highway | Automate development tasks using Swift | Automation library
kandi X-RAY | highway Summary
kandi X-RAY | highway Summary
Simply paste the following command into a terminal of your choice. This will checkout highway and run the bootstrap script. The bootstrap script is building highway using Swift Package Manager. The built highway command line tool will be placed (alongside other stuff) in ~/.highway. Make sure to add ~/.highway/bin to your $PATH after the bootstrap process is finished.
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highway Key Features
highway Examples and Code Snippets
def construct_highway(
number_of_cells: int,
frequency: int,
initial_speed: int,
random_frequency: bool = False,
random_speed: bool = False,
max_speed: int = 5,
) -> list:
"""
Build the highway following the paramet
def simulate(
highway: list, number_of_update: int, probability: float, max_speed: int
) -> list:
"""
The main function, it will simulate the evolution of the highway
>>> simulate([[-1, 2, -1, -1, -1, 3]], 2, 0.0, 3)
[
def update(highway_now: list, probability: float, max_speed: int) -> list:
"""
Update the speed of the cars
>>> update([-1, -1, -1, -1, -1, 2, -1, -1, -1, -1, 3], 0.0, 5)
[-1, -1, -1, -1, -1, 3, -1, -1, -1, -1, 4]
>
Community Discussions
Trending Discussions on highway
QUESTION
The goal for my code is to make a rough roadmap using the latitude and longitude of the exits on the pennsylvania turnpike drawing a line between each exit.
I am using a for loop to plot a line on the map every time it loops. This works if i hard code the latitude and longitude but as soon as i plug in my variables nothing gets plotted. Since the coordinates are in order I am just increasing the index every time it loops to get the next coordinates. I have printed the variables inside the loop and verified they have the desired value. I have tried putting the values in ordered pairs but the plot function didn't like me using nparrays. I'm not sure if there is something simple i am missing, but I appreciate any input.
...ANSWER
Answered 2022-Apr-17 at 19:11Okay, based on the discussion above, see below for a solution.
Notes:
- Am using pandas DataFames to easily work with the
.csv
file. thenames
field is the column names. - Am not using orthographic projection at all.
- Am iterating through the list of highway exits one exit at a time; at each index, am extracting the current and next exits' data - am sure there's a more 'pythonic' way to do this, but this is readable at least.
- edit: the final index in the loop is length-1
QUESTION
I have a road shapefile from the following location
https://www.globio.info/download-grip-dataset
I downloaded the shapefile for North America and subset the roads for Canada as follows:
...ANSWER
Answered 2022-Mar-31 at 17:00Using st_sample
works, but wasn't totally straightforward. The size of the road data makes it hard to work with in-memory.
Below the road_canada
object is simplified, uses only the geometry column, combined, sampled, and finally had to be cast to POINT
to get the coordinates to show up.
QUESTION
I am in need of help proving that an algorithm has greedy choice property and optimal substructure.
Context of the problem:
Consider a problem where a company owns n
gas stations that are connected by a highway.
Each gas station has a limited supply g_i
of gas-cans. Since the company don't know which gas station is most visited they want all of them to have the same amount of gas.
So they hire a fueling-truck to haul gas between the stations in their truck. However, truck also consumes 1 gas-can per kilometer driven.
Your task will be to help the chain calculate the largest amount of gas-cans g_bar
they can have at all Stations.
Consider the example: Here we have g = (20, 40, 80, 10, 20) and
p = (0, 5, 13, 33, 36) (in kilometers). In order to send one gas-can from station 3 to
station 4 we need to put 41 gas-cans in the truck, as the fueling-truck will consume 40 before reaching their destination (to send two gas-cans we need to put 42 in the truck).
The optimal g_bar
for the example is 21 and can be achieved as follows:
Station 2 sends 11 gas-cans towards Station 1. One gas-can arrives while ten are consumed on the way.
Station 3 sends 59 gas-cans towards Station 4. 19 arrive while 40 are consumed on the way.
Station 4 now has 29 gas-cans and send eight towards Station 5. Two of these arrive and six are consumed on the way.
