(hadoop.pig) multiple counts in single table - hadoop

Question

So, I have a data that has two values, string, and a number.
data(string:chararray, number:int)
and I am counting in 5 different rules,
1: int being 0~1.
2: int being 1~2.
~
5: int being 4~5.
So I was able to count them individually,
zero_to_one = filter avg_user by average_stars >= 0 and average_stars <= 1;
A = GROUP zero_to_one ALL;
zto_count = FOREACH A GENERATE COUNT(zero_to_one);
one_to_two = filter avg_user by average_stars > 1 and average_stars <= 2;
B = GROUP one_to_two ALL;
ott_count = FOREACH B GENERATE COUNT(one_to_two);
two_to_three = filter avg_user by average_stars > 2 and average_stars <= 3;
C = GROUP two_to_three ALL;
ttt_count = FOREACH C GENERATE COUNT( two_to_three);
three_to_four = filter avg_user by average_stars > 3 and average_stars <= 4;
D = GROUP three_to_four ALL;
ttf_count = FOREACH D GENERATE COUNT( three_to_four);
four_to_five = filter avg_user by average_stars > 4 and average_stars <= 5;
E = GROUP four_to_five ALL;
ftf_count = FOREACH E GENERATE COUNT( four_to_five);
So, this can be done, but
this only results in 5 individual table.
I want to see if there is any way (is ok to be fancy, I love fancy stuff)
T can make the result in single table.
Which means if
zto_count = 1
ott_count = 3
. = 2
. = 3
. = 5
then the table will be {1,3,2,3,5}
It just is easy to parse data, and organize them that way.
Is there any ways?

Answer 1

Using this as input:
foo 2
foo 3
foo 2
foo 3
foo 5
foo 4
foo 0
foo 4
foo 4
foo 5
foo 1
foo 5
(0 and 1 each appear once, 2 and 3 each appear twice, 4 and 5 each appear thrice)
This script:
A = LOAD 'myData' USING PigStorage(' ') AS (name: chararray, number: int);
B = FOREACH (GROUP A BY number) GENERATE group AS number, COUNT(A) AS count ;
C = FOREACH (GROUP B ALL) {
zto = FOREACH B GENERATE (number==0?count:0) + (number==1?count:0) ;
ott = FOREACH B GENERATE (number==1?count:0) + (number==2?count:0) ;
ttt = FOREACH B GENERATE (number==2?count:0) + (number==3?count:0) ;
ttf = FOREACH B GENERATE (number==3?count:0) + (number==4?count:0) ;
ftf = FOREACH B GENERATE (number==4?count:0) + (number==5?count:0) ;
GENERATE SUM(zto) AS zto,
SUM(ott) AS ott,
SUM(ttt) AS ttt,
SUM(ttf) AS ttf,
SUM(ftf) AS ftf ;
}
Produces this output:
C: {zto: long,ott: long,ttt: long,ttf: long,ftf: long}
(2,3,4,5,6)
The number of FOREACHs in C shouldn't really matter because C is going to only have 5 elements at most, but if it is then then they can be put together like this:
C = FOREACH (GROUP B ALL) {
total = FOREACH B GENERATE (number==0?count:0) + (number==1?count:0) AS zto,
(number==1?count:0) + (number==2?count:0) AS ott,
(number==2?count:0) + (number==3?count:0) AS ttt,
(number==3?count:0) + (number==4?count:0) AS ttf,
(number==4?count:0) + (number==5?count:0) AS ftf ;
GENERATE SUM(total.zto) AS zto,
SUM(total.ott) AS ott,
SUM(total.ttt) AS ttt,
SUM(total.ttf) AS ttf,
SUM(total.ftf) AS ftf ;
}