I have 2 tabular files. One file contains a mapping of 50 key values only called lookup_file.txt. The other file has the actual tabular data with 30 columns and millions of rows. data.txt I would like to replace the id column of the second file with the values from the lookup_file.txt..
How can I do this? I would prefer using awk in bash script.. Also, Is there a hashmap data-structure i can use in bash for storing the 50 key/values rather than another file?
Assuming your files have comma-separated fields and the "id column" is field 3:
awk '
BEGIN{ FS=OFS="," }
NR==FNR { map[$1] = $2; next }
{ $3 = map[$3]; print }
' lookup_file.txt data.txt
If any of those assumptions are wrong, clue us in if the fix isn't obvious...
EDIT: and if you want to avoid the (IMHO negligible) NR==FNR test performance impact, this would be one of those every rare cases when use of getline is appropriate:
awk '
BEGIN{
FS=OFS=","
while ( (getline line < "lookup_file.txt") > 0 ) {
split(line,f)
map[f[1]] = f[2]
}
}
{ $3 = map[$3]; print }
' data.txt
You could use a mix of "sort" and "join" via bash instead of having to write it in awk/sed and it is likely to be even faster:
key.cvs (id, name)
1,homer
2,marge
3,bart
4,lisa
5,maggie
data.cvs (name,animal,owner,age)
snowball,dog,3,1
frosty,yeti,1,245
cujo,dog,5,4
Now, you need to sort both files first on the user id columns:
cat key.cvs | sort -t, -k1,1 > sorted_keys.cvs
cat data.cvs | sort -t, -k3,3 > sorted_data.cvs
Now join the 2 files:
join -1 1 -2 3 -o "2.1 2.2 1.2 2.4" -t , sorted_keys.cvs sorted_data.cvs > replaced_data.cvs
This should produce:
snowball,dog,bart,1
frosty,yeti,homer,245
cujo,dog,maggie,4
This:
-o "2.1 2.2 1.2 2.4"
Is saying what columns from the 2 files you want in your final output.
It is pretty fast for finding and replacing multiple gigs of data compared to other scripting languages. I haven't done a direct comparison to SED/AWK, but it is much easier to write a bash script wrapping this than writing in SED/AWK (for me at least).
Also, you can speed up the sort by using an upgraded version of gnu coreutils so that you can do the sort in parallel
cat data.cvs | sort --parallel=4 -t, -k3,3 > sorted_data.cvs
4 being how many threads you want to run it in. I was recommended 2 threads per machine core will usually max out the machine, but if it is dedicated just for this, that is fine.
There are several ways to do this. But if you want an easy one liner, without much in the way of validation I would go with an awk/sed solution.
Assume the following:
the files are tab delimited
you are using bash shell
the id in the data file is in the first column
your files look like this:
1 one
2 two
3 three
4 four
5 five
1 col2 col3 col4 col5
2 col2 col3 col4 col5
3 col2 col3 col4 col5
4 col2 col3 col4 col5
5 col2 col3 col4 col5
I would use awk and sed to accomplish this task like this:
awk '{print "sed -i s/^"$1"/"$2"/ data"}' lookup | bash
what this is doing is going through each line of lookup and writing the following to stdout
sed -i s/^1/one/ data
sed -i s/^2/two/ data
and so on.
it next pipes each line to the shell (| bash), which will execute the sed expression. -i for inplace, you may want -i.bak to create a backup file. note you can change the extension to whatever you would like.
the sed is looking for the id at the start of the line, as indicated by the ^. You don't want to be replacing an 'id' in a column that might not contain an id.
your output would look like the following:
one col2 col3 col4 col5
two col2 col3 col4 col5
three col2 col3 col4 col5
four col2 col3 col4 col5
five col2 col3 col4 col5
of course, your ids are probably not simply 1 to one, 2 to two, etc, but this might get you started in the right direction. And I use the term right very loosely.
The way I'd do this is to use awk to write an awk program to process the larger file:
awk -f <(awk '
BEGIN{print " BEGIN{"}
{printf " a[\"%s\"]=\"%s\";",$1,$2}
END {print " }";
print " {$1=a[$1];print $0}"}
' lookup_file.txt
) data.txt
That assumes that the id column is column 1; if not, you need to change both instances of $1 in $1=a[$1]
