bash 如何排序、uniq 和显示出现超过 X 次的行
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How to sort,uniq and display line that appear more than X times
提问by Andrew Kennen
I have a file like this:
我有一个这样的文件：
80.13.178.2
80.13.178.2
80.13.178.2
80.13.178.2
80.13.178.1
80.13.178.3
80.13.178.3
80.13.178.3
80.13.178.4
80.13.178.4
80.13.178.7
I need to display unique entries for repeated line (similar to uniq -d) but only entries that occur morethan just twice (twice being an example so flexibility to define the lower limit.)
我需要显示重复行的唯一条目（类似于 uniq -d），但只显示出现两次以上的条目（两次作为示例，因此可以灵活地定义下限。）
Output for this example should be like this when looking for entries with three or more occurrences:
在查找出现三个或更多的条目时，此示例的输出应如下所示：
80.13.178.2
80.13.178.3
采纳答案by hek2mgl
With pure awk:
与纯awk：
awk '{a[sort test.file | uniq -cd | awk -v limit=2 ' > limit{print }'
]++}END{for(i in a){if(a[i] > 2){print i}}}' a.txt
It iterates over the file and counts the occurances of every IP. At the end of the file it outputs every IP which occurs more than 2 times.
它遍历文件并计算每个 IP 的出现次数。在文件的末尾，它输出出现 2 次以上的每个 IP。
回答by iruvar
Feed the output from uniq -cdto awk
将输出从uniq -cd到awk
