重复数字之间距离

addictlinux

RT
cat file01

1. 
   
2. 121
   
3. 230
   
4. 444
   
5. 142
   
6. 567
   
7. 908
   
8. 121
   
9. 132
   
10. 211
    
11. 320
    
12. 142
    
13. 121
    
14. 901
    
15. 239
    
16. 121

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for example : 121  
the first 121 between the second (121) has 6 numbers
the second 121 between the third has 4 numbers
the third 121 between the forth has 2 numbers

how to write a shell script to get the correct result !
thanks

Shell_HAT

上看下看左看右看，6是从哪里冒出来的？

addictlinux

hehe, the first 121 between the second (121) has 5 numbers

addictlinux

1. #!/bin/awk -f
   
2. {temp=$1;
   
3.   num=NR;
   
4.   for(i in a)
   
5.     if(a[i]==temp){
   
6.        print $1,num-i-1;
   
7.        delete a[i];
   
8.     }
   
9.        a[num]=$1;
   
10. }

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not good!

addictlinux

1. 
   
2. 121 5
   
3. 142 6
   
4. 121 4
   
5. 121 2

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blackold

回复 2# Shell_HAT


    多干几次就干来了，应该是行号差。

awk可以解决吧。

ywlscpl

1. awk -v RS=121 'RS==RT&&NR>1{print "The "++n" between "n+1" ("RS") has "NF" numbers"}' file

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addictlinux

回复 7# ywlscpl


    good idea RS=121 !
  but I want to find all repeat data not only 121 !
  In fact , I consider the output like this,
121 :  5,4,2
142: 6

ywlscpl

回复 8# addictlinux

1. for n in `awk 'a[$1]++==1' file`
   
2. do
   
3.    awk -v RS="$n" 'RS==RT&&NR>1{print "The "++n" between "n+1" ("RS") has "NF" numbers"}' file
   
4.    echo "-----------------------"
   
5. done

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ywlscpl

another output type

1. for n in `awk 'a[$1]++==1' file`
   
2. do
   
3.    awk -v RS="$n" 'RS==RT&&NR>1{printf !m++?RS": "NF:","NF}END{print ""}' file
   
4. done

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