I have a corpus file and the rules file. I am trying to find matching words where the word from rule appear in corpus.
# cat corpus.txt
this is a paragraph number one
second line
third line
# cat rule.txt
a
b
c
This returns 2 lines
# grep -F0 -f rule.txt corpus.txt
this is a paragraph number one
second line
But I am expecting 4 words like this...
a
paragraph
number
second
Trying to achive these results using grep or awk.
Assuming words are seperated by white spaces
awk '{print "\\S*" $1 "\\S*"}' rule.txt | grep -m 4 -o -f - corpus.txt
Related
I have a quite large fixed format file without spaces (file1):
file1:
0808563800555550000367120000500000
0005555566369330000078020000500000
01066666780000000008933600009000005635
0904251263088000000786590056500000
0000469011009904440425120444444440
I want to extract lines with fields 4-8,11-15 and 20-24 when fields 4-8 (only) are in a list of IDs in file2
file2:
55555
42512
The desired outputs are:
55555 36933 07802
42512 08800 78659
I have tried the following combination of cut | grep commands:
cut -c 4-8,11-15,20-24 file1 --output-delimiter=' ' | grep -w -F -f file2
It works fine and the speed is very good, but the problem is that I am getting columns where the lookup ID (fields 4-8) is not in the first column of the cutted data, and that is because grep checks the three columns after cut, not only the first one.
Here are the outputs of the command above:
85638 55555 36712
55555 36933 07802
66666 00000 89336
42512 08800 78659
04690 00990 42512
I know one may write the output to a file and then use, for example awk, but I thought there could be a much simpler approach to avoid longer processing time (for example, makes grep picks only the match in a specific cutted column).
Any help will be very appreciated and many thanks!
With GNU awk for FIELDWIDTHS:
$ awk -v FIELDWIDTHS='3 5 2 5 4 5 *' 'NR==FNR{a[$0]; next} $2 in a{ print $2, $4, $6 }' file2 file1
55555 36933 07802
42512 08800 78659
Would you please try the following:
cut -c 4-8,11-15,20-24 file1 --output-delimiter=' ' | grep -wf <(sed 's/^/^/' file2)
Each line in file2 is prepended by a caret ^ character to anchor to
the start of the line of the output by cut.
It may be a bit slower than before due to the lack of -F option.
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
I have a list of files, and I want to look for some specific keywords in those files. The output should be a line for each file with matches, showing the words that we found just once. For example, if I have the following file test.txt
one,two,three
four,five,six,
seven,eight,nine
and i do a grep of the words five and eight, it should return something like this:
test.txt:five,eight
I'm not interested in the lines, or the number of matches. I just want to know which words matched in each file. How can I do that?
GNU grep + awk solution:
Let's say we have file test1.txt with contents:
one,two,three
four,five,six,
seven,eight,nine
and test2.txt with contents:
one
two
three, four, five
Finding matches for words five and eight:
grep -Hwo '\(five\|eight\)' test*
| awk -F':' '{ a[$1]=(a[$1])? a[$1]","$2:$2 }END{ for(i in a) print i FS a[i] }'
The output:
test1.txt:five,eight
test2.txt:five
grep details:
-H - Print the file name for each match
-w - Select only those lines containing matches that form whole words
-o - Print only the matched (non-empty) parts of matching lines
awk details:
-F':' - field separator
a[$1]=(a[$1])? a[$1]","$2:$2 - using filename $1 as array key for accumulating all matched words
I'm trying to do a grep command that finds all lines in a file whos first word begins "as" and whos first word also ends with "ng"
How would I go about doing this using grep?
This should just about do it:
$ grep '^as\w*ng\b' file
Regexplanation:
^ # Matches start of the line
as # Matches literal string as
\w # Matches characters in word class
* # Quantifies \w to match either zero or more
ng # Matches literal string ng
\b # Matches word boundary
May have missed the odd corner case.
If you only want to print the words that match and not the whole lines then use the -o option:
$ grep -o '^as\w*ng\b' file
Read man grep for all information on the available options.
I am pretty sure this should work:
grep "^as[a-zA-Z]*ng\b" <filename>
hard to say without seeing samples from the actual input file.
sudo has already covered it well, but I wanted to throw out one more simple one:
grep -i '^as[^ ]*ng\b' <file>
-i to make grep case-insensitive
[^ ]* matches zero or more of any character, except a space
^ finds the 'first character in a line', so you can search for that with:
grep '^as' [file]
\w matches a word character, so \w* would match any number of word characters:
grep '^as\w*' [file]
\b means 'a boundary between a word and whitespace' which you can use to ensure that you're matching the 'ng' letters at the end of the word, instead of just somewhere in the middle:
grep '^as\w*ng\b' [file]
If you choose to omit the [file], simply pipe your files into it:
cat [file] | grep '^as\w*ng\b'
or
echo [some text here] | grep '^as\w*ng\b'
Is that what you're looking for?
grep (GNU grep) 2.14
Hello,
I have a log file that I want to filter on a selected word. However, it tends to filter on many for example.
tail -f gateway-* | grep "P_SIP:N_iptB1T1"
This will also find words like this:
"P_SIP:N_iptB1T10"
"P_SIP:N_iptB1T11"
"P_SIP:N_iptB1T12"
etc
However, I don't want to display anything after the 1. grep is picking up 11, 12, 13, etc.
Many thanks for any suggestions,
You can restrict the word to end at 1:
tail -f gateway-* | grep "P_SIP:N_iptB1T1\>"
This will work assuming that you have a matching case which is only "P_SIP:N_iptB1T1".
But if you want to extract from P_SIP:N_iptB1T1x, and display only once, then you need to restrict to show only first match.
grep -o "P_SIP:N_iptB1T1"
-o, --only-matching show only the part of a line matching PATTERN
More info
At least two approaches can be tried:
grep -w pattern matches for full words. Seems to work for this case too, even though the pattern has punctuation.
grep pattern -m 1 to restrict the output to first match. (Also doable with grep xxx | head -1)
If the lines contains the quotes as in your example, just use the -E option in grep and match the closing quote with \". For example:
grep -E "P_SIP:N_iptB1T1\"" file
If these quotes aren't in the text file, and there's blank spaces or endlines after the word, you can match these too:
# The word is followed by one or more blanks
grep -E "P_SIP:N_iptB1T1\s+" file
# Match lines ending with the interesting word
grep -E "P_SIP:N_iptB1T1$" file