Linux terminal command line variable using in grep - linux

I want to display the amount of words that have exactly 14, 15 and 16 unique letters. I want to use a for loop. (It has to be a one liner.)
This is what I have so far:
for i in {14..16}; do echo "There are $(cat /usr/share/dict/dutch | grep -P '^.{"$i"}$' | grep -vP -c '(.).*\1') words with exactly $i unique letters"; done
Result:
There are 0 words with exactly 14 unique letters
There are 0 words with exactly 15 unique letters
There are 0 words with exactly 16 unique letters
This means the loop works and when I run it like this:
echo "There are $(cat /usr/share/dict/dutch | grep -P '^.{14}$' | grep -vP -c '(.).*\1') words with exactly 14 unique letters" &&
echo "There are $(cat /usr/share/dict/dutch | grep -P '^.{15}$' | grep -vP -c '(.).*\1') words with exactly 15 unique letters" &&
echo "There are $(cat /usr/share/dict/dutch | grep -P '^.{16}$' | grep -vP -c '(.).*\1') words with exactly 16 unique letters"
The results are:
There are 13 words with exactly 14 unique letters
There are 2 words with exactly 15 unique letters
There are 0 words with exactly 16 unique letters
This shows that I am doing something wrong with the variable ($i) inside the grep-command. I don't know how I should do it or solve this problem.
Thanks in advance

It looks like you need to use single quotes instead of double quotes around the variable in your first regex:
for i in {14..16}; do
echo "there are $(cat /usr/share/dict/dutch | grep -P '^.{'$i'}$' | grep -vP -c '(.).*\1') words with exactly $i unique letters";
done
Because your first regex was wrapped in single quotes, it was being used literally, without any variable expansion. By putting the variable in single quotes, you're actually specifying two literal strings wrapped tightly around your actual variable (put another way, your variable isn't actually wrapped in any quotes at all).
Edit: This is what I get on my system:
$ for i in {14..16}; do echo "there are $(cat /usr/share/dict/dutch | grep -P '^.{'$i'}$' | grep -vP -c '(.).*\1') words with exactly $i unique letters"; done
there are 13 words with exactly 14 unique letters
there are 2 words with exactly 15 unique letters
there are 0 words with exactly 16 unique letters
Edit 2: This might demonstrate the issue more clearly:
$ i=12
$ echo '^.{"$i"}$'
^.{"$i"}$
$ echo '^.{'$i'}$'
^.{12}$

Related

Grepping inside a log with a threshold on the result

I have certain tags stored in a series of .log files and i would like for the grep to show me Only the values > 31, meaning different to 0 and higher than 31
I have this code:
#! /bin/bash
-exec grep -cFH "tag" {} ; | grep -v ':[0-31]$' >> file.txt
echo < file.txt
Output:
I have this result from the grep:
/opt/logs/folder2.log:31
i was expecting not to bring nothing back if the result is 31 or less but still shows the result 31
i have also tried to add:
|tail -n+31
but didn't work
[0-31] means "0 or 1 or 2 or 3 or 1".
To drop all lines with 0-9, 10-19, 20-29, 30, and 31, you could use the following:
... | grep -ve ':[0-9]$' -e ':[12][0-9]$' -e ':3[01]$'
or as single regex:
... | grep -v ':\([12]\?[0-9]\|3[01]\)$'
With extended grep:
... | grep -vE ':([12]?[0-9]|3[01])$'

Finding the Number of strings in a File

I'm trying to write a very small program that will check the number of sub strings in a large text file. All it will do is count the first 2000 lines of the text file, find any "TTT" sub-strings, count them, and set a variable to that total. I'm a bit new to shell, so any help would be amazingly appreciated!
#!/bin/bash
$counter=(head -2000 [file name] | grep TTT | grep -o TTT | wc -l)
echo $counter
For what it's worth you might awk better suited for this task:
awk -F"ttt" '{j=(NF-1)+j}END{print j}' filename
This will split each record in your file by delimiter "ttt". Then it counts the number of fields, subtracts one, and adds that to the total.
A file like:
ttt tttttt something
1 5 ttt
tt
one more ttt record
Would be split (visualizing with pipe delim) like:
| || something
1 5 |
tt
one more | record
Counting the number of fields per record:
4
2
1
2
Subtracting one from that:
3
1
0
1
Which totals to 5, which is how many "ttt" substrings are present.
To incorporate this into your script (and fixing your other issue):
#!/bin/bash
counter=$(awk -F"ttt" '{j=(NF-1)+j}END{print j}' filename)
echo $counter
The change here is that when we set a variable in Bash we don't include the $ sign at the front. Only in referencing the variable do we include the $.
You have some minor syntax errors there, probably you meant this:
counter=$(head -2000 [file name] | grep TTT | grep -o TTT | wc -l)
echo $counter
Notice the tiny changes I made there to make it work.
Btw the grep TTT in the middle is redundant, you can simply drop it, that is:
counter=$(head -2000 [file name] | grep -o TTT | wc -l)
grep can already do what you want: counter=$(grep -c TTT $infile). You can limit the number of hits (not lines) with -m NUM, --max-count=NUM, which makes grep stop at the end of the file OR when NUM occurrences are found.

