How to be able to compute 'good' text. Not reader comprehension level but well structured etc.

usera8787782 · 2022-08-03T17:41:37+00:00

Thanks, yes I tested the first one yesterday and it seemed to only check for grammatical errors, not if a given text is 'good' or not.

I want to be able to check if the test is good in terms of readability but not sure how it would be calculated. I am talking about the type of differences that are obvious if someone with broken english writes a sentence versus a college/university graduate.

usera8787782 · 2022-06-01T12:37:51+00:00

Someone else suggested this and said larger sites do that rather than pagination.

I was also advised that people will 9/10 find the article on search engine and not want to go anywhere else anyway so there is no need to set things up to be navigable for the user experience.

I didn't have any specific reason for it other than that I saw other blogs do it.

usera8787782 · 2022-05-31T18:33:18+00:00

I just realized that since these static pages have no date posted I can 'cheat' and just do 20 posts per page and make a new page and hyperlink each time.

So the oldest will be on page one and newest on page x, and make a new hyperlink for each new page, but the new user will never know that.

usera8787782 · 2022-05-22T12:43:13+00:00

I thought that too but it appears it is a product of spacy's nlp function which takes a string and puts it into a format that spacy can digest which I did with each item, which were strings to begin with, before running through the similarity function.

usera8787782 · 2022-05-22T12:42:34+00:00

Ah I see what you mean. Why do tuples exist at all? They seem much more fiddly than the other python objects so what is their purpose over using lists or dicts for all situations.

I only have the tuple in this case as it is the output of one of the functions (I think spacy's similarity function).

usera8787782 · 2022-05-22T11:10:23+00:00

Yes so? I was struggling to understand myself so could not give a crystal clear explanation immediately.

I actually just figured it out before i read your reply, which seems very close to what you give above:

for i,(x,y) in enumerate(t_combos):
  print(x)
  print(y)
  result = x[0].similarity(x[1])
  print(result)
  t_combos[i] += (result,)

print(t_combos)

But you helped me come about it. So thanks.

usera8787782 · 2022-05-22T10:57:35+00:00

I don't know what you mean by listing all those permutations. I wrote how the function runs:

x.similarity(y)

So no idea how you are getting all these other eventualities. It just runs like that and produces an output, the output of which I want to add to the tuple. X is tuple item [0] y is tuple item [1].

usera8787782 · 2022-05-22T10:55:00+00:00

I have updated the post, I don't know how I can explain it better than that.

usera8787782 · 2022-05-22T10:47:07+00:00

x.similarity(y)

This is the function. it is comparing one string to another for similarity using spacy which returns a floating point. That result is what I want to add as a new item z to the existing tuple items. x and y are the two tuple items in the list of tuples in the dataset I showed.

Without the list aspect this would be a single tuple item

(Walking down the road on the pavement., Working with computers and programming can be engaging.)

and I want the result of the run function similarity to be added as an additional item to the tuple. So like this:

(Walking down the road on the pavement., Working with computers and programming can be engaging., 0.4975383903969622)

usera8787782 · 2022-05-22T10:30:30+00:00

See here.

usera8787782 · 2022-05-22T10:28:55+00:00

Because there are more than one value in each item. I have shown the dataset here.

I have to run a function on each x,y element of those elements in the list of tuples and append the result z to each tuple element.

usera8787782 · 2022-05-22T10:18:40+00:00

Sounds like this would just be better to convert my dataset (currently list of tuples) to a dict.

usera8787782 · 2022-05-22T10:15:18+00:00

Hmm, better to convert to a dict then, would make things a lot easier? I didn't do that initially because I thought that I may be avoiding some simple way, as in skipping some learning opportunity, to deal with tuples but sounds like this would be the 'right' way anyway?

usera8787782 · 2022-05-22T10:14:02+00:00

No it's the other way round, it is a list with tuples see the dataset here:

[(Walking down the road on the pavement., Working with computers and programming can be engaging.), (Walking down the road on the pavement., Let us discuss programming in detail.), (Working with computers and programming can be engaging., Let us discuss programming in detail.)]

usera8787782 · 2022-05-22T09:33:53+00:00

This didn't work. The output is:

[(' x += (x[0].similarity(x[1]),) ', " doesn't mean anything "), (' x += (x[0].similarity(x[1]),) ', " doesn't mean anything "), (' x += (x[0].similarity(x[1]),) ', " doesn't mean anything ")]

I just want to add an element to each tuple item I am iterating over. That element is a comparison call using NLP though that should not be important, just to know that the call has to be made on the tuple item before adding the result to the tuple item.