The final distribution of gas-cans is: (21, 29, 21, 21, 22).
Given an integer g_bar
. Determine whether it is possible to get at least g_bar
gas-cans in every Gas Station.
in order for the greedy choice property and optimal substructure to make sense for a decision problem, you can define an optimal solution to be a solution with at least g_bar
gas-cans in every gas station if such a solution exists; otherwise, any solution is an optimal solution.
Input: The position p_i
and gas-can supply g_i of
each bar. Here g_i
is the supply for the bar at position p_i
. You may assume that the positions are in sorted order – i.e. p_1 < p_2 < . . . < p_n
.
Output: The largest amount g_bar
, such that each gas-station can have a gas-can supply of at least g_bar
after the truck have transferred gas-cans between the stations.
How can i prove Greedy Choice and Optimal Substructure for this?
...ANSWER
Answered 2022-Mar-20 at 06:03Let's define an optimal solution: Each station has at least X
gas cans in each station (X
= g_bar
).
Proving greedy property
Let us assume our solution is sub-optimal. There must exist a station i
such that gas[i]
< X
. Based on our algorithm, we borrow X - gas[i]
from station i+1
(which is a valid move, since we had already found a solution). Now station i
has gas = X
. This contradicts the original assumption that there must exist a station i
such that gas[i]
< X
, which means our solution isn't suboptimal. Hence, we prove the optimality.
Proving optimal substructure
Assume we have a subset of the stations of size N'
, and our solution is suboptimal. Again, if the solution is suboptimal, there must exist a station i
such that gas[i]
< X
. You can use the greedy proof again to prove that our solution isn't suboptimal. Since we have optimal solution for each arbitrary subset, we prove that we have optimal substructure.
QUESTION
I have an address 2300 S SUPER TEMPLE PL
which I expect to get 2300 S SUPER TEMPLE PLACE
as a result after spelling out the PL to PLACE. I have a dictionary of abbreviated street names:
ANSWER
Answered 2022-Mar-17 at 19:42Use a comprehension:
QUESTION
I am loading a graph of OSM roads using a polygon using the following command:
...ANSWER
Answered 2022-Mar-02 at 21:31Use network_type="all_private"
(the default argument) to also get access=private ways (the "all" network type only gets public ways). See docs for more: https://osmnx.readthedocs.io/en/stable/osmnx.html#osmnx.graph.graph_from_polygon
QUESTION
I am using background subtraction to identify and control items on a conveyor belt. The work is very similar to the typical tracking cars on a highway example. I start out with an empty conveyor belt so the computer knows the background in the beginning. But after the conveyor has been full of packages for a while the computer starts to think that the packages are the background. This requires me to restart the program from time to time. Does anyone have a workaround for this? Since the conveyor belt is black I am wondering if it maybe makes sense to toss in a few blank frames from time to time or if perhaps there is some other solution. Much Thanks
...ANSWER
Answered 2022-Jan-31 at 14:14You say you're giving the background subtractor some pure background initially.
When you're done with that and in the "running" phase, call apply()
specifically with learningRate = 0
. That ensures the model won't be updated.
QUESTION
I have to combine the Customer_address
when they have the same address_index
but different line_nbr
:
I need to combine CUSTOMER_NUM 31 and 30 to read
...ANSWER
Answered 2022-Jan-10 at 19:48On SQL Server 2017+, STRING_AGG()
makes this easy:
QUESTION
My code looks something like this at the moment:
...ANSWER
Answered 2022-Jan-06 at 17:40Perhaps there is a better solution, but you could use the parameter optionsAfterRender in the Options binding in order to modify the tag:
QUESTION
I am trying to query a varchar column called "JsonCode" from the "Weather" table in snowflake. The "JsonCode" column looks like this:
...ANSWER
Answered 2021-Dec-14 at 20:41This should work without the need of a lateral flatten:
QUESTION
So I have this data frame hotel_weather_top_ten
:
ANSWER
Answered 2021-Dec-11 at 18:54You can do that with the following.
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