Question

我想找到多次出现（ny）字的行。例如，如果输入文本是

John is a teacher, who is not highly paid.
abc abcde
James lives in Detroit.
abc abc abcde
Paul has 2 dogs and 2 cats.

输出应为

John is a teacher, who is not highly paid.
abc abc abcde
Paul has 2 dogs and 2 cats.

第一行重复is，第二行重复abc，最后一行重复2。

Answer 1

^(?=.*\b(\w+)\b.*\b\1\b).*$

试试这个。看看演示。

https://www.regex101.com/r/rG7gX4/6

与grep -P

一起使用

Answer 2

这是在awk

中执行此操作的简单方法

awk '{f=0;delete a;for (i=1;i<=NF;i++) if (a[$i]++) f=1} f' file
John is a teacher, who is not highly paid.
abc abc abcde
Paul has 2 dogs and 2 cats.

它循环遍历每个单词并在数组a中计算它们如果找到多个单词，请设置标记f
如果标志f为真，则执行默认操作，打印行。

看多少：

awk '{f=0;delete a;for (i=1;i<=NF;i++) if (a[$i]++) f=1} f {for (i in a) if (a[i]>1) printf "%sx\"%s\"-",a[i],i;print $0}' file
2x"is"-John is a teacher, who is not highly paid.
2x"abc"-abc abc abcde
2x"2"-Paul has 2 dogs and 2 cats.

一些改进：忽略大小写。移除.和,。

awk '{f=0;delete a;for (i=1;i<=NF;i++) {w=tolower($i);sub(/[.,]/,"",w);if (a[w]++) f=1}} f' file

如何查找文件中多次出现（ny）个单词的行？

2 个答案: