I have a file with a format similar to (in reality there are many more columns and rows)
ID flag
1 a
2 n
3 n
4 a
5 n
6 n
7 a
8 a
9 n
10 n
11 n
12 n
I want to find out how many rows have the value n in $2.
I have tried awk '{if ($2 == 'n') SUM += 1} END {print SUM}' filename
This does not work. Any ideas?
One way:
awk '$2=="n"{x++}END{print x}' file
Related
I have millions of records in my file, what i need to do is print columns 1396 to 1400 for specific number of rows, and if i can get this in excel or notepad.
Tried with this command
awk {print $1396,$1397,$1398,$1399,$1400}' file_name
But this is running for each row.
You need a condition to specify which rows to apply the action to:
awk '<<condition goes here>> {print $1396,$1397,$1398,$1399,$1400}' file_name
For example, to do this only for rows 50 to 100:
awk 'NR >= 50 && NR <= 100 {print $1396,$1397,$1398,$1399,$1400}' file_name
(Depending on what you want to do, you can also have much more complicated selection patterns than this.)
Here's a simpler example for testing:
awk 'NR >= 3 && NR <= 5 {print $2, $3}'
If I run this on an input file containing
1 2 3 4
2 3 4 5
3 a b 6
4 c d 7
5 e f 8
6 7 8 9
I get the output
a b
c d
e f
Guys I have a file like this
NR column
1 1
2 1
3 0
4 0
5 0
6 1
7 1
8 1
9 1
10 0
11 0
12 0
13 1
14 1
What I need is to find the NR what will tell me where there are 1.
so my ideal output should tell me from NR=1 - 2 (there are 1s, then), NR=6 - 9, NR=13 - 14
or
1
2
6
9
13
14
Since, I think is easier not consider in the output the first row and the last. I expect that the output is
2
6
9
13
I've been trying a way to use getline but unsuccessfully.
I am sure there is an easy way to do this, help?
Thanks
Assuming your output above was incorrect (and it should really be the line number where the 0/1 or 1/0 transition happens - so the lines would be: "1, 3, 6, 10, 13"), then an awk oneliner is:
awk 'prev!=$0{print NR};{prev=$0}' file
which says:
for every line that doesn't match the prev line, print the line number, and
for every line, save the prev line
$ awk 'NR>1 && $0!=prev{print NR} {prev=$0}' file
3
6
10
13
or for your updated requirements:
$ awk '$1!=prev{print NR-prev} {prev=$1} END{if (prev) print NR}' file
1
2
6
9
13
14
awk to the rescue!
$ awk '!p&&$2==1{p=$1}
p&&!$2{print p"-"($1-1);p=0}
END{if(p) print p"-"$1}' file
1-2
6-9
13-14
{
if (NR > 1 && last != $0) {
print NR;
}
last = $0;
}
Another way
awk '$2!=x{x=$2;print NR-!($2)}END{if(x)print NR}' file
1
2
6
9
13
14
I am looking for a way to select a column (e. g. eighth column) of a data file and write the first five numbers of that column in a row, the next five numbers in second row, and so on.
I have been testing with awk and printf without success.
The awk way to do this is to switch from using OFS and ORS to separate the output using the modulus function:
$ seq 1 20 | awk '{printf "%s", $1 (NR % 5 ? OFS : ORS)}'
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
Change $1 to $8 for the eigth column for example and NR % 5 to NR % 10 for rows of 10 instead of 5. The seq command just generate a single column of digits from 1 to 20 used for demonstration.
I also find using xargs useful for this kind of thing:
$ seq 1 20 | awk '{print $1}' | xargs -n5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
The awk isn't necessary for the example as seq only produces a single column however for your question change $1 to $8 to select only the eighth column from your input. With this approach you could also switch out awk with cut.
This will also produce the format requested
seq 1 20 | awk '{printf("%s ", $1); if (NR % 5 == 0) printf("\n")}'
where $1 indicates de column number which could be changed when passing an archive to the awk line.
I have a one column file composed by only integer as
1
1
4
3
3
2
I want to count how many time a number appear in the file. The output file should be:
1 2
2 1
3 2
4 1
Thanks
try this line:
awk '{a[$0]++}END{for(x in a)print x,a[x]}' file
awk '{tot[$0]++} END{for (n in tot) {print n,tot[n]}} ' numbers
I wanted to merge two files into a single one line by line using the first three columns as a key. Example:
file1.txt
a b c 1 4 7
x y z 2 5 8
p q r 3 6 9
file2.txt
p q r 11
a b c 12
x y z 13
My desired output for the above two files is:
a b c 1 4 7 12
x y z 2 5 8 13
p q r 3 6 9 11
The number of columns in each file is not fixed, it can vary from line to line. Also, I got more than 27K lines in each file.
They are not ordered. They only thing is that the first three fields are the same for both files.
You could also use join, it requires sorted input and that the first 3 fields are merged. The example below sorts each file and lets sed merge and separate the fields:
join <(sort file1.txt | sed 's/ /-/; s/ /-/') \
<(sort file2.txt | sed 's/ /-/; s/ /-/') |
sed 's/-/ /; s/-/ /'
Output:
a b c 1 4 7 12
p q r 3 6 9 11
x y z 2 5 8 13
Join on the first three fields where the number of fields are variable (four or more):
{
# get the forth field until the last
for (i=4;i<=NF;i++)
f=f$i" "
# concat fields
arr[$1OFS$2OFS$3]=arr[$1OFS$2OFS$3]f;
# reset field string
f=""
}
END {
for (key in arr)
print key, arr[key]
}
Run like:
$ awk -f script.awk file1 file2
a b c 1 4 7 12
p q r 3 6 9 11
x y z 2 5 8 13
try this:
awk 'NR==FNR{a[$1$2$3]=$4;next}$1$2$3 in a{print $0, a[$1$2$3]}' file2 file1
If the columns have varying lengths, you could try something like this using SUBSEP:
awk 'NR==FNR{A[$1,$2,$3]=$4; next}($1,$2,$3) in A{print $0, A[$1,$2,$3]}' file2 file1
For varying columns in file1 and sorted output, try:
awk '{$1=$1; i=$1 FS $2 FS $3 FS; sub(i,x)} NR==FNR{A[i]=$0; next}i in A{print i $0, A[i]}' file2 file1 | sort