Here is the code to look for all the words in a file starting with a. What about words that start with a-f?
grep -E '\ba' data.txt
Since the question may be quite ambiguous, I will try to add all the possible solutions I can find for it. Let's start with a sample file, in my case containing the following words so I can explain myself better:
ascii
affair
a-f-f-i-l-i-a-t-e
barcode
break
charset
character
delete
draft
execute
example
force
failure
grip
group
halt
held
inode
instance
joke
jekyll
That said, if you want to grep all the words starting with the letters that go from a to f (this is, every word starting by a, b, c, d, e and f) you would use:
grep -E "^[a-f]" data.txt
And the output you would see is this one:
ascii
affair
barcode
break
charset
character
delete
draft
execute
example
force
failure
Now, if you want to filter the words starting literally by the string a-f for some reason then you would use this:
grep -E "^a-f" data.txt
And you would see that the output would look like this:
a-f-f-i-l-i-a-t-e
In this case the key is not that much grep knowledge but regex stuff.
Hope you find it useful.
Related
I am trying to replace string between two strings in a file with the command below. There could be any number of such patterns in the file. This is just an example.
sed 's/word1.*word2/word1/' 1.txt
There are two instances where 'word1' followed by 'word2' occurs in the sample source file I'm testing. Content of the 1.txt file
word1---sjdkkdkjdk---word2 I want this text----word1---jhfnkfnsjkdnf----word2 I need this also
Result is as below.
word1 I need this also
Expected Output :
word1 I want this text----word1 I need this also
Can anybody help me with this please?
I looked at other stack-overflow questionnaire but they discuss about replacing only one instance of the pattern.
Regular expressions are greedy - they match the longest possible string, so everything from the first 'word1' to the last 'word2'. Not sure if any version of sed supports non-greedy regexps... you could just use perl, though, which does:
perl -pe 's/word1.*?word2/word1/g' 1.txt
should do the trick. That ? changes the meaning of the prior * from 'match as many times as possible as long as the rest of the pattern matches' to 'match as few times as possible as long as the rest of the pattern matches'.
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/word1/{/g; s/word2/}/g; s/{[^{}]*}/word1/g; s/}/word2/g; s/{/word1/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
word1 I want this text----word1 I need this also
It's lengthy and looks complicated but it's a technique that is used fairly often and is really just a series of simple steps to robustly convert word1 to { and word2 to } so you're dealing with characters instead of strings in the actual substitution s/{[^{}]*}/word1/g and so can use a negated bracket expression to avoid the greedy regexp taking up too much of the line.
See https://stackoverflow.com/a/35708616/1745001 for more info on the general approach used here to be able to turn strings into characters that cannot be present in the input by the time the real work takes place and then restore them again afterwards.
If you only have two instances of the word1-word2 pattern on a line, this should work:
sed 's/\(word1\).*word2\(.*\)\(word1\).*word2\(.*\)/\1\2\3\4/' 1.txt
I grab the parts we want to keep inside escaped brackets \( and \) then I can refer to those parts as \1 \2 and so on.
I'm having a hard time getting a grasp of using grep for a class i am in was hoping someone could help guide me in this assignment. The Assignment is as follows.
Using grep print all 5 letter lower case words from the linux dictionary that have a single letter duplicated one time (aabbe or ababe not valid because both a and b are in the word twice). Next to that print the duplicated letter followed buy the non-duplicated letters in alphabetically ascending order.
The Teacher noted that we will need to use several (6) grep statements (piping the results to the next grep) and a sed statement (String Editor) to reformat the final set of words, then pipe them into a read loop where you tear apart the three non-dup letters and sort them.
Sample Output:
aback a bck
abaft a bft
abase a bes
abash a bhs
abask a bks
abate a bet
I haven't figured out how to do more then printing 5 character words,
grep "^.....$" /usr/share/dict/words |
Didn't check it thoroughly, but this might work
tr '[:upper:]' '[:lower:]' | egrep -x '[a-z]{5}' | sed -r 's/^(.*)(.)(.*)\2(.*)$/\2 \1\3\4/' | grep " " | egrep -v "(.).*\1"
But do your way because someone might see it here.
