Useless but mildly interesting language model using compressors built-in to Python.

You can "train" it using some training data:

data = open(my_favorite_text_file).read().lower()
alphabet = "qwertyuiopasdfghjklzxcvbnm,.;1234567890 "
model = ziplm.ZipModel(alphabet, training=data)
"".join(model.sample_sequence(10)) # sample 10 characters from the alphabet

You can also run it without any training data, and just forward sample to see what kinds of patterns gzip likes:

alphabet = "abc"
model = ziplm.ZipModel(alphabet)
"".join(model.sample_sequence(100)) # I get 'ccabcabcabcabcabcabcabcabcabcabcabcabcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbcbccabcbcbcbc'

You can also get the probability for a sequence:

alphabet = "qwertyuiopasdfghjklzxcvbnm "
model = ziplm.ZipModel(alphabet)
model.sequence_logprob("this is my favorite string") # I get -83.8

You can also try using bz2 and lzma as language models by passing them as the compressor argument to the model

import lzma
model = ziplm.ZipModel(alphabet, compressor=lzma)
"".join(model.sample_sequence(100)) # I get 'cccbaaaaacccccabcacccbaaaaabaacaabaacaabaacaabaabacaaaaaaaaaaacccbabacaaaaaaaaaaaccccacaaccbaaaaaccc'

This works because of two facts:

1. A language model is nothing but a distribution on the next token given previous tokens, $p(x \mid c)$.
2. There is a general equivalence between probability distributions and codes.

The second point is what makes this interesting. Information theory tells us that we can derive codes from probability distributions. That is, if I have some datapoints $x$, and I know that they follow probability distribution $p(x)$, I can come up with a lossless binary code to encode the $x$ where the length of each code is $-\log_2 p(x)$. This code minimizes the average code length: the only way to get shorter average code length would be to go into the realm of lossy compression. This is called the Shannon Limit.

Since I can convert probability distributions to codes in this way, I can also convert codes to probability distributions. If I have a code (like gzip) that describes my datapoint with length $l(x)$ in binary, then that corresponds to a probability distribution $p(x) = 2^{-l(x)}$. If the code is $K$-ary, then the corresponding distribution is $$p(x) = K^{-l(x)}.$$

The ZipLM model works by converting code lengths to probabilities in this way. If I have a vocabulary of size $K$, and a string $c$, then the probability distribution for continuations $x$ is: $$p(x \mid c) \propto K^{-l(cx)},$$ where the proportionality reflects the fact that we have to sum over the compressed lengths of $cx^\prime$ for all $x^\prime$ in the vocabulary. That's all there is to it.

It's pretty bad, but it doesn't generate total junk. Here I trained the gzip model in Moby Dick---from the Project Gutenberg text---and the output at least has some recognizable parts:

data = open("mobydick.txt").read().lower()
alphabet = "qwertyuiopasdfghjkl;'zxcvbnm,. "
model = ziplm.Model(alphabet, data)
"".join(model.sample_sequence(100)) 

This gives me

"'theudcanvas. ;cm,zumhmcyoetter toauuo long a one aay,;wvbu.mvns. x the dtls and enso.;k.like bla.njv'"

which at least seems to have "long a one" in it.