I have a file and I want to know how many times does a word is inside that file.(NOTE: A row can have the same word)
You can use this command. Hope this wil help you.
grep -o yourWord file | wc -l
Use the grep -c option to count the number of occurences of a search pattern.
grep -c searchString file
awk solution:
awk '{s+=gsub(/word/,"&")}END{print s}' file
test:
kent$ cat f
word word word
word
word word word
kent$ awk '{s+=gsub(/word/,"&")}END{print s}' f
7
you may want to add word boundary if you want to match an exact word.
Yes, i know you want a grep solution, but my favorite perl with the rolex operator can't missing here... ;)
perl -0777 -nlE 'say $n=()=m/\bYourWord\b/g' filename
# ^^^^^^^^
if yoy want match the YourWord surrounded with another letters like abcYourWordXYZ, use
perl -0777 -nlE 'say $n=()=m/YourWord/g' filename
Related
I'm trying to reduce a .sm file1 - around 10 GB by filtering it using a fair long set of words (around 180.108 items) listed in a text file file2.
File1 is structured as follows:
word <http://internet.address.com> 1
i.e. one word followed by a blank space, an internet address, and a number.
File2 is a simple .txt file, a list of words, one on each line.
My aim is to create a third file File3 containing only those lines in file1 whose first word matches with the word-list of file2, and disregard the rest.
My attempt is the following:
grep -w -F -f file2.txt file1.sm > file3.sm
I've also attempted something along this line:
gawk 'FNR==NR {a[$1]; next } !($2 in a)' file2.txt file1.sm > file3.sm
but with no success. I understand /^ and \b might play a part here, but I don't know how to fit them in the syntax. I've looked around extensively but no solution seems to fit.
My problem is that here grep reads the entire file1's line, and it can happen that the matching word lies in the webpage address, which I'm not interested in finding out.
sed 's/^/^/' file2.txt | grep -f - file1.sm
join is the best tool for this, not grep/awk:
join -t' ' <(sort file1.sm) <(sort file2.txt) >file3.sm
Hello How can I grep only match before : mark?
If I run grep test1 file, it shows all three lines.
test1:x:29688:test1,test2
test2:x:22611:test1
test3:x:25163:test1,test3
But I would like to get an output test1:x:29688:test1,test2
I would appreciate any advice.
If the desired lines always start with test1 then you can do:
grep '^test1' file
If it's always followed by : but not the other (potential) matches then you can include it as part of the pattern:
grep 'test1:' file
As your data is in row, columns delimited by a character, you may consider awk:
awk -F: '$1 == "test1"' file
I think that you just need to add “:” after “test1”, see an example:
grep “test1:” file
I have the following issue, I need to retrieve all words that contains exactly 2 vowels (in any order) from a file. The file only contains one word per line.
My current workaround is:
Grep1: Retrieve words such as earth, over, under, one...
grep -i "^[aeiou][^aeiou]*[aeiou][^aeiou]*$" genesis.words > A.txt
and
Grep2: Retrieve words such as formless, deep, said...
grep -i "^[^aeiou][^aeiou]*[aeiou][^aeiou]*[aeiou][^aeiou]*$" genesis.words > B.txt
the above solution works but when I concatenate both regexs into a single regex then return nothing!
Mother of Grep1 & Grep2: should retrieve everything!
grep -i "^[aeiou][^aeiou]*[aeiou][^aeiou]*$|^[^aeiou][^aeiou]*[aeiou][^aeiou]*[aeiou][^aeiou]*$" genesis.words
I think issue is around my implementation of ^$ in expression but have tried diff versions with no sucess!
Any help will be highly appreciated!