Obtaining the total of coincidences with multiple pattern using grep command

I have a file in Linux contains strings:
CALLTMA
Starting
Starting
Ending
Starting
Ending
Ending
CALLTMA
Ending
I need the quantity of any string (FE. #Ending, # Starting, #CALLTMA). In my example I need obtaining:
CALLTMA : 2
Starting: 3
Ending : 4
I can obtaining this output when I execute 3 commands:
grep -i "Starting" "/myfile.txt" | wc -l
grep -i "Ending" "/myfile.txt" | wc -l
grep -i "CALLTMA" "/myfile.txt" | wc -l
I want to know if it is possible to obtain the same output using only one command.
I try running this command
grep -iE "CALLTMA|Starting|Ending" "/myfile.txt" | wc -l
But this returned the total of coincidences. I appreciate your help .
Use sort and uniq:
sort myfile.txt | uniq -c
The -c adds the counts to the unique lines. If you want to sort the output by frequency, add
| sort -n
to the end (and change to -nr if you want the descending order).
A simple awk way to handle this:
awk '{counts[$1]++} END{for (c in counts) print c, counts[c]}' file
Starting 3
Ending 4
CALLTMA 2
grep -c will work. You can put it all together in a short script:
for i in Starting CALLTMA Ending; do
printf "%-8s : %d\n" "$i" $(grep -c "$i" file.txt)
done
(to enter the search terms as arguments, just use the arguments array for the loop list, e.g. for i in "$#"; do)
Output
Starting : 3
CALLTMA : 2
Ending : 4

How to do sum from the file and move in particular way in another file in linux?

Acttualy this is my assignment.I have three-four file,related by student record.Every file have two-three student record.like this
Course Name:Opreating System
Credit: 4
123456 1 1 0 1 1 0 1 0 0 0 1 5 8 0 12 10 25
243567 0 1 1 0 1 1 0 1 0 0 0 7 9 12 15 17 15
Every file have different coursename.I did every coursename and studentid move
in one file but now i don't know how to add all marks and move to another file on same place where is id? Can you please tell me how to do it?
It looks like this:
Student# Operating Systems JAVA C++ Web Programming GPA
123456 76 63 50 82 67.75
243567 80 - 34 63 59
I did like this:
#!/bin/sh
find ~/2011/Fall/StudentsRecord -name "*.rec" | xargs grep -l 'CREDITS' | xargs cat > rsh1
echo "STUDENT ID" > rsh2
sed -n /COURSE/p rsh1 | sed 's/COURSE NAME: //g' >> rsh2
echo "GPA" >> rsh2
sed -e :a -e '{N; s/\n/ /g; ta}' rsh2 > rshf
sed '/COURSE/d;/CREDIT/d' rsh1 | sort -uk 1,1 | cut -d' ' -f1 | paste -d' ' >> rshf
Some comments and a few pointers :
It would help to add 'comments' for each line of code that is not self evident ; i.e. code like mv f f.bak doesn't need to be commented, but I'm not sure what the intent of your many lines of code are.
You insert a comment with the '#' char, like
# concatenate all files that contain the word CREDITS into a file called rsh1
find ~/2011/Fall/StudentsRecord -name "*.rec" | xargs grep -l 'CREDITS' | xargs cat > rsh1
Also note that you consistently use all uppercase for your search targets, i.e. CREDITS, when your sample files shows mixed case. Either used correct case for your search targets, i.e.
`grep -l 'Credits'`
OR tell grep to -i(gnore case), i.e.
`grep -il 'Credits'
Your line
sed -n /COURSE/p rsh1 | sed 's/COURSE NAME: //g' >> rsh2
can be reduced to 1 call to sed (and you have the same case confusion thing going on), try
sed -n '/COURSE/i{;s/COURSE NAME: //gip;}' rsh1 >> rsh2
This means (-n don't print every line by default),
`gip` = global substitute,
= ignore case in matching
print only lines where substituion was made
So you're editing out the string COURSE NAME for any line that has COURSE in it, and only printing those lines' (you probably don't need the 'g' (global) specifier given that you expect only 1 instance per line)
Your line
sed -e :a -e '{N; s/\n/ /g; ta}' rsh2 > rshf
Actually looks pretty good, very advanced, you're trying to 'fold' each 2 lines together into 1 line, right?
But,
sed '/COURSE/d;/CREDIT/d' rsh1 | sort -uk 1,1 | cut -d' ' -f1 | paste -d' ' >> rshf
I'm really confused by this, is this where you're trying to total a students score? (with a sort embedded I guess not). Why do you think you need a sort,
While it is possible to perform arithmetic in sed, it is super-crazy hard, so you can either use bash variables to calculate the values OR use a unix tool that is designed to process text AND perform logical and mathematical operations of the data presented, awk or perl come to mind here
Anyway, one solution to total each score is to use awk
echo "123456 1 1 0 1 1 0 1 0 0 0 1 5 8 0 12 10 25" |\
awk '{for (i=2;i<=NF;i++) { tot+=$i }; print $1 "\t" tot }'
Will give you a clue on how to proceed for that.
Awk has predefined variables that it populates for each file, and each line of text that it reads, i.e.
$0 = complete line of text (as defined by the internal variables RS (RecordSeparator)
which defaults to '\n' new-line char, the unix end-of-line char
$1 = first field in text (as defined by the internal variables FS (FieldSeparator)
which defaults to (possibly multiple) space chars OR tab char
a line with 2 connected spaces chars and 1 tab char has 3 fields)
NF = Number(of)Fields in current line of data (again fields defined by value of FS as
described above)
(there are many others, besides, $0, $n, $NF, $FS, $RS).
you can programatically increment for values like $1, $2, $3, by using a variable as in the example code, like $i (i is a variable that has a number between 2 and NF. The leading '$'
says give me the value of field i (i.e. $2, $3, $4 ...)
Incidentally, your problem could be easily solved with a single awk script, but apparently, you're supposed to learn about cat, cut, grep, etc, which is a very worthwhile goal.
I hope this helps.

