使用Sed或Awk根据行是否包含数值将文件分为两个

Question

I have used sed and awk for little while now, but I am having a challenge with the below problem. I am asking for an experienced sed/awk guru to help.

I have a file where some lines have numbers and some lines do not, like:

afjjdjfj.uihuihi
trfg.rtyhd
0rtgfd.tjbghhh
hbvfd4.rtgbvdgf
00fhfg.fdrgf
rtygfd.ijhniuh

etc.

I would like to have exactly two files out of this one, where every line is represented in one of the two files (none are deleted).

One containing all lines with any numbers 0-9 on them so given above file result would be:

0rtgfd.tjbghhh
hbvfd4.rtgbvdgf
00fhfg.fdrgf

and another file containing the rest of the lines that do not have any numbers 0-9 on them, so given the above, file it would be:

afjjdjfj.uihuihi
trfg.rtyhd
rtygfd.ijhniuh

I've tried different strategies in both sed and awk and nothing is giving me exactly what I need.

What would be the best sed or awk one liner to solve this problem?

Thank you for your time,

Tom

Answer 1

轻松使用Awk：

awk '/[0-9]/{print > file1; next} {print > file2}' inputfile

Answer 2

With single GNU sed command:

sed -ne '/[0-9]/w with_digits.txt' -e '//!w no_digits.txt' input

Results:

> cat no_digits.txt 
afjjdjfj.uihuihi
trfg.rtyhd
rtygfd.ijhniuh

---

> cat with_digits.txt
0rtgfd.tjbghhh
hbvfd4.rtgbvdgf
00fhfg.fdrgf

---

w filename
Write the pattern space to filename.

Answer 3

If you don't mind running twice over the input, you can use just grep:

grep    '[0-9]' input > with_digits
grep -v '[0-9]' input > without_digits

Answer 4

perl -MFile::Slurp -lpe '/\d/ ? append_file("digits.txt",$_) : append_file("no_digits.txt",$_)' input.txt