $i; delim = OFS}; printf "\n"}' inputfile
This also fixes a couple of problems in that answer.
To generalize that:
awk -v skipstart=1 -v skipend=1 '{delim = ""; for (i=skipstart+1;i<=NF-skipend;i++) {printf delim "%s", $i; delim = OFS}; printf "\n"}' inputfile
Then you can change the number of fields to skip at the beginning or end by changing the variable assignments at the beginning of the command.
You can use Bash for that:
while read -a cols; do echo ${cols[#]:0:1} ${cols[#]:1,-1}; done < file.txt
you can use awk, for example, cut off 1st,2nd and last 3 columns
awk '{for(i=3;i<=NF-3;i++} print $i}' file
if you have a programing language such as Ruby (1.9+)
$ ruby -F"\t" -ane 'print $F[2..-3].join("\t")' file
Try the following:
echo a#b#c | awk -F"#" '{$1 = ""; $NF = ""; print}' OFS=""
Use
cut -b COLUMN_N_BEGINS-COLUMN_N_UNTIL INPUT.TXT > OUTPUT.TXT
-f doesn't work if you have "tabs" in the text file.
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