adds number_of_n_syllable_words_all function

[ThHuberResearch]

This adds the number_of_n_syllable_words_all function to descriptive_statistics.py.
For a list of texts, it counts the frequency of all n-syllable words for all values of n that it finds.

At the moment it is already possible, but a bit cumbersome, to find the frequencies of all syllables that exist in a text.
Consider this example:

docs = ['This has a very long word: Pneumonoultramicroscopicsilicovolcanoconiosis']

The very long word (apparently) has 13 syllables. But I don't necessarily know that it exists in my corpus. Currently I could find it by calling the number_of_n_syllable_words in a loop, and using a large enough number:

from linguaf.descriptive_statistics import number_of_n_syllable_words

docs = ['This has a very long word: Pneumonoultramicroscopicsilicovolcanoconiosis']
for i in range(1, 14):
    freqs = number_of_n_syllable_words(docs, n=(i, i+1))
    print(i, freqs)

But this is a bit inefficient.
With the new function this becomes much easier:

from linguaf.descriptive_statistics import number_of_n_syllable_words_all

docs = ['This has a very long word: Pneumonoultramicroscopicsilicovolcanoconiosis']
freqs = number_of_n_syllable_words_all(docs)
print(freqs)
print(freqs[5])  # how many 5-syllable words?

Prints:

defaultdict(<class 'int'>, {1: 6, 13: 1})
0

This new function also allows the user to easily find the frequency of only certain n-syllable words, such as "frequency of all 3-syllable and 5-syllable words, but nothing else". Again, this is already possible with number_of_n_syllable_words but all texts need to be processed twice with this approach (due to requiring two function calls - one with n=(3,4) and one with n=(5,6) which is a bit slow.

[Perevalov]

Thanks @ThHuberSG, gonna test it locally

[Perevalov]

Looks good, tested locally