Shell Delete sed

I have a file with some data like this:

1, 3, 0, 0, 0
0, 4, 5, 0, 5
2, 6, 0, 1, 0

I would like to write a shell script to delete all the lines where the 3rd argument is 0 (here line 1 and 2)

I already know that

sed -i '/0/d' file.txt

but I don't know how I can select the 3rd arguments whatever the othe开发者_StackOverflow社区rs one.

Do you have an idea?

---

Awk may be a better tool than sed for this task:

awk -F' *, *' '$3 != 0 {print}' FILE

But sed can do it:

sed -i '/^[0-9][0-9]*, [0-9][0-9]*, 0,/d' FILE

---

sed -iE '/^([^,]+,){2} 0,/d' file

explained with explain.py:

sed -iE '/^([^,]+,){2} 0,/d' file
\_/  ||  ||| || || \_/ | |
 |   ||  ||| || ||  |  | \- delete command
 |   ||  ||| || ||  |  |
 |   ||  ||| || ||  |  \- followed by blank zero komma
 |   ||  ||| || ||  |
 |   ||  ||| || ||  \- two times this inner pattern
 |   ||  ||| || ||
 |   ||  ||| || |\- followed by a comma
 |   ||  ||| || |
 |   ||  ||| || \- at least one of them
 |   ||  ||| ||
 |   ||  ||| |\- komma
 |   ||  ||| |
 |   ||  ||| \- not
 |   ||  |||
 |   ||  ||\- start of pattern, group of characters, which are
 |   ||  ||
 |   ||  |\- at begin of line
 |   ||  |
 |   ||  \- start of pattern for delete command
 |   ||
 |   |\- Extended regexp (parenthesis and braces without masking)
 |   |
 |   \- inplace changes
 |
 \- run sed

---

awk with in place editing (similar to sed -i)

awk '$3!="0,"{print $0>FILENAME}' file

Ruby(1.9+)

ruby -ane 'print unless $F[2]=="0," ' file