I have come close to counting all occurrences of punctuation, however punctuation characters that are right next to each other get counted as one.
Like so:
cat filename.txt |
tr -sc '[:punct:]' '\n' |
sort |
uniq -c |
sort -bnr`
Which prints something like this:
15 ,
9 !
5 .
2 ;
2 !"
2 '
1 -
1 --
1 :
1 ?
It is clearly only counting punctuation, but how would I separate those that are right next to each other?
This:
tr -sc '[:punct:]' '\n'
Basically what you do here is replace all the non-punctuation characters with \n. So when there is no such character between two punctuation chars , you get them next to each other
You want something like that:
cat filename.txt | tr -cd [:punct:] | fold -w 1 | sort | uniq -c | sort -bnr
Related
I am trying to count and display only the words that are repeated more than once in a file. The basic idea is:
You are given a file with names and characters like commas, colons, slashes, etc..
Use the cut command to display only the first names in the file (other commands are also allowed).
Count and then display only the names repeated more than once.
I got to the point of counting and displaying all the names. However, I haven't found a way to display and to count only those names repeated more than once.
Here is a section of the file:
user1:x:80:200:Mia,Spurs:/home/user1:/bin/bash
user2:x:80:200:Martha,Dalton:/home/user2:/bin/bash
user3:x:80:200:Lucy,Carlson:/home/user3:/bin/bash
user4:x:80:200:Carl,Bingo:/home/user4:/bin/bash
Here is what I have been able to do:
Daniel#Daniel-MacBook-Pro Files % cut -d ":" -f 5-5 file1 | cut -d "," -f 1-1 | sort -n | uniq -c
1 Mia
3 Martha
1 Lucy
1 Carl
1 Jessi
1 Joke
1 Jim
2 Race
1 Sem
1 Shirly
1 Susan
1 Tim
You can filter out the rows with count 1 with grep.
cut -d ":" -f 5 file1 | cut -d "," -f 1 | sort | uniq -c | grep -v '^ *1 '
We have to find the difference(d) Between last 2 nos and display rows with the highest value of d in ascending order
INPUT
1 | Latha | Third | Vikas | 90 | 91
2 | Neethu | Second | Meridian | 92 | 94
3 | Sethu | First | DAV | 86 | 98
4 | Theekshana | Second | DAV | 97 | 100
5 | Teju | First | Sangamithra | 89 | 100
6 | Theekshitha | Second | Sangamithra | 99 |100
Required OUTPUT
4$Theekshana$Second$DAV$97$100$3
5$Teju$First$Sangamithra$89$100$11
3$Sethu$First$DAV$86$98$12
awk 'BEGIN{FS="|";OFS="$";}{
avg=sqrt(($5-$6)^2)
print $1,$2,$3,$4,$5,$6,avg
}'|sort -nk7 -t "$"| tail -3
Output:
4 $ Theekshana $ Second $ DAV $ 97 $ 100$3
5 $ Teju $ First $ Sangamithra $ 89 $ 100$11
3 $ Sethu $ First $ DAV $ 86 $ 98$12
As you can see there is space before and after $ sign but for the last column (avg) there is no space, please explain why its happening
2)
awk 'BEGIN{FS=" | ";OFS="$";}{
avg=sqrt(($5-$6)^2)
print $1,$2,$3,$4,$5,$6,avg
}'|sort -nk7 -t "$"| tail -3
OUTPUT
4$|$Theekshana$|$Second$|$0
5$|$Teju$|$First$|$0
6$|$Theekshitha$|$Second$|$0
I have not mentiond | as the output field separator but still it appears, why is this happening and the difference is zero too
I am just 6 days old in unix,please answer even if its easy
your field separator is only the pipe symbol, so surrounding whitespace is part of the field definitions and that's what you see in the output. In combined uses pipe has the regex special meaning and need to be escaped. In your second case it means space or space is the field separator.
$ awk 'BEGIN {FS=" *\\| *"; OFS="$"}
{d=sqrt(($NF-$(NF-1))^2); $1=$1;
print d "\t" $0,d}' file | sort -n | tail -3 | cut -f2-
4$Theekshana$Second$DAV$97$100$3
5$Teju$First$Sangamithra$89$100$11
3$Sethu$First$DAV$86$98$12
a slight rewrite will eliminate the number of fields dependency and fixes the format.
How do I edit command pr
tr [:upper:] [:lower:] < "${FILE}" | tr -d [:digit:] | sort | pr -s' ' -t3 | nl
so that the output looks like this:
1 bcxwhex jbdafn osnu
2 bijly jdofnx uriqcl
3 btgr jz uztyp
4 bwifmn kfaky wfwdz
5 bxgvs kigdo wgdax
6 cfukt lf wgfil
7 cgyqlp lnccknh ypt
8 eocbm njevos zcplnln
9 hgmbc ocndbmr znknpo
10 iawmkbh opder zyezfq
and not like this:
1 bcxwhex jbdafn osnu
2 bijly jdofnx uriqcl
3 btgr jz uztyp
4 bwifmn kfaky wfwdz
5 bxgvs kigdo wgdax
6 cfukt lf wgfil
7 cgyqlp lnccknh ypt
8 eocbm njevos zcplnln
9 hgmbc ocndbmr znknpo
10 iawmkbh opder zyezfq
Don't use -s ' ' if you don't want the columns to separated by a single space.
As an aside, don't use upper case for your private variables.
You should replace space to tab with tr, before you do nl
tr [:upper:] [:lower:] < "${FILE}" | tr -d [:digit:] | sort | pr -s' ' -t3 | tr ' ' \\t| nl
And that will tidy up the spaces for you
So I have a text being outputted that has http status codes in one column and an ip adress in the other. I wan't to sort this by number of occurances so that
1 2 1 3 4 5 4 4
Looks like
4 4 4 1 1 2 3 5
This is for the second column of status codes, the ip adresses dont need to be sorted in any particular order
Since 4 is the most common one it should be first and then 1 and so forth.
However all that I can find is how to use uniq for example in order to count the occurances, thereby removing duplicates and prefixing a number to each row.
The regular sort command does not support this as far as i can tell as well.
Any help would be appreciated
You can still use sort | uniq -c, then interpret the number of occurrences by printing the number the given times by looping:
tr ' ' '\n' < file \
| sort | uniq -c | sort -k1,1nr -k2n \
| while read times status ; do
for i in $(seq 1 $times); do
printf '%s ' $status
done
done
I have a file that include the following lines :
2 | blah | blah
1 | blah | blah
3 | blah
2 | blah | blah
1
1 | high | five
3 | five
I wanna extract only the lines that has 3 columns (3 fields, 2 seperators...)
I wanna pipe it to the following commands :
| sort -nbsk1 | cut -d "|" -f1 | uniq -d
So after all I will get only :
2
1
Any suggestions ?
It's a part of homework assignment, we are not allowed to use awk\sed and some more commands.. (grep\tr and whats written above can be used)
Thanks
since you said grep is allowed:
grep -E '^([^|]*\|){2}[^|]*$' file
grep '.*|.*|.*' will select lines with at least three fields and two separators.