OS is AIX 6100-09-04-1441
You were close. This should work:
grep -i "^[^aeiou]*[aeiou][^aeiou]*[aeiou][^aeiou]*$" genesis.words > A.txt
So it should find all eight possibilities (two vowels identify three nonvowel sequence, each possibly empty; 2^3 is 8):
[ ]I[ ]o[ ]
[ ]e[ ]a[r]
[ ]e[r]a[ ]
[ ]e[l]a[n]
[T]e[ ]a[ ]
[D]e[ ]a[r]
[D]e[w]a[r]
[D]a[w]a[ ]
[H]a[w]a[y]
As for concatenation, | needs escaping. You can use a single anchoring:
^(regexp1\|regexp2)$
Since the * can match 0 times or more you should be able to start the string with [^aeiou]*: try
"^[^aeiou]*[aeiou][^aeiou]*[aeiou][^aeiou]*$"
As for fixing your regex, I think you need to escape the bar as \|, so
grep -i "^[aeiou][^aeiou]*[aeiou][^aeiou]*$\|^[^aeiou][^aeiou]*[aeiou][^aeiou]*[aeiou][^aeiou]*$" genesis.words
If you don't mind Perl, you could use this:
perl -lne '$m=$_; tr/[aeiou]//cd; print $m if length()==2;' /usr/share/dict/words
That says... "save the current line (word) in $m. Delete everything that is not a vowel. Print the original word if there are two things (i.e vowels) left."
Note that I am using the system dictionary as input for my tests.
You could do pretty much the same thing in awk.
If you're able to use an alternative to grep tr with wc works well:
words=/path/to/words.txt
while read -e word ; do
v=$(echo $word | tr -cd 'aeiou' | wc -c)
[[ ! $v -eq "2" ]] || echo $word >> output.txt
done < $words
This reads the original file line by line, counts the vowels & returns results with only 2 to output.txt.
I am trying to grep the output of a command that outputs unknown text and a directory per line. Below is an example of what I mean:
.MHuj.5.. /var/log/messages
The text and directory may be different from time to time or system to system. All I want to do though is be able to grep the directory out and send it to a variable.
I have looked around but cannot figure out how to grep to the end of a word. I know I can start the search phrase looking for a "/", but I don't know how to tell grep to stop at the end of the word, or if it will consider the next "/" a new word or not. The directories listed could change, so I can't assume the same amount of directories will be listed each time. In some cases, there will be multiple lines listed and each will have a directory list in it's output. Thanks for any help you can provide!
If your directory paths does not have spaces then you can do:
$ echo '.MHuj.5.. /var/log/messages' | awk '{print $NF}'
/var/log/messages
It's not clear from a single example whether we can generalize that e.g. the first occurrence of a slash marks the beginning of the data you want to extract. If that holds, try
grep -o '/.*' file
To fetch everything after the last space, try
grep -o '[^ ]*$' file
For more advanced pattern matching and extraction, maybe look at sed, or Awk or Perl or Python.
Your line can be described as:
^\S+\s+(\S+)$
That's assuming whitespace is your delimiter between the random text and the directory. It simply separates the whitespace from the non-whitespace and captures the second part.
Or you might want to look into the word boundary character class: \b.
I know you said to use grep, but I can't help to mention that this is trivially done using awk:
awk '{ print $NF }' input.txt
This is assuming that a whitespace is the delimiter and that the path does not contain any whitespaces.
I do the following in order to get all WORD in file but not in lines that start with "//"
grep -v "//" file | grep WORD
Can I get some other elegant suggestion to find all occurrences of WORD in the file except lines that begin with //?
Remark: "//" does not necessarily exist at the beginning of the line; there could be some spaces before "//".
For example
// WORD
AA WORD
// ss WORD
grep -v "//" file | grep WORD
This will also exclude any lines with "//" after WORD, such as:
WORD // This line here
A better approach with GNU Grep would be:
grep -v '^[[:space:]]*//' file | grep 'WORD'
...which would first filter out any lines beginning with zero-or-more spaces and a comment string.
Trying to put these two conditions into a single regular expression is probably not more elegant.
awk '!/^[ \t]*\/\// && /WORD/{m=gsub("WORD","");total+=m}END{print total}' file