polyline | polyline encoding and decoding in javascript | Map library
kandi X-RAY | polyline Summary
kandi X-RAY | polyline Summary
A simple google-esque polyline implementation in Javascript. Compatible with nodejs (npm install @mapbox/polyline and the browser (copy src/polyline.js)). Encodes from / decodes into [lat, lng] coordinate pairs. Use fromGeoJSON() to encode from GeoJSON objects, or toGeoJSON to decode to a GeoJSON LineString.
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Top functions reviewed by kandi - BETA
- Encode a single coordinate .
- Converts a string to a polyline .
- Flips an array of coords .
- Prints the exit message
- Rounds value to positive
polyline Key Features
polyline Examples and Code Snippets
Lines of Code : 58License : Permissive (MIT)
Copy/**
* Encode pairs of lat and lon into a polyline encoded string
* @param points {Array.Array[lat, lon]}
* @return encoded polyline {String}
*/
function encode(points)
/**
* Decode a polyline string into an array of coordinates.
* @see This is # copy an existing test case
cp -r test/functional/basic test/functional/potsdamerplatz
# extract the relevant polylines from a large polyline source
grep -Pia "potsdamer\s?(platz|strasse|straße)" /data/planet.polylines > test/functional/potsdame var myPolyline = L.polyline( latlngs, options?)
weight : 2, //The width of the line
colorName : '', //Class-name to give the fill color
fillColorName : '', //Same as colorName
borderColorName: '', //C 
License : No LicenseCopypublic boolean isPalindrome(ListNode head) {
ListNode mid = middleNode(head);
ListNode headSecond = reverseList(mid);
ListNode rereverseHead = headSecond;
// compare both the halves
while (head != null &&a private boolean recursivePalindrome(String text, int forward, int backward) {
if (forward == backward)
return true;
if ((text.charAt(forward)) != (text.charAt(backward)))
return false;
if (forward < var path = [];
for (var i=0; i// This example creates a 2-pixel-wide red polyline showing the path of
// the first trans-Pacific flight between Oakland, CA, and Brisbane,
// Australia which was made by Charles Kingimport pyqtgraph as pg
from pyqtgraph.Qt import QtCore
class GraphPolyLine(pg.PolyLineROI):
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
def checkPointMove(self, handle, pos, modifiers):
import fiona
import json
import numpy as np
# sudo pip3 install git+https://github.com/ulikoehler/UliEngineering.git
from UliEngineering.Math.Coordinates import BoundingBox
import polyline as pl
import csv
# https://gis.stackexchange.com/
speed:
Start Animation
> fabric.Polyline.fromObject = function(object, callback) {
> return fabric.Object._fromObject('Polyline', object, callback, 'points'); };
var canvas = new fabric.Canvas('c', {
});
// my custom TreeNCommunity Discussions
Trending Discussions on polyline
QUESTION
I'm coming across an issue where once my test card number is typed into the first iframe the test tries to target the 2nd iframe (CVC) & it times out whilst trying to target the element. All info is listed below. Any help to why this is failing is appreciated!
Custom commands used:
...ANSWER
Answered 2022-Feb-15 at 19:59@BillBaily Thanks but I was looking for a larger piece of HTML, covering both iframes.
But another suggestion - I have seen on another project that uses WorldPay has nested iframes, so you would need something like
QUESTION
I am using CPython and I saw in one example file with a star symbol.
Could you give an explanation what in this context the * symbol means?
Here the pointsets is a numpy array that comes from pybind11, because it is an output of C++ code.
Does the Point(*point) has something to do with the pointers in C++?
ANSWER
Answered 2022-Jan-07 at 12:33It's called unpacking operator. Here is what's written in the docs:
An asterisk
*denotes iterable unpacking. Its operand must be an iterable. The iterable is expanded into a sequence of items, which are included in the new tuple, list, or set, at the site of the unpacking.
