bash - join two csv files with key value

Question

I have two csv files, I want to join them using a key value, the column of the city.

One csv file, d01.csv has this form,

Barcelona, 19.5, 29.5
Tarragona, 20.4, 31.5 
Girona, 17.2, 32.5
Lleida, 16.5, 33.5 
Vic, 17.5, 31.4

The other one, d02.csv, has the next structure,

City, Data, TMax, TMin
Barcelona, 20140916, 19.9, 28.5
Tarragona, 20140916, 21.4, 30.5  
Lleida, 20140916, 17.5, 32.5 
Tortosa, 20140916, 20.5, 30.4

I need a new csv file, with a column of cities which appear in the 2 csv files.

City, Tmin, Tmax, Date, Tmin1, Tmax1
Barcelona, 19.5, 29.5, 20140916, 19.9, 28.5
Tarragona, 20.4, 31.5, 20140916, 21.4, 30.5
Girona, 17.2, 32.5, 20140916, 17.5, 32.5
Lleida, 16.5, 33.5, 20140916, 20.5, 30.4

I tried to do that with

join -j 2 -t ',' d01.csv d02.csv | awk -F "," '{print $1, $2, $3, $4, $5} > d03.csv

but it is not complete...how can I order the key value?

Answer 1

Here's how to use join in bash:

{
  echo "City, Tmin, Tmax, Date, Tmin1, Tmax1"
  join -t, <(sort d01.csv) <(sed 1d d02.csv | sort)
} > d03.csv
cat d03.csv

City, Tmin, Tmax, Date, Tmin1, Tmax1
Barcelona, 19.5, 29.5, 20140916, 19.9, 28.5
Lleida, 16.5, 33.5 , 20140916, 17.5, 32.5 
Tarragona, 20.4, 31.5 , 20140916, 21.4, 30.5  

Note that join only outputs records where the key exists in both files. To get all of them, specify that you want missing records from both files, specify the fields you want, and give a default value for the missing fields:

join -t, -a1 -a2 -o 0,1.2,1.3,2.2,2.3,2.4 -e '?' <(sort d01.csv) <(sed 1d d02.csv | sort)

Barcelona, 19.5, 29.5, 20140916, 19.9, 28.5
Girona, 17.2, 32.5,?,?,?
Lleida, 16.5, 33.5 , 20140916, 17.5, 32.5 
Tarragona, 20.4, 31.5 , 20140916, 21.4, 30.5  
Tortosa,?,?, 20140916, 20.5, 30.4
Vic, 17.5, 31.4,?,?,?