I get results like below from a pipeline in linux:
1 test1
1 test2
2 test3
1 test4
3 test5
1 test6
1 test7
How can I use grep to retrieve only the lines where the first column is > 1?
Don't use grep for this. Try awk instead:
<pipeline> | awk '$1>1 {print $0}'
grep -v "^1"
-v selects non-matching lines
^ is the start of a line
EDIT: As pointed out in the comments, this solution does not filter out lines starting with multi-digit numbers. Adding a space after the 1 solves the problem:
grep -v "^1 "
use the "^" char, it marks the beginning of a line
-v will not include lines starting with 1
include the extra space, so it will exclude lines like "1 asd" but not "12 asd"
grep -v "^1 "
Related
I have a file with thousands of rows. I want to print the rows which do not contain a period.
awk '{print$2}' file.txt | head
I have used this to print the column I am interested in, column 2 (The file only has two columns).
I have removed the head and then did
awk '{print$2}' file.txt | grep -v "." | head
But I only get blank lines not any actual values which is expected, I think it has included the spaces between the rows but I am not sure.
Is there an alternative command?
As suggested by Jim, I did-
awk '{print$2}' file.txt | grep -v "\." | head
However the number of lines is greater than before, is this expected? Also, my output is a list of numbers but with spaces in between them (Vertical), is this normal?
file.txt example below-
120.4 3
270.3 7.9
400.8 3.9
200.2 4
100.2 8.7
300.2 3.4
102.3 6
49.0 2.3
38.0 1.2
So the expected (and correct) output would be 3 lines, as there is 3 values in column 2 without the period:
$ awk '{print$2}' file.txt | grep -v "\." | head
3
4
6
However, when running the code as above, I instead get 5, which is also counting the spaces between the rows I think:
$ awk '{print$2}' file.txt | grep -v "\." | head
3
4
6
You seldom need to use grep if you're already using awk
This would print the second column on each line where that second column doesn't contain a dot:
awk '$2 !~ /\./ {print $2}'
But you also wanted to skip empty lines, or perhaps ones where the second column is not empty. So just test for that, too:
awk '$2 != "" && $2 !~ /\./ {print $2}'
(A more amusing version would be awk '$2 ~ /./ && $2 !~ /\./ {print $2}' )
As you said, grep -v "." gives you only blank lines. That's because the dot means "any character", and with -v, the only lines printed are those that don't contain, well, any characters.
grep is interpreting the dot as a regex metacharacter (the dot will match any single character). Try escaping it with a backslash:
awk '{print$2}' file.txt | grep -v "\." | head
If I understand well, you can try this sed
sed ':A;N;${s/.*/&\n/};/\n$/!bA;s/\n/ /g;s/\([^ ]*\.[^ ]* \)//g' file.txt
output
3
4
6
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
Given input such as:
1
1a
1.1b
2.0c
How to extract the integer/decimal number at beginning of each input line, using only Linux/Unix command line utilities?
Using awk, you could say:
awk '{print $0+0}'
Awk is available in Linux, BSD, and many other Unix-like operating systems. It helps in this way:
echo "1" | awk '{a+=$0; print a}' # output 1
echo "1a" | awk '{a+=$0; print a}' # output 1
echo "1.1b" | awk '{a+=$0; print a}' # output 1.1
echo "2.0c" | awk '{a+=$0; print a}' # output 2
Some more awk
For extracting only digits
$ awk 'gsub(/[[:alpha:]].*/,x,$1) + 1' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1.1
2.0
For integer
$ awk '{print int($0)}' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1
2
---edit---
If there is any blank line in file, you can avoid printing zero from following
$ awk 'NF{$0+=0}1' << EOF
1
1a
1.1b
2foot4c
2
EOF
1
1
1.1
2
2
Here is a way to do this with sed:
echo "12.3abc" | sed -n 's/^\([0-9.][0-9.]*\).*/\1/p'
Output:
12.3
The block in parentheses matches all numbers or periods '.' that occur at the beginning of the line. Everything after that is match by the '.*'.
The \1 says to replace the entire line with just the portion that was matched in the parentheses.
Assuming your version of grep supports -o:
grep -o '^[0-9.]\+' data.in
NB: This will match any sequence of digits and decimal points at the start of the line.
I want to find the last appearance of a string in a text file with linux commands. For example
1 a 1
2 a 2
3 a 3
1 b 1
2 b 2
3 b 3
1 c 1
2 c 2
3 c 3
In such a text file, i want to find the line number of the last appearance of b which is 6.