Sorting space delimited numbers with Linux/Bash

Is there a Linux utility or a Bash command I can use to sort a space delimited string of numbers?
Here's a simple example to get you going:
echo "81 4 6 12 3 0" | tr " " "\n" | sort -g
tr translates the spaces delimiting the numbers, into carriage returns, because sort uses carriage returns as delimiters (ie it is for sorting lines of text). The -g option tells sort to sort by "general numerical value".
man sort for further details about sort.
This is a variation from #JamesMorris answer:
echo "81 4 6 12 3 0" | xargs -n1 | sort -g | xargs
Instead of tr, I use xargs -n1 to convert to new lines. The final xargs is to convert back, to a space separated sequence of numbers.
This is a variation on ghostdog74's answer that's too big to fit in a comment. It shows digits instead of names of numbers and both the original string and the result are in space-delimited strings (instead of an array which becomes a newline-delimited string).
$ s="3 2 11 15 8"
$ sorted=$(echo $(printf "%s\n" $s | sort -n))
$ echo $sorted
2 3 8 11 15
$ echo "$sorted"
2 3 8 11 15
If you didn't use the echo when setting the value of sorted, then the string has newlines in it. In that case echoing it without quotes puts it all on one line, but, as echoing it with quotes would show, each number would appear on its own line. This is the case whether the original is an array or a string.
# demo
$ s="3 2 11 15 8"
$ sorted=$(printf "%s\n" $s | sort -n)
$ echo $sorted
2 3 8 11 15
$ echo "$sorted"
2
3
8
11
15
$ s=(one two three four)
$ sorted=$(printf "%s\n" ${s[#]}|sort)
$ echo $sorted
four one three two
Using Bash parameter expansion (to replace spaces with newlines) we can do:
str="3 2 11 15 8"
sort -n <<< "${str// /$'\n'}"
# alternative
NL=$'\n'
str="3 2 11 15 8"
sort -n <<< "${str// /${NL}}"
If you actually have a space-delimited string of numbers, then one of the other answers provided would work fine. If your list is a bash array, then:
oldIFS="$IFS"
IFS=$'\n'
array=($(sort -g <<< "${array[*]}"))
IFS="$oldIFS"
might be a better solution. The newline delimiter would help if you want to generalize to sorting an array of strings instead of numbers.
Improving on Evan Krall's nice Bash "array sort" by limiting the scope of IFS to a single command:
printf "%q\n" "${IFS}"
array=(3 2 11 15 8)
array=($(IFS=$'\n' sort -n <<< "${array[*]}"))
echo "${array[#]}"
printf "%q\n" "${IFS}"
$ awk 'BEGIN{split(ARGV[1], numbers);for(i in numbers) {print numbers[i]} }' \
"6 7 4 1 2 3" | sort -n
I added this to my .zshrc (or .bashrc) file:
#sort a space-separated list of words (e.g. a list of HTML classes)
sortwords() {
echo $1 | xargs -n1 | sort -g | xargs
}
Call it from the terminal like this:
sortwords "banana date apple cherry"
# apple banana cherry date
Thanks to #FranMowinckel and others for inspiration.

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