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23:14
If this is asking about bigram frequency distributions in English, it should be a lot more clearly worded. It sounds like one needs a programmer and a substantial corpus to check. However, I’m pretty sure this is a solved problem.
@aăâ If you're working on learning English, you might also want to take a look at ell.stackexchange.com. It's a site similar to this one where you can find lots of answers to common problems non-native speakers have!
/shameless plug
Also, questions relating to computational linguistics tend to be frowned on in ELU as being “too hard for regular people”, and get summarily shuffled off to Linguistics.SE.
@WendiKidd I actually think this one is on the high not the low end, at least, if it is to be professionally answered.
It’s a very very very easy programming problem, but you need a corpus to run off.
So, Is that question OK to ask here? On this site. If not, I'll delete it.
You need to clarify what you want.
I am just guessing.
@tchrist Oh I agree! I just meant, learners in general might like to know ELL exists :) Sorry if I didn't make that clear.
23:17
I think you are saying, for each letter, what are the most frequent next letter following, and in what order/percentage.
So say A might be followed by N 10% of the time, by S 8% of the time, by T 6 % of the time, etc.
Repeat for all letters.
Yes, @tchrist
It’s a programming problem. Not a hard one, again, but you need the data to start with.
Yeah, that sounds like a linguistics/programming problem to me. Assuming correct interpretation of the question
@aăâ Are you a programmer?
Yes, I'm iOS developer on Stackoverflow.
I join here by stack exchange.
23:19
Ah, then you could figure this out for yourself then! Good.
If you get a good enough model, you can use Markov chains to generate English-looking nonsense. This is useful for memorable passwords, amongst other things.
We have a migrated question about this.
2
Q: Common English bigrams / trigrams - recognising that a jumble of letters contain only valid English words

StuRI need to figure out the best method of ranking strings against one another so that I can tell which contain meaningful English words / sentences. This string contains English words (with no spaces or punctutation): THISISASENTENCETHATCONTAINSENGLISHWORDS This contains a jumble of random l...

Which landed here:
5
Q: Common English bigrams / trigrams - recognising that a jumble of letters contain only valid English words

StuRI have a database of one million strings which I want to rank against one another so that I can tell which contain meaningful English words / sentences. These strings contain no spaces or punctuation. Some contain real English words: THISISASENTENCETHATCONTAINSENGLISHWORDS Some are a jumbl...

Thanks! I'll find and learn about Markov chains.
Well, Markov chains are used for generating things.
Imagine doing it with words.
So given a pair of words "the big", what is the most likely next word?
Suppose the most common word is "red", so you output "red". If you use trigrams, 3-word moving windows, so would then ask what is the most likely word to follow "big red".
Is that what you want to do, but with letters?
Something like that.
You must read about this, then: :)
Mark V Shaney is a synthetic Usenet user whose postings in the net.singles newsgroups were generated by Markov chain techniques, based on text from other postings. The username is a play on the words "Markov chain". Many readers were fooled into thinking that the quirky, sometimes uncannily topical posts were written by a real person. Bruce Ellis did the code and Rob Pike did the design. Don P. Mitchell wrote the Markov chain code, demonstrating it to Rob and Bruce on the Tao Te Ching at first. They chose to apply it to the net.singles netnews group. Examples A classic example, from 19...
It was really funny.
Rob is such a prankster.
I just want to get the percentage rate of 7 letters After Consonant and Vowels.
23:25
You can’t do that.
It won’t be meaningful.
Each letter has its own frequency distribution for the letter to follow it.
You can’t very well collect it and average it like that.
Imagine the letter Q.
In English, it has a 100% chance of being followed by a U.
Including that with other letters produces nonsense.
See why?
You mean no rule for it.
No.
I mean you are creating nonsense.
Let us assume that the letter T is 50% followed by R, 30% by S, and 20% by E.
Now average T and Q together: you just added a bogus 50% chance of U!
That is hopelessly wrong.
It is not how English, or any language, works.
Just go write a little program to process /usr/share/dict/words and calculate all the numbers yourself. It is trivial.
That is not the same as frequency in a corpus, but it is still very useful and meaningful, and easy to write.
Thanks!
I could have written it in less time than the time I have spent writing about the question here.
I'll try by myself. Thanks for your time!
23:30
Good luck.
You should edit your question to make it clear what you are really looking for.
Whoa.
I guess that makes sense. For the longest time here, cops would just confiscate their stashes and send the college kids along with a warning. I guess now they will only get the warning, but keep the pot.
Does anyone else caption their photos in the third person? Like, "aedia uses the Pentax Spotmatic in the park. Photo taken by [husband] with the K1000."
I don't write "Me and the blah do stuff," but maybe that's because I'm also almost never in the pictures.
caption as in text under picture on a blog?
or metadata?
Physical photo album, mostly.
I don't think I caption that many digital photos.
ok I was going to recommend adobe lightroom for managing digital photos but it does not apply then
Have it.
I just also have some film cameras.
23:43
a: n=16% l=15% t=14% r=10% c=7% s=5% b=5% m=4% p=4% d=4% g=3% e=3% u=2% i=2% v=1% k=1%
b: l=18% e=15% a=15% i=14% r=12% o=10% u=8% b=2% s=1%
c: o=21% a=17% h=14% e=10% i=7% r=7% t=6% u=5% l=4% k=4% y=3% c=1%
d: e=28% i=25% o=11% a=11% r=7% u=4% l=3% y=2% d=2% n=2%
e: r=21% n=14% s=11% d=8% l=7% t=6% a=5% c=4% m=4% p=3% x=3% e=2% o=2% u=2% i=1% v=1% f=1% g=1%
f: o=17% i=17% e=13% l=12% u=11% a=10% r=9% f=5% t=2% y=2%
g: e=18% a=14% r=13% i=13% l=10% o=9% u=6% n=5% y=4% h=4% g=3% m=1%
h: e=23% o=17% i=16% a=15% y=13% u=3% r=3% t=3% l=2% n=1% m=1%
That discards info below 1%.
@aăâ That’s running off /usr/share/dict/words.
And rounding, of course.

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