打印唯一的行，比较不超过N个字符

Question

使用uniq，您可以选择仅比较第一个N字符

$ cat foo.txt
The quick brown fox jumps over the lazy dog.
The quick brown fox jumps over the lazy cat.
The quick brown fox jumps over the lazy mouse.

$ uniq -w 40 foo.txt
The quick brown fox jumps over the lazy dog.

使用awk可以达到同样的效果吗？我读 this example

awk '!a[$0]++'

但它比较整行。

Answer 1

awk有substr()函数：

awk '!a[substr($0,1,40)]++'

以你的例子：

kent$  echo "The quick brown fox jumps over the lazy dog.
The quick brown fox jumps over the lazy cat.
The quick brown fox jumps over the lazy mouse."|awk '!a[substr($0,1,40)]++'
The quick brown fox jumps over the lazy dog

Answer 2

使用FIELDWIDTHS和FPAT的两种选择：

awk '!a[$1]++' FIELDWIDTHS=40

awk '!a[$1]++' FPAT='.{40}'