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Shell Programming and Scripting

Divide numbers into intervals

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Top Forums Shell Programming and Scripting Divide numbers into intervals
# 1  
Old 08-24-2010
Divide numbers into intervals
divide input values into specified number (-100 or -200) according to the key (a1 or a2 ....)
For ex: if we give -100 in the command line it would create 100 number intervals (1-100, 100-200, 200-300) untill it covers the value 300 in a1.

Note: It should work the same even with huge numbers 239349456667 with huge limit( -10000000 )out having the float number problem.

Thanx a billion!!!!

input

Code:
a1	300
a2	400
a3	198
a4	312

Code:
./script input -100 >>output1

output1

Code:
a1	1	100
a1	100	200
a1	200	300
a2	1	100
a2	100	200
a2	200	300
a2	300	400
a3	1	100
a3	100	200
a4	100	200
a4	200	300
a4	300	400

Code:
./script input -200 >>output2

output2

Code:
a1	1	200
a1	200	400
a2	1	200
a2	200	400
a3	1	200
a4	1	200
a4	200	400

# 2  
Old 08-24-2010
Code:
perl -lane'BEGIN {
  $l = shift; $, = "\t";
  }
print $_, "out of range" and next 
  if $F[1] < $l;
while (1) {  
  last if (($i += $l) + $l) > $F[1];
  print $F[0], $i ? $i : 1, $i + $l;
  }' 100 infile

# 3  
Old 08-24-2010
Quote:
Originally Posted by radoulov
Code:
perl -lane'BEGIN {
  $l = shift; $, = "\t";
  }
print $_, "out of range" and next 
  if $F[1] < $l;
while (1) {  
  last if (($i += $l) + $l) > $F[1];
  print $F[0], $i ? $i : 1, $i + $l;
  }' 100 infile


The code is not giving the desired o/p see below

Code:
o/p
a1      100     200
a1      200     300

BR

---------- Post updated at 14:21 ---------- Previous update was at 14:13 ----------

radoulov, you forget to reset the $i counter.

after modification on your code, it worked , see below:-


Code:
 perl -lane'BEGIN {
  $l = shift; $, = "\t";
  }
print $_, "out of range" and next 
  if $F[1] < $l;
while (1) {  
  last if ($i + $l) > $F[1];        
  print $F[0], $i ? $i : 1, $i+$l; $i+=$l ; 
  } ; $i=0 ;' 100 test1

# 4  
Old 08-24-2010
Quote:
Originally Posted by ahmad.diab
The code is not giving the desired o/p see below

Code:
o/p
a1      100     200
a1      200     300

BR
Yep,
sorry:
Code:
perl -lane'BEGIN {
  $l = shift; $, = "\t";
  }
print $_, "out of range" and next 
  if $F[1] < $l;
for ($i = 0; $i < $F[1]; $i += $l) {  
  print $F[0], $i ? $i : 1, $i + $l;
  }' 100 infile



---------- Post updated at 01:25 PM ---------- Previous update was at 01:21 PM ----------

With awk:

Code:
awk '{
  if (l > $2) {
    print $0, "out of range"; next
    }
  for (i = 0; i < $2; i += l)
    print $1, i ? i : 1, i + l 
  }' OFS='\t' l=100 infile

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