Please how can I count the list of numbers in a file

awk '{for(i=1;i<=NF;i  ){if($i>=0 && $i<=25){print $i}}}'

Using the command above I can display the range of numbers on the terminal but if there are so many it will be difficult to count them. Please how can I show the count of the numbers on the terminal for example

1-20, 
2-22, 
3-23, 
4-24,
etc

I know I can use wc but I don't know how to infuse it into the command above

CodePudding user response：

Pipe the output to sort -n and uniq -c

awk '{for(i=1;i<=NF;i  ){if($i>=0 && $i<=25){print $i}}}' filename | sort -n | uniq -c

You need to sort first because uniq requires all the same elements to be consecutive.

CodePudding user response：

While I'm personally an awk fan, you might be glad to learn about grep -o functionality. I'm using grep -o to match all numbers in the file, and then awk can be used to pick all the numbers between 0 and 25 (inclusive). Last, we can use sort and uniq to count the results.

grep -o  "[0-9][0-9]*" file  |  awk ' $1 >= 0 && $1 <= 25 ' |  sort -n  | uniq -c

Of course, you could do the counting in awk with an associative array as Ed Morton suggests:

egrep -o  "\d " file |  awk ' $1 >= 0 && $1 <= 25 ' |  awk 'cnt[$1]   END { for (i in cnt) printf("%s-%s\n",  i,cnt[i] ) } ' 

I modified Ed's code (typically not a good idea - I've been reading his code for years now) to show a modular approach - an awk script for filtering numbers in the range of 0 and 25 and another awk script for counting a list (of anything).

I also provided another subtle difference from my first script with egrep instead of grep.

To be honest, the second awk script generates some unexpected output, but I wanted to share an example of a more general approach.

CodePudding user response：

awk '
    { for(i=1;i<=NF;i  ) if (0<=$i && $i<=25) cnts[$i]   }
    END { for (n in cnts) print n, cnts[n] }
' file