AWK命令从第3列打印到第n列

Question

How to print from 3rd column to till last columns using awk command in unix, if there are 'n' columns in a file. I am getting with cut command but I need awk command. I am trying to do with awk -F " " '{ for{i=3;i<=NF;i++) print $i}' , I am getting the output but it is not in the correct format. Can anyone suggest me the proper command.

Answer 1

Combining Ed Morton's answers in:

• Print all but the first three columns
• delete a column with awk or sed

We get something like this:

awk '{sub(/^(\S+\s*){2}/,""); sub(/(\s*\S+){2}$/,"")}1'
#     ^^^^^^^^^^^^^^^^^^^^^^  ^^^^^^^^^^^^^^^^^^^^^^
#     remove 2 first cols      remove 2 last cols

Which you can adapt to your exact needs in terms of columns.

See an example given this input:

$ paste -d ' ' <(seq 5) <(seq 2 6) <(seq 3 7) <(seq 4 8) <(seq 5 9) 
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9

Let's just print the 3rd column:

$ awk '{sub(/^(\S+\s*){2}/,""); sub(/(\s*\S+){2}$/,"")}1' <(paste -d ' ' <(seq 5) <(seq 2 6) <(seq 3 7) <(seq 4 8) <(seq 5 9))
3
4
5
6
7

Answer 2

Assuming your fields are space-separated then with GNU awk for gensub():

$ cat file
a b c d e f
g h i j k l

$ awk '{print gensub(/(\S\s){2}/,"",1)}' file
c d e f
i j k l

In general to print from, say, field 3 to field 5 if they are blank separated using GNU awk again with gensub():

$ awk '{print gensub(/(\S\s){2}((\S\s){2}\S).*/,"\\2",1)}' file
c d e
i j k

or the 3rd arg to match():

$ awk 'match($0,/(\S\s){2}((\S\s){2}\S)/,a){print a[2]}' file
c d e
i j k

or in general if they are separated by any single character:

$ awk '{print gensub("([^"FS"]"FS"){2}(([^"FS"]"FS"){2}[^"FS"]).*","\\2",1)}' file
c d e
i j k

$ awk 'match($0,"([^"FS"]"FS"){2}(([^"FS"]"FS"){2}[^"FS"])",a){print a[2]}' file
c d e
i j k

If the fields are separated by a string instead of a single-character but the RS is a single character then you should temporarily change FS to RS (which by definition you KNOW can't be present in the record) so you can negate it in the bracket expressions:

$ cat file
aSOMESTRINGbSOMESTRINGcSOMESTRINGdSOMESTRINGeSOMESTRINGf
gSOMESTRINGhSOMESTRINGiSOMESTRINGjSOMESTRINGkSOMESTRINGl

$ awk -F'SOMESTRING' '{gsub(FS,RS)} match($0,"([^"RS"]"RS"){2}(([^"RS"]"RS"){2}[^"RS"])",a){gsub(RS,FS,a[2]); print a[2]}' file
cSOMESTRINGdSOMESTRINGe
iSOMESTRINGjSOMESTRINGk

If both the FS and the RS are multi-char then there's various options but the simplest is to use the NUL character or some other character you know can't appear in your input file instead of RS as the temporary replacement FS:

$ awk -F'SOMESTRING' '{gsub(FS,"\0")} match($0,/([^\0]\0){2}(([^\0]\0){2}[^\0])/,a){gsub("\0",FS,a[2]); print a[2]}' file
cSOMESTRINGdSOMESTRINGe
iSOMESTRINGjSOMESTRINGk

Obviously change FS to OFS in the final gsub()s above if desired.

If the FS was a regexp instead of a string and you want to retain it in the output then you need to look at GNU awk for the 4th arg for split().

Answer 3

Your attempt was close but appears that it would print each and every column on a new line. To correct this we create a variable called 'line' and initialize it to an empty string. The first time we are in the loop we just add the column to 'line'. From that point on we will append to 'line' with the field separator and the next column. Finally, we print 'line'. This will happen for each line in the file.

awk '{line="";for(i=3;i<=NF;i++) if(i==3) line=$i; else line=line FS $i; print line}'

In this example I assume to use awk's default field separator. Also any lines that are less than three will print blank lines.

Answer 4

If you don't mind normalizing the space, the most straightforward way is

$ awk '{$1=$2=""}1' | sed -r 's/^ +//'

in action

$ seq 11 40 | pr -6ts' ' | awk '{$1=$2=""}1' | sed -r 's/^ +//'

21 26 31 36
22 27 32 37
23 28 33 38
24 29 34 39
25 30 35 40

for the input

$ seq 11 40 | pr -6ts' '

11 16 21 26 31 36
12 17 22 27 32 37
13 18 23 28 33 38
14 19 24 29 34 39
15 20 25 30 35 40

Answer 5

要从第三列打印到最后，请输入cat filename | awk'{for（i = 1; i <3; i ++）$ i =“”; print $ 0}'