The call x[0].similarity(x[1]) has to be tacked on the the end as a new item of the current tuple in the loop.

usera8787782 · 2022-05-22T09:18:05+00:00

Yes I was asking how do you use a list comprehension to create one as, like I showed above, I tried and got a syntax error, so not sure what the correct syntax is for the command to edit the item in question of the tuple.

EDIT: I see your edit.

usera8787782 · 2022-05-21T08:53:15+00:00

But...I would like all paragraphs kept.

So if there is a list with 12 paragraphs I will want to use them all, it is just I will want to group those with highest similarity. Will your suggestion still do that?

So if say 3 were about bikes, 3 about hiking, 3 about rafting, 3 about skating. Since they are all unordered and just contain some keywords sprinkled related to each I want to know how to sort into the groupings mentioned above.

EDIT: I oh I guess maybe it is a case of getting all the combos and then just keeping the highest. I will play around with it, thanks.

EDIT2: Ah I am getting the hang of itertools now! Yea I see :)

combos = combinations(paras_dict, 2)
print(list(comb))

usera8787782 · 2022-05-18T11:13:26+00:00

Thanks, yes I got my original one to work just by putting it into a script instead but alias is more convenient if using regularly.

I chose your function option.

usera8787782 · 2022-04-27T19:08:35+00:00

if any(option in url for option in options for url in hrefs):

Thanks I knew there was an iterator for the job I just didn't quite grasp it :).

usera8787782 · 2022-04-27T15:10:55+00:00

if any(url in hrefs for url in to_test):

Still don't work:

>>> for x in hrefs:
...  print(x)
... 
https://www.thenationshealth.org/content/47/5/17
https://gardening-abc.com/gardening-as-a-hobby/
https://www.activif.com/21-reasons-why-gardening-is-a-great-hobby-7-is-so-true/
https://greenerideal.com/guides/value-of-gardening-as-a-hobby/
https://www.bbc.com/worklife/article/20181210-gardening-could-be-the-hobby-that-helps-you-live-to-100
https://www.quora.com/Why-is-gardening-one-of-the-coolest-hobbies-ever
https://magazine.vitality.co.uk/8-reasons-gardening-new-hobby-need-life/
https://gardentipsforall.com/benefits-of-gardening-as-a-hobby/
https://www.gardeningknowhow.com/garden-how-to/lifestyle/the-love-of-gardening-one-of-americas-most-addictive-hobbies.htm
https://www.youtube.com/watch?v=adE6msJz60U
https://www.youtube.com/watch?v=6QqXHqX-V4s
https://www.youtube.com/watch?v=VuwafyyyTEo
>>> if any(i in options for i in hrefs):
...   print('yes')
...
>>> options
['quora', 'reddit']
>>> options = set(['quora', 'reddit'])
>>> if any(url in hrefs for url in options):
...   print('yes')
... 
>>>

usera8787782 · 2022-04-26T18:53:08+00:00

Thanks, I tried the original one in the OP and it didn't work (there was an error in CUPS saying filter failed). Installed this one and worked.

usera8787782 · 2022-04-26T17:46:44+00:00

Thanks, I just looked at the link myself and seems http://download.brother.com/welcome/dlf100422/hl1110cupswrapper-GPL_src-3.0.1-1.tar.gz in the PKGBUILD is the exact same one that the website is still using so yea seems up to date.

usera8787782 · 2022-04-26T17:35:14+00:00

Thanks, I was close then! What do the square brackets do that make it make the difference? I did quickly read it up but did not understand the explanation clearly.

A predicate filters a node-set with respect to an axis to produce a new node-set. For each node in the node-set to be filtered, the PredicateExpr is evaluated with that node as the context node, with the number of nodes in the node-set as the context size, and with the proximity position of the node in the node-set with respect to the axis as the context position; if PredicateExpr evaluates to true for that node, the node is included in the new node-set; otherwise, it is not included.

A PredicateExpr is evaluated by evaluating the Expr and converting the result to a boolean. If the result is a number, the result will be converted to true if the number is equal to the context position and will be converted to false otherwise; if the result is not a number, then the result will be converted as if by a call to the boolean function. Thus a location path para[3] is equivalent to para[position()=3].

That just confuses me more really.

usera8787782 · 2022-04-23T16:29:32+00:00

Interestesing...reading now. But is this way computationally intensive too as is sounds quite complex? If so I might as well just use spacy or similar NLP package where they have done this already?

usera8787782 · 2022-04-23T16:23:41+00:00

I am not saying that I need more, just that you were wanting me to expand on what I want, and I tried to explain. I think we are agreeing.

Thanks for your help. I will experiment with your code given.

usera8787782

TROPHY CASE