这似乎可行，但请自行检查:

This seems to work , but make a check on your own:

cat <<EOF >file1
foo1 foo2 foo3 foo4 
foo3 foo4 foo1 foo2
foo2 foo1 foo3 foo4
fooA fooB fooC fooD
fooC fooD fooA fooB
fooD fooC fooA fooB
fooD fooB fooC fooA
EOF
awk '!f1[$1$2$3$4]++ && !f1[$3$4$1$2]++' file1
#Output
foo1 foo2 foo3 foo4 
foo2 foo1 foo3 foo4
fooA fooB fooC fooD
fooD fooC fooA fooB
fooD fooB fooC fooA

正如评论所指出的那样，为避免可能的字段不必要的连接以及避免foob ar和foo bar字段之间的混淆，***使用字段分隔符FS(无论该FS设置了什么值-默认情况下为空格)作为数组的一部分表示:

As pointed out on comments, to avoid possibly unwanted concatenating of the fields and avoid confusion between foob ar and foo bar fields, is better to use the field separator FS (in whatever value this FS has been set - space by default) as part of the array indeces :

awk '!f1[$1FS$2FS$3FS$4]++ && !f1[$3FS$4FS$1FS$2]++' file1