ural | A helper library full of URL-related heuristics | AWS library
kandi X-RAY | ural Summary
kandi X-RAY | ural Summary
TL;DR: a LRU is a hierarchical reordering of a URL so that one can perform meaningful prefix queries on URLs.
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Top functions reviewed by kandi - BETA
- Normalize a URL
- Normalize url path
- Strip country names from a netloc
- Check if the fragment should be stripped
- Decode a hostname
- Convert a query value to a string
- Attempt to decode a string
- Fix common query mistakes
- Determine if a query item should be stripped
- Parse a YouTube URL
- Return True if url is a YouTube URL
- Return True if value is a youtube video id
- Return a list of urlpath components
- Ensure the URL is valid
- Infer redirection from url
- Adds a key to the registry
- Tokenize a hostname
- Print the most common value for a given counter
ural Key Features
ural Examples and Code Snippets
from ural.facebook import has_facebook_comments
has_facebook_comments('https://www.facebook.com/permalink.php?story_fbid=1354978971282622&id=598338556946671')
>>> True
has_facebook_comments('https://www.facebook.com/108824017345866/vid
from ural.youtube import is_youtube_url
is_youtube_url('https://lemonde.fr')
>>> False
is_youtube_url('https://www.youtube.com/watch?v=otRTOE9i51o')
>>> True
is_youtube_url('https://youtu.be/otRTOE9i51o)
>>> True
from u
from ural.lru import LRUTrie
trie = LRUTrie()
# To respect tlds
trie = LRUTrie(tld_aware=True)
from ural.lru import LRUTrie
trie = LRUTrie()
trie.set('http://www.lemonde.fr', {'type': 'general press'})
trie.match('http://www.lemonde.fr')
>>
Community Discussions
Trending Discussions on ural
QUESTION
how to get class member by string? have .h:
...ANSWER
Answered 2021-Feb-03 at 23:32how i can get speed class member by string?
You generally cannot. There is no mechanism in the language to achieve this. In other words, the C++ language doesn't have sufficient introspection features. You cannot access things based on the names of variables using runtime strings because those names aren't stored anywhere in the program. The names are just used for compilation and have no meaning when the program runs.
You could create an associative map data structure, which maps a string to a pointer to data member. The standard library provides containers that implements such maps: std::map
and std::unordered_map
:
QUESTION
I have a list of file in my R environment. I want to merge some of them together using a mapping file.
The mapping file is named map_rule1, and it looks like following.
...ANSWER
Answered 2020-Nov-23 at 00:35Here's what I think might work. Tested on a sanitized version of the map_rule1
set of rules: It had two sources of error that you probably will need to trap or pre-sanitize against: 1) e6
was undefined, and 2) I decided that figuring out how to deal with the missing merge-by
columns was an additional level of complexity that I didn't feel up to:
QUESTION
I am speaking to those who know Cartopy well ... because I use Cartopy to produce a map, but I do not know very well how it works.
First, I created a map of Europe (in the broadest sense, from the Atlantic to the Urals), as shown in the attached figure.
Then, I have a separate file, say dft0
, indicating for each European country the time of appearance (Time0
) of a certain phenomenon, counted in number of days with respect to an arbitrary date D
and sorted from min
to max
; as an example of the first rows:
ANSWER
Answered 2020-Apr-30 at 14:38This solution is based on your posted code sample and draws heavily on this answer
QUESTION
My code merges csv files and removes duplicates with pandas. Is it possible to add an additional header with values to the single merged file?
The additional header should be called Host Alias
and should correspond to Host Name
E.g. Host Name
is dpc01n1
and the corresponding Host Alias
should be dev_dom1
Host Name
is dpc02n1
and the corresponding Host Alias
should be dev_dom2
etc.
Here is my code
...ANSWER
Answered 2020-Apr-13 at 06:13def func(row):
if row['Host Name'] == "dpc01n1":
return 'dev_dom1'
#do your Host Alias generate logic here,and return
combo["Host Alias"]=combo.apply(func, axis=1)
QUESTION
I would like to merge several csv files (delimeter ";") in a directory and output them into a single csv file with either another ";" delimeter or with a ",". All csv files have the same amount of headers (the headers must stay) and they're called the same throughout all csv files. But their content might have duplicates which I want to have removed.
The files can have a size up to 20 MB.
The files start with the same name but vary at the end (in which a * replaces them)
...ANSWER
Answered 2020-Apr-12 at 23:11this is me shooting in the dark; have a go at this and lemme know if it works
QUESTION
So basically this code connects to servers and downloads files matching this pattern from a directory into another. However, if a server is not reachable, it stops the whole process. I want it to skip it to the next server instead. How can I do this?
Also it downloads every file containing pc_dblatmonstat_
in it. While this is partially correct, I only need those files in which the names are like pc_dblatmonstat_x_x
where x replaces the actual value of that file in the directory.
E.g. I want the file pc_dblatmonstat_tpc01n1_scl000101018.log instead of pc_dblatmonstat_tpc01n1.log
Here is what I have got
...ANSWER
Answered 2020-Apr-08 at 11:19In your for loop, you can try, except
the error when a server is unreachable and skip to the next one like this:
QUESTION
I would like to copy several files with specific, identical name from the server to local using paramiko (for a school project). However I would like to have a list of servers for the script to go through and execute the same code and also detect if the server is online or not. How do I do this?
There are several identical files that I do not need. I need to pull the specific "dblatmonstat" files only.
Example of file name pc_dblatmonstat_dpc01n1_scl000101014.log
Like: first go through...
dpc01n1.sccloud.xxx.com
And then the same code through... dpc02n1.sccloud.xxx.com
...and so on and so forth.
Here is what I have so far:
...ANSWER
Answered 2020-Apr-08 at 04:34I have not tested the following, but it should work and give you an idea of how to approach the problem.
- Convert script to function
- Create a list of servers
- Iterate through the list with the function to retrieve the files
This is reflected in the following
QUESTION
I've been trying to create a mobile version of my slider which is 320px, but when I open it in the developer tools and set the width to 320 I see empty spaces from the sides as well as bellow the wrapper. What am I doing wrong?
Sorry if it's a damn question, just this issue has really got me and I don't know what to do.
...ANSWER
Answered 2019-Aug-30 at 16:01You should check in the inspector which element is causing the margins to be added.
Doing this, I can see that the tag itself has a margin of 8px, likely causing your problem.
Try adding this to your CSS:
QUESTION
i've a simple question.
Is it possible to update all images on a website wich has the same image-url?
For example i have the url 'http://www.example.com/mynewimage_new.png'
Now i want to set all images on the site how has the image-ural "http://www.example.com/mynewimage_old.png" to the new url.
Thanks a lot.
...ANSWER
Answered 2019-Feb-07 at 17:26Sure. As with any jQuery operation it mainly comes down to your selector and your operation on the matched elements. In this case the selector is all img
elements with a given src
value:
QUESTION
I am trying to 1st divide up four-letter words based upon the last two letters of the word (suffix) and 2nd count up how many words I have for each of these endings.
I have a list containing 3,164 words called filtered and I have sorted them by their suffixes, which doesn't seem much of a help.
(I want to create a dictionary that takes the suffix as a key and the words as a list but I don't know where to begin!) It would be something like:
OUTPUT:
dic = {'ab': ['Ahab', 'Arab', 'Saab, ...]; 'al': ['Aral', 'Baal', ...]}
and so on. Would that be possible?
...ANSWER
Answered 2018-Dec-19 at 21:26Assuming that suffixes are always two letters long and are case-sensitive, you can iterate through the word list and append each word to the dict of lists with the last two letters of the word as the key:
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