Python3 regex: Keep some Emojis, discard the rest - regex

Question

noob here. I have strings where I want to keep some emoji and to discard the rest.
INPUT:
This book is so funny❤️. This book 📖 is the bomb(AS IN THE BEST
IN THE WORLD 🌎 🌍🗺 )I love 💗 💕 💖 it!I definitely recommend it!'
DESIRED OUTPUT:
This book is so funny❤️. This book is the bomb(AS IN THE BEST
IN THE WORLD )I love 💗 💕 💖 it!I definitely recommend it!'
I have the re.compile that matches:
my emoji
all emoji Removing Emoticons from..... See David Mabodo answer
I don't know how to put it together in re.compile that excludes one from the other. Alternatively keep alphanumeric, punctuation, and my emoji, and substitute the rest to "".
mytext = This book is so funny❤️. This book 📖 is the bomb(AS IN THE BEST
IN THE WORLD 🌎 🌍🗺 )I love 💗 💕 💖 it!I definitely recommend it!'
# Desired out put:
# u'This book is so funny❤️. This book is the bomb(AS IN THE BEST
IN THE WORLD )I love 💗 💕 💖 it!I definitely recommend it!'
print ("Original text:")
print (mytext, "\n")
# Strip out emoticon modifiers, leaving a simplified emoticon to work with.
# https://en.wikipedia.org/wiki/Variation_Selectors_(Unicode_block)
# https://en.wikipedia.org/wiki/Variation_Selectors_Supplement
Emoji_Modifiers = re.compile(u'([\U0000FE00-\U0000FE0F])|([\U000E0100-\U000E0100])')
mytext_mod_gone = Emoji_Modifiers.sub(r'', mytext)
print ("Modifiers Removed:")
print (mytext_mod_gone, "\n")
# All emoticons
find_regex = re.compile(u'([\U00002600-\U000027BF])|([\U0001f300-\U0001f64F])|([\U0001f680-\U0001f6FF])')
# Heart emoticons
#find_regex = re.compile(u"([\U00002619])|([\U00002661])|([\U00002665])|([\U00002763])|([\U00002764])|([\U00002765])|([\U00002766])|([\U00002767])|([\U00002E96])|([\U00002E97])|([\U00002F3C])|([\U0001F394])|([\U0001F48C])|([\U0001F48F])|([\U0001F491])|([\U0001F493])|([\U0001F494])|([\U0001F495])|([\U0001F496])|([\U0001F497])|([\U0001F498])|([\U0001F499])|([\U0001F49A])|([\U0001F49B])|([\U0001F49C])|([\U0001F49D])|([\U0001F49E])|([\U0001F49F])|([\U0001F4D6])|([\U0001F5A4])|([\U0001F60D])|([\U0001F618])|([\U0001F63B])|([\U0001F970])|([\U0001F9E1])")
# Alphanumeric + punctuation for an alternative solution
#find_regex = re.compile(r"[^a-zA-Z0-9!,.?!#&'()*+,-./:;<=>?#\^_`{|}~\s]") #
mytext_emoji_gone = find_regex.sub(r'', mytext)
I am falling down at:
Negating unicode with a Negative Lookbehind (?<!...). I don't understand the operands well enough, and regex101.com only works with r', not u'.
Combining multiple regex together in a re.compile. Say if I wanted to keep alphanumeric and my emoji, it complains when I do re.compile(u'(\Uxxxx)' | r'(regex)' ). unsupported operand type(s) for |: 'str' and 'str', so a OR type statement does not work here...and an OR gives undesired results.
Could I have some help with either:
Ignoring a subset of emoticons and deleting the rest (my preferred solution)
Keeping (alphanumeric, punctuation, and my emoticons), and deleting the rest.
A specific question: Can you 'stack' re.compiles? IE create 2 different re.compiles to match (or not match) things, then join them together.

Answer 1

regex101 has a Unicode option, it is a flag you can turn on from the right side of the regex box.
I think the easiest thing to do is to find all the emojis in the string except for the ones you want to keep and replace them with an empty string like you wanted to do. To do that you can use a regex that will find any emoji (for this example I'll use [\U00010000-\U0010ffff] but I'm sure there are better ones out there so use one of those) and add a negative look ahead to ignore the emoji you wish to keep.
The finale regex should look similar to this:
(?![\u2764])[\U00010000-\U0010ffff]
The first part (?![\u2764]) will make sure the match is not an emoji you wish to keep and the second part [\U00010000-\U0010ffff] will make sure it's an emoji
You can add all the other emojis you wish to keep in the square brackets (?![\u2764
here ])

Answer 2

I went with:
find_regex = re.compile(u"(?![\U00002619])(?![\U00002661])(?![\U00002665])(?![\U00002763])(?![\U00002764])(?![\U00002765])(?![\U00002766])(?![\U00002767])(?![\U00002E96])(?![\U00002E97])(?![\U00002F3C])(?![\U0001F394])(?![\U0001F48C])(?![\U0001F48F])(?![\U0001F491])(?![\U0001F493])(?![\U0001F494])(?![\U0001F495])(?![\U0001F496])(?![\U0001F497])(?![\U0001F498])(?![\U0001F499])(?![\U0001F49A])(?![\U0001F49B])(?![\U0001F49C])(?![\U0001F49D])(?![\U0001F49E])(?![\U0001F49F])(?![\U0001F4D6])(?![\U0001F5A4])(?![\U0001F60D])(?![\U0001F618])(?![\U0001F63B])(?![\U0001F970])(?![\U0001F9E1])"r"[^a-zA-Z0-9!,.?!#&'()*+,-./:;<=>?#\^_`{|}~\s]")
mytext_emoji_gone = find_regex.sub(r'', mytext)
which stripped out all other emoji, leaving only the heart and book emojis, and alphanumeric and punctuation.
As part of my original question, is there a way to stack those? Currently, that is one huge long line of code. Could we do something like:
regex = re.compile(a)
regex += re.compile(b)
That would use vertial real estate but I am ok with that