join multiple files

Question

I am using the standard join command to join two sorted files based on column1. The command is simple join file1 file2 > output_file.

But how do I join 3 or more files using the same technique ? join file1 file2 file3 > output_file Above command gave me an empty file. I think sed can help me but I am not too sure how ?

Answer 1

man join :

NAME
       join - join lines of two files on a common field

SYNOPSIS
       join [OPTION]... FILE1 FILE2

it only works with two files.

if you need to join three, maybe you can first join the first two, then join the third.

try:

join file1 file2 | join - file3 > output

that should join the three files without creating an intermediate temp file. - tells the join command to read the first input stream from stdin

Answer 2

One can join multiple files (N>=2) by constructing a pipeline of join s recursively:

#!/bin/sh

# multijoin - join multiple files

join_rec() {
    if [ $# -eq 1 ]; then
        join - "$1"
    else
        f=$1; shift
        join - "$f" | join_rec "$@"
    fi
}

if [ $# -le 2 ]; then
    join "$@"
else
    f1=$1; f2=$2; shift 2
    join "$f1" "$f2" | join_rec "$@"
fi

Answer 3

I know this is an old question but for future reference. If you know that the files you want to join have a pattern like in the question here eg file1 file2 file3 ... fileN Then you can simply join them with this command

cat file* > output

Where output will be the series of the joined files which were joined in alphabetical order.

Answer 4

I created a function for this. First argument is the output file, rest arguments are the files to be joined.

function multijoin() {
    out=$1
    shift 1
    cat $1 | awk '{print $1}' > $out
    for f in $*; do join $out $f > tmp; mv tmp $out; done
}

Usage:

multijoin output_file file*

Answer 5

While a bit an old question, this is how you can do it with a single awk :

awk -v j=<field_number> '{key=$j; $j=""}  # get key and delete field j
                         (NR==FNR){order[FNR]=key;} # store the key-order
                         {entry[key]=entry[key] OFS $0 } # update key-entry
                         END { for(i=1;i<=FNR;++i) {
                                  key=order[i]; print key entry[key] # print
                               }
                         }' file1 ... filen

This script assumes:

• all files have the same amount of lines
• the order of the output is the same order of the first file.
• files do not need to be sorted in field <field_number>
• <field_number> is a valid integer.

Answer 6

The man page of join states that it only works for two files. So you need to create and intermediate file, which you delete afterwards, ie:

> join file1 file2 > temp
> join temp file3 > output
> rm temp

Answer 7

Join joins lines of two files on a common field. If you want to join more - do it in pairs. Join first two files first, then join the result with a third file etc.

Answer 8

Assuming you have four files A.txt, B.txt, C.txt and D.txt as:

~$ cat A.txt
x1 2
x2 3
x4 5
x5 8

~$ cat B.txt
x1 5
x2 7
x3 4
x4 6

~$ cat C.txt
x2 1
x3 1
x4 1
x5 1

~$ cat D.txt
x1 1

Join the files with:

firstOutput='0,1.2'; secondOutput='2.2'; myoutput="$firstOutput,$secondOutput"; outputCount=3; join -a 1 -a 2 -e 0 -o "$myoutput" A.txt B.txt > tmp.tmp; for f in C.txt D.txt; do firstOutput="$firstOutput,1.$outputCount"; myoutput="$firstOutput,$secondOutput"; join -a 1 -a 2 -e 0 -o "$myoutput" tmp.tmp $f > tempf; mv tempf tmp.tmp; outputCount=$(($outputCount+1)); done; mv tmp.tmp files_join.txt

Results:

~$ cat files_join.txt 
x1 2 5 0 1
x2 3 7 1 0
x3 0 4 1 0
x4 5 6 1 0
x5 8 0 1 0