print column name that matches string from commandline
Question
I want to input a string name (i.e. "COL2") to an awk or cut command and print the column that matches that column header string.
the datafile looks like this:
COL1 COL2 COL3 COL4 COL5 COL6
a a b d c f
a d g h e f
c v a s g a
If I pass in COL3, I want it to print the third column, etc. I'm thinking awk might be the easiest thing to use, but cut may also work. I'm just not sure how to go about doing this.
Solution
Awk 1 liner for above problem (if you are interested):
awk -v col=COL2 'NR==1{for(i=1;i<=NF;i++){if($i==col){c=i;break}} print $c} NR>1{print $c}' file.txt
awk -v col=COL3 'NR==1{for(i=1;i<=NF;i++){if($i==col){c=i;break}} print $c} NR>1{print $c}' file.txt
Just pass your column name COL1, COL2, COL3 etc with -vcol=
flag.
OTHER TIPS
a slight modification of anubhava post on top, for multiple columns
awk -vcol1="COL2" -vcol2="COL6" 'NR==1{for(i=1;i<=NF;i++){if($i==col1)c1=i; if ($i==col2)c2=i;}} NR>0{print $c1 " " $c2}' file.txt
when NR>1 does not print the column headers. This was modified to NR>0 which should print the columns with header names.
Note that the first solution prints out the whole file if the named column does not exist. To output a warning message if this occurs try
awk -v col=NoneSuch 'NR==1{for(i=1;i<=NF;i++){if($i==col){c=i;break}} if (c > 0) {print $c}} else {print "Column " col "does not exist"} NR>1 && c > 0 {print $c}' file1.txt
It's a little unclear what you're trying to do.
If you want to get the single column from the data, use substr()
.
If you want to use an argument to choose the column use something like
BEGIN { mycol = ARGV[1] ; }
{ print $mycol }
Update
Hmmm, so you want generalized column names?
Okay, we'll assume that your data is organized like this:
XXXXX YYYYY ZZZZZ
and you want to name the columns "harpo", "groucho" and "zeppo", and the column name is in ARGV[1]
:
BEGIN { cols["harpo"] = 1; cols["groucho"] = 2; cols["zeppo"] = 3; }
{ print $cols[ARGV[1]] }
Second update
Yup, this trick will do it. Replace "harpo" etc with "COL1", "Col2", and so on.
say column
is the variable you declared that is the column you want from the shell. You pass it in using awk's
-v
option
column=3
awk -vcol="$column" '{print $col}' file
When you say "pass a string" to awk, I guess you want to give the string on the command line. One option is to use the -v
feature for defining variables
$ gawk -f columnprinter.awk -v col=thecolumnnameyouwant
Alternately you can use the built-in variable ARGV
as Charlie explains.
That only leaves the matter for forming an array to associate column names with column numbers. If the first line of the input contains the column names (a common convention) this becomes pretty easy.
Use
NR==1{...}
to process the first column to get the mapping
NR==1{
colnum=-1;
for(i=1; i<=NF; i++)
if ($i == col) {
colnum=i
break
}
}
which you can use like
{
print $colnum
}