I have the following text file:
A,B,C
A,B,C
A,B,C
Is there a way, using standard *nix tools (cut, grep, awk, sed, etc), to process such a text file and get the following output:
A
A
A
B
B
B
C
C
C
You can do:
tr , \\n
and that will generate
A
B
C
A
B
C
A
B
C
which you could sort.
Unless you want to pull the first column then second then third, in which case you want something like:
awk -F, '{for(i=1;i<=NF;++i) print i, $i}' | sort -sk1 | awk '{print $2}'
To explain this, the first part generates
1 A
2 B
3 C
1 A
2 B
3 C
1 A
2 B
3 C
the second part will stably sort (so the internal order is preserved)
1 A
1 A
1 A
2 B
2 B
2 B
3 C
3 C
3 C
and the third part will strip the numbers
You could use a shell for-loop combined with cut if you know in advanced the number of columns. Here is an example using bash syntax:
for i in {1..3}; do
cut -d, -f $i file.txt
done
Try:
awk 'BEGIN {FS=","} /([A-C],)+([A-C])?/ {for (i=1;i<=NF;i++) print $i}' YOURFILE | sort
Related
I have two files that I am joining on one column. After the join, I just want the output to be all of the columns, in the original order, from only one of the files. For example:
cat file1.tsv
1 a ant
2 b bat
3 c cat
8 d dog
9 e eel
cat file2.tsv
1 I
2 II
3 III
4 IV
5 V
join -1 1 -2 1 file1.tsv file2.tsv -t $'\t' -o 1.1,1.2,1.3
1 a ant
2 b bat
3 c cat
I know I an use -o 1.1,1.2.. notation but my file has over two dozen columns. Is there some wildcard that I can use to say -o 1.* or something?
I'm not aware of wildcards in the format string.
From your desired output I think that what you want may be achievable like so without having to specify all the enumerations:
grep -f <(awk '{print $1}' file2.tsv ) file1.tsv
1 a ant
2 b bat
3 c cat
Or as an awk-only solution:
awk '{if(NR==FNR){a[$1]++}else{if($1 in a){print}}}' file2.tsv file1.tsv
1 a ant
2 b bat
3 c cat
I want to select top 3 results for every line that has the same first two column.
For example the data will look like,
cat data.txt
A A 10
A A 1
A A 2
A A 5
A A 8
A B 1
A B 2
A C 6
A C 5
A C 10
A C 1
B A 1
B A 1
B A 2
B A 8
And for the result I want
A A 10
A A 8
A A 5
A B 2
A B 1
A C 10
A C 6
A C 5
B A 1
B A 1
B A 2
Note that some of the "groups" do not contain 3 rows.
I have tried
sort -k1,1 -k2,2 -k3,3nr data.txt | sort -u -k1,1 -k2,2 > 1.txt
comm -23 <(sort data.txt) <(sort 1.txt)| sort -k1,1 -k2,2 -k3,3nr| sort -u -k1,1 -k2,2 > 2.txt
comm -23 <(sort data.txt) <(cat 1.txt 2.txt | sort)| sort -k1,1 -k2,2 -k3,3nr| sort -u -k1,1 -k2,2 > 3.txt
It seems like it's working but since I am learning to code better was wondering if there was a better way to go about this. Plus, my code will generate many files that I will have to delete.
You can do:
$ sort -k1,1 -k2,2 -k3,3nr file | awk 'a[$1,$2]++<3'
A A 10
A A 8
A A 5
A B 2
A B 1
A C 10
A C 6
A C 5
B A 8
B A 2
B A 1
Explanation:
There are two key items to understand the awk program; associative arrays and fields.
If you reference an empty awk array element, it is an empty container -- ready for anything you put into it. You can use that as a counter.
You state If first two columns are equal...
The sort puts the file in order desired. The statement a[$1,$2] uses the values of the first two fields as a unique entry into an associative array.
You then state ...select top 3 based on descending order of 3rd column...
Once again, the sort put the file into the desired order, and the statement a[$1,$2]++ counts them. Now just count up to three.
awk is organized into blocks of condition {action} The statement a[$1,$2]++<3 is true until there are more than 3 of the same pattern seen.
A wordier version of the program would be:
awk 'a[$1,$2]++<3 {print $0}'
But the default action if the condition is true is to print $0 so it is not needed.
If you are processing text in Unix, you should get to know awk. It is the most powerful tool that POSIX guarantees you will have, and is commonly used for these tasks.
