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}[^|]*$' file
grep '.*|.*|.*' will select lines with at least three fields and two separators.
Related
I am trying to count and display only the words that are repeated more than once in a file. The basic idea is:
You are given a file with names and characters like commas, colons, slashes, etc..
Use the cut command to display only the first names in the file (other commands are also allowed).
Count and then display only the names repeated more than once.
I got to the point of counting and displaying all the names. However, I haven't found a way to display and to count only those names repeated more than once.
Here is a section of the file:
user1:x:80:200:Mia,Spurs:/home/user1:/bin/bash
user2:x:80:200:Martha,Dalton:/home/user2:/bin/bash
user3:x:80:200:Lucy,Carlson:/home/user3:/bin/bash
user4:x:80:200:Carl,Bingo:/home/user4:/bin/bash
Here is what I have been able to do:
Daniel#Daniel-MacBook-Pro Files % cut -d ":" -f 5-5 file1 | cut -d "," -f 1-1 | sort -n | uniq -c
1 Mia
3 Martha
1 Lucy
1 Carl
1 Jessi
1 Joke
1 Jim
2 Race
1 Sem
1 Shirly
1 Susan
1 Tim
You can filter out the rows with count 1 with grep.
cut -d ":" -f 5 file1 | cut -d "," -f 1 | sort | uniq -c | grep -v '^ *1 '
We have to find the difference(d) Between last 2 nos and display rows with the highest value of d in ascending order
INPUT
1 | Latha | Third | Vikas | 90 | 91
2 | Neethu | Second | Meridian | 92 | 94
3 | Sethu | First | DAV | 86 | 98
4 | Theekshana | Second | DAV | 97 | 100
5 | Teju | First | Sangamithra | 89 | 100
6 | Theekshitha | Second | Sangamithra | 99 |100
Required OUTPUT
4$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.
I am dealing with the analysis of big number of dlg text files located within the workdir. Each file has a table (usually located in different positions of the log) in the following format:
File 1:
CLUSTERING HISTOGRAM
____________________
________________________________________________________________________________
| | | | |
Clus | Lowest | Run | Mean | Num | Histogram
-ter | Binding | | Binding | in |
Rank | Energy | | Energy | Clus| 5 10 15 20 25 30 35
_____|___________|_____|___________|_____|____:____|____:____|____:____|____:___
1 | -5.78 | 11 | -5.78 | 1 |#
2 | -5.53 | 13 | -5.53 | 1 |#
3 | -5.47 | 17 | -5.44 | 2 |##
4 | -5.43 | 20 | -5.43 | 1 |#
5 | -5.26 | 19 | -5.26 | 1 |#
6 | -5.24 | 3 | -5.24 | 1 |#
7 | -5.19 | 4 | -5.19 | 1 |#
8 | -5.14 | 16 | -5.14 | 1 |#
9 | -5.11 | 9 | -5.11 | 1 |#
10 | -5.07 | 1 | -5.07 | 1 |#
11 | -5.05 | 14 | -5.05 | 1 |#
12 | -4.99 | 12 | -4.99 | 1 |#
13 | -4.95 | 8 | -4.95 | 1 |#
14 | -4.93 | 2 | -4.93 | 1 |#
15 | -4.90 | 10 | -4.90 | 1 |#
16 | -4.83 | 15 | -4.83 | 1 |#
17 | -4.82 | 6 | -4.82 | 1 |#
18 | -4.43 | 5 | -4.43 | 1 |#
19 | -4.26 | 7 | -4.26 | 1 |#
_____|___________|_____|___________|_____|______________________________________
The aim is to loop over all the dlg files and take the single line from the table corresponding to wider cluster (with bigger number of slashes in Histogram column). In the above example from the table this is the third line.
3 | -5.47 | 17 | -5.44 | 2 |##
Then I need to add this line to the final_log.txt together with the name of the log file (that should be specified before the line). So in the end I should have something in following format (for 3 different log files):
"Name of the file 1": 3 | -5.47 | 17 | -5.44 | 2 |##
"Name_of_the_file_2": 1 | -5.99 | 13 | -5.98 | 16 |################
"Name_of_the_file_3": 2 | -4.78 | 19 | -4.44 | 3 |###
A possible model of my BASH workflow would be:
#!/bin/bash
do
file_name2=$(basename "$f")
file_name="${file_name2/.dlg}"
echo "Processing of $f..."
