寻找序号中的空白 [英] Finding gaps in sequential numbers

问题描述

我不是为了谋生而做这些事情，因此，如果这是一个简单的问题(或者比我想的要复杂的多)，请原谅我.我一直在浏览档案，发现了一些很接近的技巧，但是作为一个新手，我不确定如何调整自己的需求，或者它们超出了我的理解范围.

I don’t do this stuff for a living so forgive me if it’s a simple question (or more complicated than I think). I‘ve been digging through the archives and found a lot of tips that are close but being a novice I’m not sure how to tweak for my needs or they are way beyond my understanding.

我有一些大的数据文件，我可以解析这些文件以生成大多数是连续的坐标列表

I have some large data files that I can parse out to generate a list of coordinate that are mostly sequential

5
6
7
8
15
16
17
25
26
27

我想要的是差距列表

1-4
9-14
18-24

我不知道 perl ， SQL 或其他任何奇特的东西，但我认为我可以做一些可以从下一个减去一个数字的事情.然后，我至少可以将差异不为 1 或 -1 的输出grep并使用该输出来获得差距.

I don’t know perl, SQL or anything fancy but thought I might be able to do something that would subtract one number from the next. I could then at least grep the output where the difference was not 1 or -1 and work with that to get the gaps.

推荐答案

带有 awk :

awk '$1!=p+1{print p+1"-"$1-1}{p=$1}' file.txt

解释

• $1是当前输入行的第一列
• p是最后一行的上一个值
• 所以($1!=p+1)是一个条件:如果$1与先前的值+1不同，则:
• 执行此部分:{print p+1 "-" $1-1}:打印先前的值+1，-字符和第一列+ 1
每行执行
• {p=$1}:将p分配给当前的第一列

explanations

• $1 is the first column from current input line
• p is the previous value of the last line
• so ($1!=p+1) is a condition : if $1 is different than previous value +1, then :
• this part is executed : {print p+1 "-" $1-1} : print previous value +1, the - character and fist columns + 1
• {p=$1} is executed for each lines : p is assigned to the current 1st column

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