How to count number frequency from specific column?

Question

Below I have given my file (filename = 1.txt) structure. I want to count the number frequency from the first column and it should be started from line 3. Because the first line contains 411 and second line contain some text which is not the interest of mine.

I can count a specific number using below:

awk '($1==15){ ++count } END{ print count }' 1.txt> output.txt

my file structure:

411
Lattice="156.0 0.0 0.0 0.0 156.0 0.0 0.0 0.0 156.0" 
1 410.0 2 1
2 1059.0 2 2
2 1060.0 2 3
3 2117.0 2 4
4 4726.0 2 5
5 3219.0 2 6
6 4744.0 2 7
7 4918.0 2 8
8 10686.0 2 9
9 11055.0 2 10
10 16475.0 2 11
11 14698.0 2 12
11 17430.0 2 13
12 15235.0 2 14
13 15799.0 2 15
14 21476.0 2 16
15 18561.0 2 17
15 18562.0 2 18
15 21595.0 2 19
15 21636.0 2 20
15 21684.0 2 21
16 24262.0 2 22
14 21475.0 2 23
17 24674.0 2 24

my desired output

 1 1
 2 2
 3 1
 4 1
 .
 .
14 2
15 5

Answer 1

You can use an associative array that is keyed on the column value, assigning values only from the third record (line) onward (NR>2):

$ awk 'NR>2 {count[$1]++} END {for (i in count) print i, count[i]}' 1.txt
1 1
2 2
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
11 2
12 1
13 1
14 2
15 5
16 1
17 1

Note that the order of array traversal is not guaranteed - you may need additional sorting if output order is important.

Answer 2

Tried with below script and it worked fine

for i in `awk 'NR >2 {print $1}' p.txt| sort -k1 -n -u`; do  echo $i; awk 'NR >2 {print $1}' p.txt|awk -v i="$i" '$1 == i {print $1}'| awk '{print NR}'| sed -n '$p'; done| sed "N;s/\n/ /g"

output

1 1
2 2
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
11 2
12 1
13 1
14 2
15 5
16 1
17 1