gaga | A Git-Backed Key/Value Store | Identity Management library
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kandi X-RAY | gaga Summary
A Git-Backed Key/Value Store
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QUESTION
I have a "base" DataFrame a
that contains an identifier seq
and join keys, as well as a "value" DataFrame b
which will be merged:
ANSWER
Answered 2021-May-19 at 18:41Here is an idea using itertools.combinations
to create all the possible combinations of different lengths in the columns to use for the merge
. Then for each loop, you need to select the rows from a that are nan for the columns to drop and not nan for the one to merge (what you do by hand in the example above). Finally, concat
all the dataframes
QUESTION
ANSWER
Answered 2021-Mar-03 at 15:10Check out this video (https://www.youtube.com/watch?v=rNy6aQQYbuY), very easy to implement.
QUESTION
Link : https://www.codechef.com/LRNDSA01/problems/LAPIN
Program description : Lapindrome is defined as a string which when split in the middle, gives two halves having the same characters and same frequency of each character. If there are odd number of characters in the string, we ignore the middle character and check for lapindrome. For example gaga is a lapindrome, since the two halves ga and ga have the same characters with same frequency. Also, abccab, rotor and xyzxy are a few examples of lapindromes. Note that abbaab is NOT a lapindrome. The two halves contain the same characters but their frequencies do not match. Your task is simple. Given a string, you need to tell if it is a lapindrome.
My code :
...ANSWER
Answered 2021-Mar-01 at 15:00The algorithm you implemented is both incorrect and very slow (even if it was correct). Since the input characters are limited to a..z, you could implement a fast algorithm by comparing the frequencies of letters in the left and right portions of the input text. An example implementation is below. The code assumes that the character codes of a..z are contiguous, which isn't guaranteed by the C standard.
QUESTION
I have a music app that has a job to find music recommendations based on a tag id.
There are two entities involved:
Song
- a song record contains its name and a list of music tag ids (genres) this song belongs toMusicTag
- the music tag itself, includes id, name etc.
Data is currently stored in MongoDB.
The Songs
collections in mongo have millions of songs, and each song has an average of 7 tag ids.
The MusicTags
has about 30K records.
The Songs
collection looks like that:
ANSWER
Answered 2021-Feb-02 at 09:02Elasticsearch doesn't offer a dedicated array type so what you'd typically do is define the mapping based on the type of the individual array items -- in your case a keyword
:
QUESTION
I'm learning to create data tables for a ReactJS / NextJS application without libraries.
I'm having a hard time paginating the table, can you please provide a code sample to handle this.
This is the table code I used:
...ANSWER
Answered 2021-Jan-08 at 11:53Use Array#slice
with an incremented page
state. Something like this should get you started:
QUESTION
I'm working on a DataFrame, like this one :
...ANSWER
Answered 2020-Dec-27 at 16:53IIUC, you could do:
QUESTION
I am trying to make a php curl request to my laravel api
Here's the curl function
...ANSWER
Answered 2020-Dec-04 at 15:30Sooo, turns out the syntax/version of curl I was using was defective/uncomplete.
Went over to PostMan and executed the same request then copipasted the php -curl code which looks like this :
QUESTION
I try to print the 3 columns:
- Artists.artist
- Songs.Song
- Songs.views
This is my code:
...ANSWER
Answered 2020-Nov-29 at 13:48It looks like this line will select only the first column of each row:
QUESTION
I have created a class that checks if any item of the list self._name is in the string song1. I use assert to check if it is True. If I change assert with print, I get that the result is None, which is why I the assert does not work. It should come out as True, but I am unsure what I am doing wrong.
...ANSWER
Answered 2020-Oct-30 at 15:00class Song:
def __init__(self, artist, title):
self._artist = artist
self._title = title
def checkArtist(self, name):
return any(artistname in self._artist for artistname in name.split())
QUESTION
I had a specific question about turning my data into two columns so I can make an edgelist. I've attached a screenshot of the data. There's up to V10, and each row represents artists that have worked on the same song. I wanted to create an edgelist with the artist names. For example, for rows that have person A, B, C, D, I wanted to create:
A B
A C
A D
B C
B D
C D
The code I used so far is:
reltest <- t(do.call(cbind, lapply(cleanartists[sapply(cleanartists, length) >= 2], combn, 2)))
But this gives me all possible combinations among the artist names, not just the ones that have existing relationships. This is what my data looks like:
...ANSWER
Answered 2020-Jun-11 at 19:36You can use apply
to apply your function to every row, and then only take the elements that are not NA
. And with the approach from here you can get rid of duplicates.
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On a UNIX-like operating system, using your system’s package manager is easiest. However, the packaged Ruby version may not be the newest one. There is also an installer for Windows. Managers help you to switch between multiple Ruby versions on your system. Installers can be used to install a specific or multiple Ruby versions. Please refer ruby-lang.org for more information.
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