How can I find and filter a specific column in a .csv file?

Question

I have .csv files with the following structure:

cat,dog,mouse,pig,bird,cow,...
21,34,54,566,78,1,...
23,10,12,569,56,4,...
32,20,13,123,56,3,...
34,30,44,322,66,2,...

I want to filter the column related to the mouse, for instance:

54
12
13
44

How do I do it? Please keep in mind that I do not know in which column the mouse is found (my files are quite large, there are several files to filter, and the positions of the columns vary).

If I knew the exact position, I could use, for instance:

cat $file | awk '{printf("%s\n", $3);}' > filtered_file

What if I do not know the mouse is in column 3?

I really appreciate any help.

Solution

You could do it like that:

#!/bin/bash

file=$1
column=$2
seperator=','

# Check if a csv file and a column name is given.
if [[ -z $file || -z $column ]]; then
  echo "Usage: $0 csvfile column"
  exit 1
fi

# Iterate through the first row and try to find the requested column.
field=1
for column_name in $(head -n 1 $file | tr $seperator ' '); do
  [[ $column_name == $column ]] && break
  field=$((field+1))
done

# Finally print the output.
cat $file | cut -d $seperator -f $field | sed "1d"

(Credits: I got the idea of how to get the first line from this post on stackoverflow and the idea of how to delete the first line from this post on unix.com).