calculate the percentage of not null recs in a file in unix - unix

Question

How do i figure out the percentage of not null records in my file in UNIX?
My file like this: I wanted to know the amount of records & the percentage of not null rec's. Tried whole lot of grep n cut commands but nothing seems to be working out. Can anyone help me here please...
"name","country","age","place"
"sam","US","30","CA"
"","","",""
"joe","UK","34","BRIS"
,,,,
"jake","US","66","Ohio"

Answer 1

Perl solution:
#!/usr/bin/perl
use warnings;
use strict;
use 5.012; # say, keys #arr
use Text::CSV_XS qw{ csv };
my ($count_all, #count_nonempty);
csv(in => shift,
out => \ 'skip',
headers => 'skip',
on_in => sub {
my (undef, $columns) = #_;
++$count_all;
length $columns->[$_] and $count_nonempty[$_]++
for 0 .. $#$columns;
},
);
for my $column (keys #count_nonempty) {
say "Column ", 1 + $column, ": ",
100 * $count_nonempty[$column] / $count_all, '%';
}
It uses Text::CSV_XS to read the CSV file. It skips the header line, and for each subsequent line, it calls the callback specified in on_in, which increments the count of all lines and also the count of empty fields per column if the length of a field is zero.

Answer 2

Along with choroba, I would normally recommend using a CSV parser on CSV data.
But in this case, all we want to look for is that a record contains any character that is not a comma or quote: if a record contains only commas and/or quotes, it is a "null" record.
awk '
/[^",]/ {nonnull++}
END {printf "%d / %d = %.2f\n", nonnull, NR, nonnull/NR}
' file
To handle leading/trailing whitespace
awk '
{sub(/^[[:blank:]]+/,""); sub(/[[:blank:]]+$/,"")}
/[^",]/ {nonnull++}
END {printf "%d / %d = %.2f\n", nonnull, NR, nonnull/NR}
' file
If allowing fields containing only whitespace, such as
" ","",,," "
is also a null record, we can simple ignore all whitespace
awk '
/[^",[:blank:]]/ {nonnull++}
END {printf "%d / %d = %.2f\n", nonnull, NR, nonnull/NR}
' file