CODEWITHSUNDEEP
bash
cianius
cianius

Reputation: 2412

Increment in bash loop by set amount

I know how to do a loop in bash that increases by one each time, but say I have a range 1 to 773 and I want to output a range from a loop so that I get two variables in each iteration. The first will be 1 and the second will be say 19. In the second iteration the first would be 20 and the second 39.

Ive been playing with something like: 

for start in {1..773}
do    
start=$(($start+20))
end=$(($start+20))
echo $start ## 
echo $end
done

Desired loop outcome: 

 1. $start = 1 and $end = 19
 2. $start = 20 and $end = 39
 3. $start = 40 and $end = 59 
etc

But it's not right. I want to output these two variables to a series of scripts to make R run faster, so if non bash (eg awk) solutions are easier then that's cool too if a simple > will send it the file.

Thanks!

Upvotes: 21

Views: 128080

Answers (7)

ztik
ztik

Reputation: 3612

You could use seq command

for start in `seq 1 20 700`
do    
  echo $start $(($start+19))
done

The usage of seq is:

$ seq --help
Usage: seq [OPTION]... LAST
  or:  seq [OPTION]... FIRST LAST
  or:  seq [OPTION]... FIRST INCREMENT LAST
Print numbers from FIRST to LAST, in steps of INCREMENT.

Upvotes: 21

glenn jackman
glenn jackman

Reputation: 246744

Nobody has suggested a while loop

start=0 step=20 end=$((step - 1))
while (( end < 773 )); do
    echo "$start - $end"
    (( start += step, end += step ))
done

Produces

0 - 19
20 - 39
40 - 59
...
720 - 739
740 - 759

Of course, the while loop can be written as a for loop:

for ((start=0, step=20, end=step-1; end < 773; start+=step, end+=step)); do ...

Upvotes: 6

konsolebox
konsolebox

Reputation: 75458

This is one way to do it with consistent outputs:

for ((i = 1, start = 1, end = 19; i <= 773; ++i, start += 20, end += 20)); do
    echo "$i. \$start = $start and \$end = $end"
done

Output:

1. $start = 1 and $end = 19
2. $start = 21 and $end = 39
3. $start = 41 and $end = 59
4. $start = 61 and $end = 79

Or

for ((i = 1, start = 1, end = 19; i <= 773; ++i, start += 20, end += 20)); do
    echo "$i. \$start = $start and \$end = $end"
done

Output:

1. $start = 1 and $end = 20
2. $start = 21 and $end = 40
3. $start = 41 and $end = 60
4. $start = 61 and $end = 80

Another:

for ((i = 1, start = 0, end = 19; i <= 773; ++i, start += 20, end += 20)); do
    echo "$i. \$start = $start and \$end = $end"
done

Output:

1. $start = 0 and $end = 19
2. $start = 20 and $end = 39
3. $start = 40 and $end = 59
4. $start = 60 and $end = 79

That way you can have two variables.

Upvotes: 9

anubhava
anubhava

Reputation: 784918

Unless I'm missing something you can simply do:

for ((s=0,e=19; e<773; s+=20,e+=20)); do
    echo $s "-" $e
done

OUTPUT:

0 - 19
20 - 39
40 - 59
60 - 79
80 - 99
100 - 119
120 - 139
140 - 159
160 - 179
180 - 199
200 - 219
220 - 239
240 - 259
260 - 279
280 - 299
300 - 319
320 - 339
340 - 359
360 - 379
380 - 399
400 - 419
420 - 439
440 - 459
460 - 479
480 - 499
500 - 519
520 - 539
540 - 559
560 - 579
580 - 599
600 - 619
620 - 639
640 - 659
660 - 679
680 - 699
700 - 719
720 - 739
740 - 759

Upvotes: 7

thefourtheye
thefourtheye

Reputation: 239443

If you want to print the ranges within 773, you can do like this

#!env bash
start=1
end=19
for counter in {1..773}
do
   echo $counter. "\$start = " $start " and \$end = " $end
   if [[ $start -eq 1 ]];
   then
      start=0
   fi
   start=$(($start+20))
   end=$(($end+20))
   if [[ $end -ge 773 ]];
   then
      break
   fi
done

Output

1. $start =  1  and $end =  19
2. $start =  20  and $end =  39
3. $start =  40  and $end =  59
4. $start =  60  and $end =  79
5. $start =  80  and $end =  99
6. $start =  100  and $end =  119
7. $start =  120  and $end =  139
8. $start =  140  and $end =  159
9. $start =  160  and $end =  179
10. $start =  180  and $end =  199
11. $start =  200  and $end =  219
12. $start =  220  and $end =  239
13. $start =  240  and $end =  259
14. $start =  260  and $end =  279
15. $start =  280  and $end =  299
16. $start =  300  and $end =  319
17. $start =  320  and $end =  339
18. $start =  340  and $end =  359
19. $start =  360  and $end =  379
20. $start =  380  and $end =  399
21. $start =  400  and $end =  419
22. $start =  420  and $end =  439
23. $start =  440  and $end =  459
24. $start =  460  and $end =  479
25. $start =  480  and $end =  499
26. $start =  500  and $end =  519
27. $start =  520  and $end =  539
28. $start =  540  and $end =  559
29. $start =  560  and $end =  579
30. $start =  580  and $end =  599
31. $start =  600  and $end =  619
32. $start =  620  and $end =  639
33. $start =  640  and $end =  659
34. $start =  660  and $end =  679
35. $start =  680  and $end =  699
36. $start =  700  and $end =  719
37. $start =  720  and $end =  739
38. $start =  740  and $end =  759

Upvotes: 18

pgl
pgl

Reputation: 7981

Here's one way to do it:

step=20
last=773

for ((i = 0; i <= $last; i += $step))
do
    start=$i

    end=$(($i + (($step - 1))))

    if (($end > $last))
    then
        end=$last
    fi

    echo "\$start: $start"
    echo "\$end: $end"
done

It's basically just a simple for loop.

Upvotes: 5

mnagel
mnagel

Reputation: 6854

your requirement are not perfectly clear, but you are re-using variable names.

if i do this:

for index in {1..773}
do    
  start=$(($index+20))
  end=$(($start+20))
  echo $start ## 
  echo $end
done

i get something that resembles your desired result. observe how i renamed the loop variable from start to index.

PS: if you want to change the step size (a.k.a. "increment") in your loop, simply do it like this:

#!/bin/bash
for i in {0..10..2}
  do
     echo "Welcome $i times"
done

This will increment in steps of 2, you would want to use a 20 here. That would give you 1, 21, 41, ... as value. See http://www.cyberciti.biz/faq/bash-for-loop/ for more details.

Upvotes: 26

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