bash - How to exclude files from list

Question

I have ~100k files in directory. I need to delete some of them files excluding list of(15k different pattern) pattern:

Directory: /20210111/
Example files: 
/20210111/xxx_yyy_zzz.zip
/20210111/aaa_bbb_ccc.zip
/20210111/ddd_eee_fff.zip
...

Exclude.list 
ddd
aaa
...

I tried with find:

find /20210111/ -type f -iname "*.zip" ! -iname "*$(cat Exclude.list)*" -exec ...

Getting error: arguments too long. Because exclude.list have a lot of lines.

How can I do that?

Answer 1

You can use grep to filter the output of find, then use xargs to process the resulting list.

find /20210111/ -type f -iname '*.zip' -print0 
| grep -zvFf Exclude.list - 
| xargs -0 rm

• The -print0, -z, and -0 are used to separate the filenames by the null byte, so filenames can contain any valid character (you can't store patterns containing literal newlines in your Exclude.list, anyway).
• grep's -F interprets the patterns as fixed strings instead of regexes.