14

我知道如何在 bash 中做一个每次增加一的循环，但是说我有一个 1 到 773 的范围，我想从一个循环中输出一个范围，以便在每次迭代中得到两个变量。第一个是 1，第二个是 19。在第二次迭代中，第一个是 20，第二个是 39。

我一直在玩类似的东西：

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

期望的循环结果：

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

但这是不对的。我想将这两个变量输出到一系列脚本以使 R 运行得更快，所以如果非 bash（例如 awk）解决方案更容易，那么如果一个简单的 > 将其发送文件，那也很酷。

谢谢！

4

7 回答 7

19

您的要求并不十分清楚，但您正在重用变量名。

如果我这样做：

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

我得到了类似于您想要的结果的东西。观察我如何将循环变量从开始重命名为索引。

PS：如果您想更改循环中的步长（也称为“增量”），只需这样做：

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

这将以 2 的步长递增，您可能希望在此处使用 20。那会给你带来1, 21, 41, ...价值。有关详细信息，请参阅http://www.cyberciti.biz/faq/bash-for-loop/ 。

于 2013-09-16T12:40:12.200 回答
15

你可以使用seq命令

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

的用法seq是：

$ 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.
于 2013-09-16T12:47:01.953 回答
13

如果要打印 773 内的范围，可以这样做

#!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

输出

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
于 2013-09-16T12:50:13.250 回答
7

这是使用一致输出的一种方法：

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

输出：

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

或者

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

输出：

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

其他：

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

输出：

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

这样你就可以有两个变量。

于 2013-09-16T12:52:01.807 回答
5

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 ...
于 2013-09-16T13:22:22.033 回答
5

除非我遗漏了一些你可以简单地做的事情：

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

输出：

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
于 2013-09-16T12:52:03.713 回答
2

这是一种方法：

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

它基本上只是一个简单的 for 循环。

于 2013-09-16T12:46:33.737 回答