Question: ```
echo -e 'one two three\nfour five six\nseven eight nine'
one two three
four five six
seven eight nine

```

how can I do some "MAGIC" do get this output?:

```
three
six
nine

```

UPDATE:
I don't need it in this specific way, I need a general solution so that no matter how many columns are in a row, e.g.: awk always displays the last column.

Answer: 
It can even be done only with `'bash'`, without `'sed'`, `'awk'` or `'perl'`:

```
echo -e 'one two three\nfour five six\nseven eight nine' |
  while IFS=" " read -r -a line; do
    nb=${#line[@]}
    echo ${line[$((nb - 1))]}
  done

```
Using `awk` you can first check if there is at least one column.

```
echo | awk '{if (NF >= 1) print $NF}'

echo 1 2 3 | awk '{if (NF >= 1) print $NF}'

```