I can find the first appearance with
awk '/ b / {print NR;exit}' textFile.txt
but I have no idea how to do it for the last occurrence.
cat -n textfile.txt | grep " b " | tail -1 | cut -f 1
cat -n prints the file to STDOUT prepending line numbers.
grep greps out all lines containing "b" (you can use egrep for more advanced patterns or fgrep for faster grep of fixed strings)
tail -1 prints last line of those lines containing "b"
cut -f 1 prints first column, which is line # from cat -n
Or you can use Perl if you wish (It's very similar to what you'd do in awk, but frankly, I personally don't ever use awk if I have Perl handy - Perl supports 100% of what awk can do, by design, as 1-liners - YMMV):
perl -ne '{$n=$. if / b /} END {print "$n\n"}' textfile.txt
This can work:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
We check every second file being "b" and we record the number of line. It is appended, so by the time we finish reading the file, it will be the last one.
Test:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
6
Update based on sudo_O advise:
$ awk '{if ($2=="b") a=NR} END{print a}' your_file
to avoid having some abc in 2nd field.
It is also valid this one (shorter, I keep the one above because it is the one I thought :D):
$ awk '$2=="b" {a=NR} END{print a}' your_file
Another approach if $2 is always grouped (may be more efficient then waiting until the end):
awk 'NR==1||$2=="b",$2=="b"{next} {print NR-1; exit}' file
or
awk '$2=="b"{f=1} f==1 && $2!="b" {print NR-1; exit}' file
How can I find the unique lines and remove all duplicates from a file?
My input file is
1
1
2
3
5
5
7
7
I would like the result to be:
2
3
sort file | uniq will not do the job. Will show all values 1 time
uniq has the option you need:
-u, --unique
only print unique lines
$ cat file.txt
1
1
2
3
5
5
7
7
$ uniq -u file.txt
2
3
Use as follows:
sort < filea | uniq > fileb
You could also print out the unique value in "file" using the cat command by piping to sort and uniq
cat file | sort | uniq -u
While sort takes O(n log(n)) time, I prefer using
awk '!seen[$0]++'
awk '!seen[$0]++' is an abbreviation for awk '!seen[$0]++ {print}', print line(=$0) if seen[$0] is not zero.
It take more space but only O(n) time.
I find this easier.
sort -u input_filename > output_filename
-u stands for unique.
you can use:
sort data.txt| uniq -u
this sort data and filter by unique values
uniq -u has been driving me crazy because it did not work.
So instead of that, if you have python (most Linux distros and servers already have it):
Assuming you have the data file in notUnique.txt
#Python
#Assuming file has data on different lines
#Otherwise fix split() accordingly.
uniqueData = []
fileData = open('notUnique.txt').read().split('\n')
for i in fileData:
if i.strip()!='':
uniqueData.append(i)
print uniqueData
###Another option (less keystrokes):
set(open('notUnique.txt').read().split('\n'))
Note that due to empty lines, the final set may contain '' or only-space strings. You can remove that later. Or just get away with copying from the terminal ;)
#
Just FYI, From the uniq Man page:
"Note: 'uniq' does not detect repeated lines unless they are adjacent. You may want to sort the input first, or use 'sort -u' without 'uniq'. Also, comparisons honor the rules specified by 'LC_COLLATE'."
One of the correct ways, to invoke with:
#
sort nonUnique.txt | uniq
Example run:
$ cat x
3
1
2
2
2
3
1
3
$ uniq x
3
1
2
3
1
3
$ uniq -u x
3
1
3
1
3
$ sort x | uniq
1
2
3
Spaces might be printed, so be prepared!
uniq -u < file will do the job.
uniq should do fine if you're file is/can be sorted, if you can't sort the file for some reason you can use awk:
awk '{a[$0]++}END{for(i in a)if(a[i]<2)print i}'
sort -d "file name" | uniq -u
this worked for me for a similar one. Use this if it is not arranged.
You can remove sort if it is arranged
This was the first i tried
skilla:~# uniq -u all.sorted
76679787
76679787
76794979
76794979
76869286
76869286
......
After doing a cat -e all.sorted
skilla:~# cat -e all.sorted
$
76679787$
76679787 $
76701427$
76701427$
76794979$
76794979 $
76869286$
76869286 $
Every second line has a trailing space :(
After removing all trailing spaces it worked!
thank you
Instead of sorting and then using uniq, you could also just use sort -u. From sort --help:
-u, --unique with -c, check for strict ordering;
without -c, output only the first of an equal run