All in one sed
sed -n '
# filter 5 letter word
/[a-zA-Z]\{5\}/ {
# lower letters
y/ABCDEFGHIJKLMNOPQRSTUVWXYZ/abcdefghijklmnopqrstuvwxya/
# filter non single double letter
/\(.\).*\1/ !b
/\(.\).*\(.\).*\1.*\1/ b
/\(.\).*\(.\).*\1.*\2/ b
/\(.\).*\(.\).*\2.*\1/ b
# extract peer and single
s/\(.\)*\(.\)\(.*\)\2\(.*\)/a & \2:\1\3\4/
# sort singles
:sort
s/:\([^a]*\)a\(.*\)$/:\1\2a/
y/abcdefghijklmnopqrstuvwxyz/zabcdefghijklmnopqrstuvwxy/
/^a/ !b sort
# clean and print
s/..//
s/:/ /p
}' YourFile
posix sed so --posix on GNU sed
The first bit, obviously, is to use grep to get it down to just the words that have a single duplication in. I will give you some clues on how to do that.
The key is to use backreferences, which allow you to specify that something that matched a previous expression should appear again. So if you write
grep -E "^(.)...\1...\1$"
then you'll get all the words that have the starting letter reappearing in fifth and ninth positions. The point of the brackets is to allow you to refer later to whatever matched the thing in brackets; you do that with a \1 (to match the thing in the first lot of brackets).
You want to say that there should be a duplicate anywhere in the word, which is slightly more complicated, but not much. You want a character in brackets, then any number of characters, then the repeated character (with no ^ or $ specified).
That will also include ones where there are two or more duplicates, so the next stage is to filter them out. You can do that by a grep -v invocation. Once you've got your list of 5-character words that have at least one duplicate, pipe them through a grep -v call that strips out anything with two (or more) duplicates in. That'll have a (.), and another (.), and a \1, and a \2, and these might appear in several different orders.
You'll also need to strip out anything that has a (.) and a \1 and another \1, since that will have a letter with three occurrences.
That should be enough to get you started, at any rate.
Your next step should be to find the 5-letter words containing a duplicate letter. To do that, you will need to use back-references. Example:
grep "[a-z]*\([a-z]\)[a-z]*\$1[a-z]*"
The $1 picks up the contents of the first parenthesized group and expects to match that group again. In this case, it matches a single letter. See: http://www.thegeekstuff.com/2011/01/advanced-regular-expressions-in-grep-command-with-10-examples--part-ii/ for more description of this capability.
You will next need to filter out those cases that have either a letter repeated 3 times or a word with 2 letters repeated. You will need to use the same sort of back-reference trick, but you can use grep -v to filter the results.
sed can be used for the final display. Grep will merely allow you to construct the correct lines to consider.
Note that the dictionary contains capital letters and also non-letter characters, plus that strange characters used in Southern Europe. say "รจ".
If you want to distinguish "A" and "a", it's automatic, on the other hand if "A" and "a" are the same letter, in ALL grep invocations you must use the -i option, to instruct grep to ignore case.
Next, you always want to pass the -E option, to avoid the so called backslashitis gravis in the regexp that you want to pass to grep.
Further, if you want to exclude the lines matching a regexp from the output, the correct option is -v.
Eventually, if you want to specify many different regexes to a single grep invocation, this is the way (just an example btw)
grep -E -i -v -e 'regexp_1' -e 'regexp_2' ... -e 'regexp_n'
The preliminaries are after us, let's look forward, use the answer from chiastic-security as a reference to understand the procedings
There are only these possibilities to find a duplicate in a 5 character string
(.)\1
(.).\1
(.)..\1
(.)...\1
grep -E -i -e 'regexp_1' ...
