Delete empty files - Improve performance of logic - bash

Question

I am i need to find & remove empty files. The definition of empty files in my use case is a file which has zero lines.
I did try testing the file to see if it's empty However, this behaves strangely as in even though the file is empty it doesn't detect it so.
Hence, the best thing I could write up is the below script which i way too slow given it has to test several hundred thousand files
#!/bin/bash
LOOKUP_DIR="/path/to/source/directory"
cd ${LOOKUP_DIR} || { echo "cd failed"; exit 0; }
for fname in $(realpath */*)
do
if [[ $(wc -l "${fname}" | awk '{print $1}') -eq 0 ]]
then
echo "${fname}" is empty
rm -f "${fname}"
fi
done
Is there a better way to do what I'm after or alternatively, can the above logic be re-written in a way that brings better performance please?

Answer 1

Your script is slow beacuse wc reads every file to the end, which is not needed for your purpose. This might be what you're looking for:
#!/bin/bash
lookup_dir='/path/to/source/directory'
cd "$lookup_dir" || exit
for file in *; do
if [[ -f "$file" && -r "$file" && ! -L "$file" ]]; then
read < "$file" || echo rm -f -- "$file"
fi
done
Drop the echo after making sure it works as intended.
Another version, calling the rm only once, could be:
#!/bin/bash
lookup_dir='/path/to/source/directory'
cd "$lookup_dir" || exit
for file in *; do
if [[ -f "$file" && -r "$file" && ! -L "$file" ]]; then
read < "$file" || files_to_be_deleted+=("$file")
fi
done
rm -f -- "${files_to_be_deleted[#]}"
Explanation:
The core logic is in the line
read < "$file" || rm -f -- "$file"
The read < "$file" command attempts to read a line from the $file. If it succeeds, that is, a line is read, then the rm command on the right-hand side of the || won't be executed (that's how the || works). If it fails then the rm command will be executed. In any case, at most one line will be read. This has great advantage over the wc command because wc would read the whole file.
if ! read < "$file"; then rm -f -- "$file"; fi
could be used instead. The two lines are equivalent.

Answer 2

To check a "$fname" is a file and is empty or not, use [ -s "$fname" ]:
#!/usr/bin/env sh
LOOKUP_DIR="/path/to/source/directory"
for fname in "$LOOKUP_DIR"*/*; do
if ! [ -s "$fname" ]; then
echo "${fname}" is empty
# remove echo when output is what you want
echo rm -f "${fname}"
fi
done
See: help test:
File operators:
...
-s FILE True if file exists and is not empty.

Answer 3

Yet another method
wc -l ~/tmp/* 2>/dev/null | awk '$1 == 0 {print $2}' | xargs echo rm
This will break if any of your files have whitespace in the name.
To work around that, with awk still
wc -l ~/tmp/* 2>/dev/null \
| awk 'sub(/^[[:blank:]]+0[[:blank:]]+/, "")' \
| xargs echo rm
This works because the sub function returns the number of substitutions made, which can be treated as a boolean zero/not-zero condition.
Remove the echo to actually delete the files.