Great place to start is the online book Effective AWK Programming by Arnold D. Robbins
#Dawg has the best answer. This one will be a little lighter on memory, which probably won't be a concern for your data:
sort -k1,2 -k3,3nr file |
awk '
{key = $1 FS $2}
prev != key {prev = key; count = 1}
count <= 3 {print; count++}
'
You can sort the file by first two columns primarily and by the 3rd one numerically secondarily, then read the output and only print the first three lines for each combination of the first two columns.
sort -k1,2 -k3,3rn data.txt \
| while read c1 c2 n ; do
if [[ $c1 == $l1 && $c2 == $l2 ]] ; then
((c++))
else
c=0
fi
if (( c < 3 )) ; then
echo $c1 $c2 $n
l1=$c1
l2=$c2
fi
done
This question already has answers here:
Find duplicate lines in a file and count how many time each line was duplicated?
(7 answers)
Closed 7 years ago.
I have file file.txt which look like this
a
b
b
c
c
c
I want to know the command to which get file.txt as input and produces the output
a 1
b 2
c 3
I think uniq is the command you are looking for. The output of uniq -c is a little different from your format, but this can be fixed easily.
$ uniq -c file.txt
1 a
2 b
3 c
If you want to count the occurrences you can use uniq with -c.
If the file is not sorted you have to use sort first
$ sort file.txt | uniq -c
1 a
2 b
3 c
If you really need the line first followed by the count, swap the columns with awk
$ sort file.txt | uniq -c | awk '{ print $2 " " $1}'
a 1
b 2
c 3
You can use this awk:
awk '!seen[$0]++{ print $0, (++c) }' file
a 1
b 2
c 3
seen is an array that holds only uniq items by incrementing to 1 first time an index is populated. In the action we are printing the record and an incrementing counter.
Update: Based on comment below if intent is to get a repeat count in 2nd column then use this awk command:
awk 'seen[$0]++{} END{ for (i in seen) print i, seen[i] }' file
a 1
b 2
c 3
I've got an input file (input.txt) like this:
name value1 value2
A 3 1
B 7 4
C 2 9
E 5 2
And another file with a list of names (names.txt) like so:
B
C
Using grep -f, I can get all the lines with names "B" and "C"
grep -wFf names.txt input.txt
to get
B 7 4
C 2 9
However, I want to keep the header at the top of the output file, and also rename the column name "name" with "ID". And using grep, to keep the rows with names B and C, the output should be:
**ID** value1 value2
B 7 4
C 2 9
I'm thinking awk should be able to accomplish this, but being new to awk I'm not sure how to approach this. Help appreciated!
While it is certainly possible to do this in awk, the fastest way to solve your actual problem is to simply prepend the header you want in front of the grep output.
echo **ID** value1 value2 > Output.txt && grep -wFf names.txt input.txt >> Output.txt
Update Since the OP has multiple files, we can modify the above line to pull the first line out of the input file instead.
head -n 1 input.txt | sed 's/name/ID/' > Output.txt && grep -wFf names.txt input.txt >> Output.txt
Here is how to do it with awk
awk 'FNR==NR {a[$1];next} FNR==1 {$1="ID";print} {for (i in a) if ($1==i) print}' name input
ID value1 value2
B 7 4
C 2 9
Store the names in an array a
Then test filed #1 if it contains data in array a
I know that with sed I can print
cat current.txt | sed 'N;s/\n/,/' > new.txt
A
B
C
D
E
F
to
A,B
C,D
E,F
What I would like to do is following:
A
B
C
D
E
F
to
A,D
B,E
C,F
I'd like to join 1 with 4, 2 with 5, 3 with 6 and so on.
Is this possible with sed? Any idea how it could be achieved?
Thank you.
Try printing in columns:
pr -s, -t -2 current.txt
This is longer than I was hoping, but:
$ lc=$(( $(wc -l current.txt | sed 's/ .*//') / 2 ))
$ paste <(head -"$lc" current.txt) <(tail -"$lc" current.txt) | column -t -o,
The variable lc stores the number of lines in current.txt divided by two. Then head and tail are used to print lc first and lc last lines, respectively (i.e. the first and second half of the file); then paste is used to put the two together and column changes tabs to commas.
An awk version
awk '{a[NR]=$0} NR>3 {print a[NR-3]","$0}' current.txt
A,D
B,E
C,F
This solution is easy to adjust if you like other interval.
Just change NR>3 and NR-3 to desired number.