I need help with two hings
1)the file.txt has the format of a list of films
, in which they are authors in different lines, year of publication, title, e.g.
author1
year1
title1
author2
year2
title2
author3
year3
title3
author4
year4
title4
I need to show only book titles whose author is "Joanne Rowling"
2)
one.txt contains numbers and letters for example like:
dada4dawdaw54 232dawdawdaw 53 34dadasd
77dkwkdw
65 23 laka 23
I need to sum all of them and receive score - here it should 561
I tried something like that:
awk '{for(i=1;i<=NF;i++)s+=$i}END{print s}' plik2.txt
but it doesn't make sense
For the 1st question, the solution of okulkarni is great.
For the 2nd question, one solution is
sed 's/[^0-9]/ /g' one.txt | awk '{for(i=1;i<=NF;i++) sum+= $i} END { print sum}'
The sed command converts all non-numeric characters into spaces, while the awk command sums the numbers, line by line.
For the first question, you just need to use grep. Specifically, you can do grep -A 2 "Joanne Rowling" file.txt. This will show all lines with "Joanne Rowling" and the two lines immediately after.
For the second question, you can also use grep by doing grep -Eo '[0-9]+' | paste -sd+ | bc. This will put a + between every number found by grep and then add them up using bc.
Related
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 $1} prints the name in that case.
How do I get a string after the first occurrence of a number?
For example, I have a file with multiple lines:
34 abcdefg
10 abcd 123
999 abc defg
I want to get the following output:
abcdefg
abcd 123
abc defg
Thank you.
You could use Awk for this, loop through all the columns in each line upto NF (last column in each line) and once matching the first word, print the column next to it. The break statement would exit the for loop after the first iteration.
awk '{ for(i=1;i<=NF;i++) if ($i ~ /[[:digit:]]+/) { print $(i+1); break } }' file
It is not clear what you exactly want, but you can try to express it in sed.
Remove everything until the first digit, the next digits and any spaces.
sed 's/[^0-9]*[0-9]\+ *//'
Imagine the following two input files :
001 ham
03spam
3 spam with 5 eggs
A quick solution with awk would be :
awk '{sub(/[^0-9]*[0-9]+/,"",$0); print $1}' <file>
This line substitutes the first string of anything that does not contain a number followed by a number by an empty set (""). This way $0 is redefined and you can reprint the first field or the remainder of the field. This line gives exactly the following output.
ham
spam
spam
If you are interested in the remainder of the line
awk '{sub(/[^0-9]*[0-9]+ */,"",$0); print $0}' <file>
This will have as an output :
ham
spam
spam with 5 eggs
Be aware that an extra " *" is needed in the regular expression to remove all trailing spaces after the number. Without it you would get
awk '{sub(/[^0-9]*[0-9]+/,"",$0); print $0}' <file>
ham
spam
spam with 5 eggs
You can remove digits and whitespaces using sed:
sed -E 's/[0-9 ]+//' file
grep can do the job:
$ grep -o -P '(?<=[0-9] ).*' inputFIle
abcdefg
abcd 123
abc defg
For completeness, here is a solution with perl:
$ perl -lne 'print $1 if /[0-9]+\s*(.*)/' inputFIle
abcdefg
abcd 123
abc defg
I have a text file with two columns.
Product Cost
Abc....def 10
Abc.def 20
ajsk,,lll 04
I want to search for product starts from "Abc" and ends with "def" then for those entries I want to add Cost.
I have used :
grep "^Abc|def$" 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'm using grep to display all lines that have ONLY 4,5,6,7 and 9 in the zipcode column.
How do i display only the lines of the file that contain the numbers 4,5,6,7 and 9 in the zipcode field?
A sample row is:
15 m jagger mick 41 4th 95115
Thanks
I am going to assume you meant "How do I use grep to..."
If all of the lines in the file have a 5 digit zip at the end of each line, then:
egrep "[45679]{5}$" filename
Should give you what you want.
If there might be whitespace between the zip and the end of the line, then:
egrep "[45679]{5}[[:space:]]*$" filename
would be more robust.
If the problem is more general than that, please describe it more accurately.
Following regex should fetch you desired result:
egrep "[45679]+$" file
If by "grep" you mean, "the correct tool", then the solution you seek is:
awk '$7 ~ /^[45679]*$/' input
This will print all lines of input in which the 7th field consists only of the characters 4,5,6,7, and 9. If you want to specify 'the last column' rather than the 7th, try
awk '$NF ~ /^[45679]*$/' input