comb | one stop shop for frequently needed utilities | Runtime Evironment library
kandi X-RAY | comb Summary
kandi X-RAY | comb Summary
Framework for node that provides a one stop shop for frequently needed utilities, including:.
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Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of comb
comb Key Features
comb Examples and Code Snippets
def comb_sort(data: list) -> list:
"""Pure implementation of comb sort algorithm in Python
:param data: mutable collection with comparable items
:return: the same collection in ascending order
Examples:
>>> comb_sort([
@Override
public > T[] sort(T[] arr) {
int size = arr.length;
// initialize gap
int gap = size;
// Initialize swapped as true to make sure that loop runs
boolean swapped = true;
// Keep runnin
function combSort (list) {
if (list.length === 0) {
return list
}
const shrink = 1.3
let gap = list.length
let isSwapped = true
let i = 0
while (gap > 1 || isSwapped) {
// Update the gap value for a next comb
gap = parse
Community Discussions
Trending Discussions on comb
QUESTION
I am crunching large amounts of data without a hitch until I added more data. The results are written to file as strings, but I received this error message and I am unable to find programming error after combing my codes for 2 days; my codes have been working fine before new data were added.
...ANSWER
Answered 2021-Jun-15 at 07:04First of all: a Rat
with a denominator of 0
is a perfectly legal Rational value. So creating a Rat
with a 0 denominator will not throw an exception on creation.
I see two issues really:
- how do you represent a
Rat
with a denominator of0
as a string? - how do you want your program to react to such a
Rat
?
When you represent a Rat
s as a string, there is a good chance you will lose precision:
QUESTION
I am cleaning a huge dataset made up of tens of thousands of texts using R. I know regular expression will do the job conveniently but I am poor in using it. I have combed stackoverflow but could not find solution. This is my dummy data:
...ANSWER
Answered 2021-Apr-22 at 07:37Using stringr try this:
QUESTION
I have two sheets Sheet Main and Sheet Unique
On sheet Main, I have a range of data with headers
On sheet Unique in cell A1 I have a drop-down of all the header names from sheet Main
In cell B1 of sheet Unique, I have the formula
=SUBSTITUTE(ADDRESS(A1,MATCH(A1,Main !A1:AA1,0),4),1,"")
Which gets the Column letter of the header on sheet Main from the dropdown on sheet Unique
Is cell A2 on sheet Unique I have the formula
=SORT(UNIQUE(TRANSPOSE(ARRAYFORMULA(TRIM(SPLIT(JOIN("|",Main!F2:F),"|"))))))
Which gets the unique list
of elements from the column on sheet Main that was selected from the drop-down on sheet Unique.
In this case, F2:F
is the range on sheet Main of the selected header from sheet Unique
I want to update the formula
=SORT(UNIQUE(TRANSPOSE(ARRAYFORMULA(TRIM(SPLIT(JOIN("|",Main!F2:F),"|"))))))
automatically every time I make a new selection from the Drop-down on the sheet Unique
Combing the two formulas I get and adding INDIRECT
=SORT(UNIQUE(TRANSPOSE(ARRAYFORMULA(TRIM(SPLIT(JOIN("|",Main!&INDIRECT(SUBSTITUTE(ADDRESS(Y1,MATCH(A1,Main!$A$1:$AA$1,0),4),1,""),"|"))))))))
I get the error formula parse error
I can not get what I am doing wrong!
Thanks for any assistance
...ANSWER
Answered 2021-Jun-08 at 22:48Try this in cell Unique!A2
:
=unique(sort(indirect("Main!"&substitute(address(1,match(A1,Main!1:1,0),4),"1","")&"2:"&substitute(address(1,match(A1,Main!1:1,0),4),"1","")),1,1))
QUESTION
I'm working on a habitat occupancy prediction encompassing the entire state of Wyoming. Certain site covariate rasters work in the prediction while others with matched resolution, extent, etc. do not.
A short reproduceable example of my code is below. After extensive troubleshooting I've found I have 3 rasters of the 5 I need to use that cause this script to fail, all with the same error. I'm assuming my rasters have somehow become corrupted(?) but wanted to see if anyone has another idea on what could be happening.