Now you have all the doubles, but this doesn't exclude triples etc that are identified by the following patterns (Edit added a cople of additional matching triples patterns)
(.)\1\1
(.).\1\1
(.)\1.\1
(.)..\1\1
(.).\1.\1
(.)\1\1\1
(.).\1\1\1
(.)\1\1\1\1\
you want to exclude these patterns, so grep -E -i -v -e 'regexp_1' ...
at his point, you have a list of words with at least a couple of the same character, and no triples, etc and you want to drop double doubles, these are the regexes that match double doubles
(.)(.)\1\2
(.)(.)\2\1
(.).(.)\1\2
(.).(.)\2\1
(.)(.).\1\2
(.)(.).\2\1
(.)(.)\1.\2
(.)(.)\2.\1
and you want to exclude the lines with these patterns, so its grep -E -i -v ...
A final hint, to play with my answer copy a few hundred lines of the dictionary in your working directory, head -n 3000 /usr/share/dict/words | tail -n 300 > ./300words so that you can really understand what you're doing, avoiding to be overwhelmed by the volume of the output.
And yes, this is not a complete answer, but it is maybe too much, isn't it?
I'm trying to find the code that searches all palindromes in a dictionary file
this is what I got atm which is wrong :
sed -rn '/^([a-z])-([a-z])\2\1$/p' /usr/share/dict/words
Can somebody explain the code as well.
Found the right answer.
sed -n '/^\([a-z]\)\([a-z]\)\2\1$/p' /usr/share/dict/words
I have no idea why I used -
I also don't have an explenation for the \ ater each group
You can use the grep command as explained here
grep -w '^\(.\)\(.\).\2\1'
explanation The grep command searches for the first any three letters by using (.)(.). after that we are searching the same 2nd character and 1st character is occuring or not.
The above grep command will find out only 5 letters palindrome words.
extended version is proposed as well on that page; and works correctly for the first line but then crashes... there is surely some good to keep and maybe to adapt...
Guglielmo Bondioni proposed a single RE that finds all palindromes up to 19 characters long using 9 subexpressions and 9 back-references:
grep -E -e '^(.?)(.?)(.?)(.?)(.?)(.?)(.?)(.?)(.?).?\9\8\7\6\5\4\3\2\1' file
You can extend this further as much as you want :)
Perl to the rescue:
perl -lne 'print if $_ eq reverse' /usr/share/dict/words
Hate to say it, but while regex may be able to cook your breakfast, I don't think it can find a palindrome. According to the all-knowing Wikipedia:
In the automata theory, a set of all palindromes in a given alphabet is a typical example of a language that is context-free, but not regular. This means that it is impossible for a computer with a finite amount of memory to reliably test for palindromes. (For practical purposes with modern computers, this limitation would apply only to incredibly long letter-sequences.)
In addition, the set of palindromes may not be reliably tested by a deterministic pushdown automaton which also means that they are not LR(k)-parsable or LL(k)-parsable. When reading a palindrome from left-to-right, it is, in essence, impossible to locate the "middle" until the entire word has been read completely.
So a regular expression won't be able to solve the problem based on the problem's nature, but a computer program (or sed examples like #NeronLeVelu or #potong) will work.
explanation of your code
sed -rn '/^([a-z])-([a-z])\2\1$/p' /usr/share/dict/words
select and print line that correspond to :
A first (starting the line) small alphabetic character followed by - followed by another small alaphabetic character (could be the same as the first) followed by the last letter of the previous group followed by the first letter Letter1-Letter2Letter2Letter1 and the no other element (end of line)
sample:
a-bba
a is first letter
b second letter
b is \2
a is \1
But it's a bit strange for any work unless it came from a very specific dictionnary (limited to combination by example)
This might work for you (GNU sed):
sed -r 'h;s/[^[:alpha:]]//g;H;x;s/\n/&&/;ta;:a;s/\n(.*)\n(.)/\n\2\1\n/;ta;G;/\n(.*)\n\n\1$/IP;d' file
This copies the original string(s) to the hold space (HS), then removes everything but alpha characters from the string(s) and appends this to the HS. The second copy is then reversed and the current string(s) and the reversed copy compared. If the two strings are equal then the original string(s) is printed out otherwise the line is deleted.
I have a list with English words (1 in each line, around 100.000)-> a.txt and a b.txt contains strings (around 50.000 line, one string in each line, can contain pure words, word+something, garbage). I would like to know which strings from b.txt contains English words only (without any additional chars).
