Shell Programming and Scripting

Hello to all,

I have show below a file separated by commas. In first column has numbers where the last number is 13.

Code:

1,4
2,6
3,7
5,2
6,5
7,5
8,65
9,10
11,78
13,2

What I want to know is which numbers are missing from 1 to 13 (in this case 13 is last number in column 1). My real file
has more than 5 million lines.

The output expected is as below:

Code:

4
10
12

Thanks in advance for any help.

Try

Code:

$ awk -F, '{while(++x<$1)print x}' file
4
10
12

More awk... More awk please...

Code:

awk -F, '$1>a+1{print NR}{a=$1}' file
4
9
10

9 is there in file

Hello Akshay Hegde,

Thank you for your help.

It works with sample file but is not working with real file.

Real file has 5,440,177 lines and the last number in column 1 is 5440255. So, substracting both there are 78 numbers that are missing in column 1.
But trying your script I get more than 33 million of lines and I stopped since it seems enters in an infinite loop.

Is there a way to preload an array from 1 to N (N=13 in this case, in real file N=5440255)? in order to compare array which values from column 1 are not in array?

Thanks again

Oops, thanks Akshay. Here's the remix:

Code:

awk '($1!=a+1){print a+1}{a=$1}' file
4
10
12

Hello in2nix4life,

Thanks for the help.

I've tried your script, the expected number of missing values in real file are 78 but is printing 108 numbers.

Is there a way to compare column 1 with a preloaded array that contains elements from 1 to N? in order to print values of array that are not present in column1?

Thanks again for the help.