bash - How to delete rows from a csv file based on a list values from another file?

Question

I have two files:

candidates.csv:

id,value
1,123
4,1
2,5
50,5

blacklist.csv:

1
2
5
3
10

I'd like to remove all rows from candidates.csv in which the first column (id) has a value contained in blacklist.csv. id is always numeric. In this case I'd like my output to look like this:

id,value
4,1
50,5

So far, my script for identifying the duplicate lines looks like this:

cat candidates.csv | cut -d , -f 1 | grep -f blacklist.csv -w

This gives me the output

1
2

Now I somehow need to pipe this information back into sed/awk/gawk/... to delete the duplicates, but I don't know how. Any ideas how I can continue from here? Or is there a better solution altogether? My only restriction is that it has to run in bash.

Answer 1

What about the following:

 awk -F, '(NR==FNR){a[$1];next}!($1 in a)' blacklist.csv candidates.csv

How does this work?

An awk program is a series of pattern-action pairs, written as:

condition { action }
condition { action }
...

where condition is typically an expression and action a series of commands. Here, the first condition-action pairs read:

• (NR==FNR){a[$1];next} if the total record count NR equals the record count of the file FNR (i.e. if we are reading the first file), store all values in array a and skip to the next record (do not do anything else)
• !($1 in a) if the first field is not in the array a then perform the default action which is print the line. This will only work on the second file as the condition of the first condition-action pair does not hold.