Bash的第一个元素列表识别 [英] Bash First Element in List Recognition

问题描述

我很新打坏，所以我很抱歉，如果这个问题其实很简单。我处理包含由2计数数2-32的许多垂直列表的文本文件，每个数字都有它后面行等文字。的问题是，一些清单的是丢失号码。任何指针为code，可以通过和检查，看看是否每个号码是存在的，如果没有添加一行，放在数。

I'm very new to Bash so I'm sorry if this question is actually very simple. I am dealing with a text file that contains many vertical lists of numbers 2-32 counting up by 2, and each number has a line of other text following it. The problem is that some of the lists are missing numbers. Any pointers for a code that could go through and check to see if each number is there, and if not add a line and put the number in.

一个列表看起来像：

2      djhfbadsljfhdsalkfjads;lfkjs
4      dfhadslkfjhasdlkfjhdsalfkjsahf
6      dsa;fghds;lfhsdalfkjhds;fjdsklj
8      daflgkdsakfjhasdlkjhfasdjkhf
12     dlsagflakdjshgflksdhflksdahfl

到32一路我怎么会在这种情况下，使它所以10被认为是失踪，然后加在12的上方？谢谢！

All the way down to 32. How would I in this case make it so the 10 is recognized as missing and then added in above the 12? Thanks!

推荐答案

这里有一个 AWK 为基础的解决方案（格式化的可读性，不一定是你将如何键入）：

Here's one awk-based solution (formatted for readability, not necessarily how you would type it):

awk '    { value[0 + $1] = $2 }
     END { for (i = 2; i < 34; i+=2)
             print i, value[i]
         }' input.txt

这基本上只是记录在一个键/值对（关联数组）现有生产线，然后在最后，打印所有你关心的记录，与先前保存的（可能为空）值一起。

It basically just records the existing lines in a key/value pair (associative array), then at the end, prints all the records you care about, along with the (possibly empty) value saved earlier.

注意：如果第一列需要被看作是一个字符串而不是一个整数，此变体应该工作：

Note: if the first column needs to be seen as a string instead of an integer, this variant should work:

awk '    { value[$1] = $2 }
     END { for (i = 2; i < 34; i+=2)
             print i, value[i ""]
         }' input.txt

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