# take a name of the file and save it in the log
echo "$file_name" >> $PWD/final_results.log
# search of the beginning of the table inside of each file and save it after its name
cat $f |grep 'CLUSTERING HISTOGRAM' >> $PWD/final_results.log
# check whether it works
gedit $PWD/final_results.log
done
Here I need to substitute combination of echo and grep in order to take selected parts of the table.
You can use this one, expected to be fast enough. Extra lines in your files, besides the tables, are not expected to be a problem.
grep "#$" *.dlg | sort -rk11 | awk '!seen[$1]++'
grep fetches all the histogram lines which are then sorted in reverse order by last field, that means lines with most # on the top, and finally awk removes the duplicates. Note that when grep is parsing more than one file, it has -H by default to print the filenames at the beginning of the line, so if you test it for one file, use grep -H.
Result should be like this:
file1.dlg: 3 | -5.47 | 17 | -5.44 | 2 |##########
file2.dlg: 3 | -5.47 | 17 | -5.44 | 2 |####
file3.dlg: 3 | -5.47 | 17 | -5.44 | 2 |#######
Here is a modification to get the first appearence in case of many equal max lines in a file:
grep "#$" *.dlg | sort -k11 | tac | awk '!seen[$1]++'
We replaced the reversed parameter in sort, with the 'tac' command which is reversing the file stream, so now for any equal lines, initial order is preserved.
Second solution
Here using only awk:
awk -F"|" '/#$/ && $NF > max[FILENAME] {max[FILENAME]=$NF; row[FILENAME]=$0}
END {for (i in row) print i ":" row[i]}' *.dlg
Update: if you execute it from different directory and want to keep only the basename of every file, to remove the path prefix:
awk -F"|" '/#$/ && $NF > max[FILENAME] {max[FILENAME]=$NF; row[FILENAME]=$0}
END {for (i in row) {sub(".*/","",i); print i ":" row[i]}}'
Probably makes more sense as an Awk script.
This picks the first line with the widest histogram in the case of a tie within an input file.
#!/bin/bash
awk 'FNR == 1 { if(sel) print sel; sel = ""; max = 0 }
FNR < 9 { next }
length($10) > max { max = length($10); sel = FILENAME ":" $0 }
END { if (sel) print sel }' ./"$prot"/*.dlg
This assumes the histograms are always the tenth field; if your input format is even messier than the lump you show, maybe adapt to taste.
In some more detail, the first line triggers on the first line of each input file. If we have collected a previous line (meaning this is not the first input file), print that, and start over. Otherwise, initialize for the first input file. Set sel to nothing and max to zero.
The second line skips lines 1-8 which contain the header.
The third line checks if the current line's histogram is longer than max. If it is, update max to this histogram's length, and remember the current line in sel.
The last line is spillover for when we have processed all files. We never printed the sel from the last file, so print that too, if it's set.
If you mean to say we should find the lines between CLUSTERING HISTOGRAM and the end of the table, we should probably have more information about what the surrounding lines look like. Maybe something like this, though;
awk '/CLUSTERING HISTOGRAM/ { if (sel) print sel; looking = 1; sel = ""; max = 0 }
!looking { next }
looking > 1 && $1 != looking { looking = 0; nextfile }
$1 == looking && length($10) > max { max = length($10); sel = FILENAME ":" $0 }
END { if (sel) print sel }' ./"$prot"/*.dlg
This sets looking to 1 when we see CLUSTERING HISTOGRAM, then counts up to the first line where looking is no longer increasing.
I would suggest processing using awk:
for i in $FILES
do
echo -n \""$i\": "
awk 'BEGIN {
output="";
outputlength=0
}
/(^ *[0-9]+)/ { # process only lines that start with a number
if (length(substr($10, 2)) > outputlength) { # if line has more hashes, store it
output=$0;
outputlength=length(substr($10, 2))
}
}
END {
print output # output the resulting line
}' "$i"
done
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
I am new to shell scripting, can anyone give me shell script for the condition below
My Input:
id | name | values
----+------+--------
1 | abc | 2
1 | abc | 3
1 | abc | 4
1 | abc | 5
1 | abc | 6
1 | abc | 7
Expected Output:
1,abc,2
"
"
1 million records
You can use awk for this:
awk -F '[[:blank:]]*\\|[[:blank:]]*' -v OFS=, 'NF==3 && NR>1{sub(/^[[:blank:]]*/, "", $1); print}' file
1,abc,2
1,abc,3
1,abc,4
1,abc,5
1,abc,6
1,abc,7