Data is at this link. The data is the unmarked object (saved as .rds) and 2 very small clips off of: 1. the raster that works, and 2. one of the rasters that does not work
Steps I took to originally align the rasters for stacking - for information purposes ...ANSWER
Answered 2021-Jun-07 at 08:27Answer
The error arises because you have missings in sNoJoy
. Had those not been missing, it would have worked just fine.
Question rewritten
Your problem has nothing to do with your parallel code. It boils down to this:
QUESTION
I have the following data frame
...ANSWER
Answered 2021-May-31 at 18:26The m
in combn
determines the number of combinations as well
QUESTION
I'm looking to write a script that monitors all files/folders tracked by git to notify me if any of them have been changed. I'm looking for file changes and not necessarily staged changes.
I've been combing through various SO questions and documentations, but I can't find any relevant examples for implementing a file system watcher based off of git that would be efficient, since I might want to implement this across a multitude of repositories and possibly hundreds of files all on one system without being extremely inefficient.
Is git even the right solution for this? Would I be better off using fswatch
even across many, many files?
Many thanks
...ANSWER
Answered 2021-May-31 at 10:08Git itself doesn't watch the file system (*) and as such also doesn't provide any hooks when files change. It only inspects the file system whenever a git command is executed.
Therefore there aren't (and can't be) any hooks that execute on file change. You'll need to find another way to watch the file system, if you need this.
(*) Apparently there is a way to actually use file system monitoring in git, but that only seems to be used to speed up git status
(and other operations) and still doesn't mean that git actively monitors changes. It just replaces "check each file for modification" with "check the list of changed files since the last time for modification".
QUESTION
Given the following code from Leetcode:
...ANSWER
Answered 2021-May-29 at 06:09The list comprehension in itself is quite readable once you get used to them. For instance, I would prefer it, but would add at least one line break there. And consistent spaces around '+'.
QUESTION
I have a dataframe consisting of Wikipedia articles with geocoordinates and some statistics. The column 'Availability' contains a tuple of the languages that article is available in (out of a selection).
What I'm trying to do is plot a bubble map with plotly, and the legend being the availability in those languages. For example, out of ['ca','es']
you would have [],['ca'],['es'],['ca','es']
meaning not available, only in catalan, only in spanish or available in both respectively.
The problem is that when trying to use those combinations to create a dataframe with only the matching rows using Dataframe.isin(), it always returns an empty df.
The columns of the dataframe are:
Columns: [French Title, Qitem, Pageviews, page_title_1, page_title_2, Availability, Lat, Lon, Text]
Here is my code:
...ANSWER
Answered 2021-May-28 at 13:26You can use Series.apply()
to achieve your goal:
QUESTION
Context:
My data analysis involves manipulating ~100 different trials separately, and each trial has >1000 rows. Eventually, one step requires me to combine each trial with a column value from a different dataset. I plan to combine this dataset with each trial within an array using left_join() and "ID" as the key.
I want to mutate() the trial name into a new column labeled "ID". I feel like this should be a simple task, but I'm still a novice when working with lists and arrays.
I don't know how to share .csv files, but you can save the example datasets as .csv files within a practice folder named "data".
...ANSWER
Answered 2021-May-28 at 04:13library(tidyverse)
# Create practice dataset
df1 <- tibble(Time = seq(1, 5, by = 1),
Point = seq(6, 10, by = 1)) %>% print()
#> # A tibble: 5 x 2
#> Time Point
#>
#> 1 1 6
#> 2 2 7
#> 3 3 8
#> 4 4 9
#> 5 5 10
df2 <- tibble(Time = seq(6, 10, by = 1),
Point = seq(1, 5, by = 1)) %>% print()
#> # A tibble: 5 x 2
#> Time Point
#>
#> 1 6 1
#> 2 7 2
#> 3 8 3
#> 4 9 4
#> 5 10 5
write_csv(df1, "21May27_CtYJ10.csv")
write_csv(df2, "21May27_HrOW07.csv")
rm(df1, df2)
QUESTION
Is there a way for me to to run the outer loops simultaneously without mess with the inner loops?
I'm loosing the data.table performance if I use loops this way?
Note: I know there is some data.table functions that makes combinations like ones that I made, but they don't work for what I'm planning to do.
...ANSWER
Answered 2021-May-28 at 01:47Here is an approach using foreach
& doParallel
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