Shell Programming and Scripting

Of course,
line by line.

Code:

NR == 1 { print "lead1   07q4    07q1    06q4    06q3    sept    oct     nov     tot" }

Just print the header (you may prefere to use BEGIN, instead of NR == 1, but in that case you should use the -v syntax for the users variable:

Code:

awk -v users="tim|tad"

Code:

$1 ~ "^("users")$" && NF > 1

+ for every record that matches the pattern:

- first field is one of the names given (the variable users is defined at the end: users="tim|tad" , the pipe | means alternation, tim OR tad.
- AND (logical and) the number of fields is greater than 1, this is specific for the OP input file format, we want to skip the following lines:

Code:

        lead1    2008q1
        tom
        pat
        tim
        tad

+ do the following:

Code:

{ 
x[$1]++
if (x[$1] == 1) {
  p[$1] = sprintf ("%s\t%s\t%s\t%s\t%s", $1, $2, $5, $6, $7)
  t[$1] = $2 + $5
}

- x is an associative array, the keys are the names ($1), the value/element is their count (how many times they appear in the matched lines). So when the name appears for the first time: x[$1] == 1:

Code:

tim    357    03     04     25     3020   3120   20
or
tad    1734   0      0      0      5213   5213   0

Save fields 1, 2, 5, 6 and 7 as value of another associative array p with key $1 (the name), sprintf is needed for formatting (values separated by tabs).
- the t array is the sum of $2 and $5 and uses the same key.

Code:

if (x[$1] == 3)

- if x[$1] is 3, the following lines:

Code:

tim    0      21     1800   0
or
tad    0      0      672    0

Code:

{
  printf "%s\t%s\t%s\t%s\t%d\n", p[$1], $2, $3, $4, t[$1] 
 }

print the previously saved (when p[$1] was 1 ) record (p[$1]) plus $2 (sept), $3 (oct), $4 (nov) and the sum.