Can I do this with grep?
Example:
a.txt:
apple
pie
b.txt:
applepie
applebs
bspie
bsabcbs
Output:
c.txt:
applepie
Since your question is underspecified, maybe this answer can help as a shot in the dark to clarify your question:
c='cat b.txt'
while IFS='' read -e line
do
c="$c | grep '$line'"
done < a.txt
eval "$c" > c.txt
But this would also match a line like this is my apply on a pie. I don't know if that's what you want.
This is another try:
re=''
while IFS='' read -e line
do
re="$re${re:+|}$line"
done < a.txt
grep -E "^($re)*$" b.txt > c.txt
This will let pass only the lines which have nothing but a concatenation of these words. But it will also let pass things like 'appleapplepieapplepiepieapple'. Again, I don't know if this is what you want.
Given your latest explanation in the question I would propose another approach (because building such a list out of 100000+ words is not going to work).
A working approach for this amount of words could be to remove all recognized words from the text and see which lines get emptied in the process. This can easily be done iteratively without exploding the memory usage or other resources. It will take time, though.
cp b.txt inprogress.txt
while IFS='' read -e line
do
sed -i "s/$line//g" inprogress.txt
done < a.txt
for lineNumber in $(grep -n '^$' inprogress.txt | sed 's/://')
do
sed -n "${lineNumber}p" b.txt
done
rm inprogress.txt
But this still would not really solve your issue; consider if you have the words to and potato in your list, and removing the to would occur first, then this would leave a word pota in your text file, and pota is not a word which would then be removed.
You could address that issue by sorting your word file by word length (longest words first) but that still would be problematic in some cases of compound words, e. g. redart (being red + art) but dart would be removed first, so re would remain. If that is not in your word list, you would not recognize this word.
Actually, your problem is one of logical programming and natural language processing and probably does not fit to SO. You should have a look at the language Prolog which is designed around such problems as yours.
I will post this as an answer as well since I feel this is the correct answer to your specific question.
Your requirement is to find non-English words in a file (b.txt) based on a word list ( a.txt) which contains a list of English words. Based on the example in your question said word list does not contain compound words (e.g. applepie) but you would still like to match the file against compound words based on words in your word list (e.g. apple and pie).
There are two problem you are facing:
Not every permutation of words in a.txt will be a valid English compound word so just based on this your problem is already impossible to solve.
If you, nonetheless, were to attempt building a list of compound words yourself by compiling a list of all possible permutations you cannot easily do this because of the size of your wordlist (and resulting memory problems). You would most probably have to store your words in a more complex data structure, e.g. a tree, and build permutations on the fly by traversing the tree which is not doable in shell scripting.
Because of these points and your actual question being "can this be done with grep?" the answer is no, this is not possible.
I'm trying to write a grep (or egrep) command that will find and print any lines in "words.txt" which contain the same lower-case letter three times in a row. The three occurrences of the letter may appear consecutively (as in "mooo") or separated by one or more spaces (as in "x x x") but not separated by any other characters.
words.txt contains:
The monster said "grrr"!
He lived in an igloo only in the winter.
He looked like an aardvark.
Here's what I think the command should look like:
grep -E '\b[^ ]*[[:alpha:]]{3}[^ ]*\b' 'words.txt'
Although I know this is wrong, but I don't know enough of the syntax to figure it out. Using grep, could someone please help me?
Does this work for you?
grep '\([[:lower:]]\) *\1 *\1'
It takes a lowercase character [[:lower:]] and remembers it \( ... \). It than tries to match any number of spaces _* (0 included), the rememberd character \1, any number of spaces, the remembered character. And that's it.
You can try running it with --color=auto to see what parts of the input it matched.
Try this. Note that this will not match "mooo", as the word boundary (\b) occurs before the "m".
grep -E '\b([[:alpha:]]) *\1 *\1 *\b' words.txt
[:alpha:] is an expression of a character class. To use as a regex charset, it needs the extra brackets. You may have already known this, as it looks like you started to do it, but left the open bracket unclosed.