It is much like the "..." operator from Javascript ES6. (the spread operator)
https://docs.python.org/3/reference/expressions.html#expression-lists
QUESTION
How can this code produce a similar Graph Diagram as the other one? I have posted my entire code, if someone is an expert on this, share your input. I don't understand drawing fully in WPF! I am still reading on the topic.
My Code:
...ANSWER
Answered 2021-Dec-17 at 11:53You were on the right path to resolve it, I used your code as guideline.
So, 1st here is the code that provide your desired result,
This goes to the MainWindow xaml:
QUESTION
Lets assume that there is the following BigQuery database:
asset_id latitude longitude trip_status timestamp 2 52.1052016 10.141829999999999 false 1638274080 2 52.10512 10.1420266 false 1638274081 2 52.104774899999995 10.1427066 true 1638274085 2 52.1044833 10.1431966 true 1638274088 2 52.104156599999996 10.143821599999999 true 1638274092 2 52.10398 10.1441433 true 1638274094 2 52.1038016 10.1444783 true 1638274096 2 52.1036183 10.144823299999999 true 1638274098 2 52.1034333 10.1451783 true 1638274100 2 52.1032483 10.1455383 false 1638274102 2 52.1030533 10.145886599999999 true 1638274104 2 52.1028666 10.146175 true 1638274106 2 52.10279 10.1463266 true 1638274108 2 52.1026616 10.1466566 true 1638274110 2 52.102464999999995 10.147016599999999 true 1638274112 2 52.102215 10.1474083 true 1638274114 2 52.101968299999996 10.147795 true 1638274116 2 52.101756599999995 10.148195 false 1638274117 2 52.101538299999994 10.14864 false 1638274119 2 52.1013583 10.149076599999999 false 1638274121In provided data there is a flag - trip_status, indicating, whether the given coordinates have been captured during the trip mode.
trip_statusvalue changing fromfalsetotrueindicates the start of the trip.trip_statusvalue changing fromtruetofalseindicates the end of the trip.- All consecutive rows with
trip_status = trueare rows belonging to the same trip
Question:
Is there a way in BigQuery to extract separate trips from such dataset? Maybe somehow group data where trip_status flag is true and return as a separate datasets?
For example, from given data, I need to retrieve something like:
Trip 1:
asset_id latitude longitude trip_status timestamp 2 52.104774899999995 10.1427066 true 1638274085 2 52.1044833 10.1431966 true 1638274088 2 52.104156599999996 10.143821599999999 true 1638274092 2 52.10398 10.1441433 true 1638274094 2 52.1038016 10.1444783 true 1638274096 2 52.1036183 10.144823299999999 true 1638274098 2 52.1034333 10.1451783 true 1638274100Trip 2:
asset_id latitude longitude trip_status timestamp 2 52.1030533 10.145886599999999 true 1638274104 2 52.1028666 10.146175 true 1638274106 2 52.10279 10.1463266 true 1638274108 2 52.1026616 10.1466566 true 1638274110 2 52.102464999999995 10.147016599999999 true 1638274112 2 52.102215 10.1474083 true 1638274114 2 52.101968299999996 10.147795 true 1638274116Or, even better, something like:
row asset_id origin.latitude origin.longitude destination.latitude destination.longitude polyline start_timestamp end_timestamp 0 2 52.104774899999995 10.1427066 52.1034333 10.1451783 ST_MAKELINE from all trip coordinates 1638274085 1638274100 1 2 52.1030533 10.145886599999999 52.101968299999996 10.147795 ST_MAKELINE from all trip coordinates 1638274104 1638274116 ...ANSWER
Answered 2021-Dec-01 at 00:22Consider below approach
QUESTION
Not to fill a region bounded by a bezier curve and a line.
E.g. here's a very simple stroked Bezier curve
...ANSWER
Answered 2021-Nov-25 at 15:10As pointed out in the comment @Mike 'Pomax' Kamermans
create a pattern definition with an id, then point to that by fragment identifier in the attribute
I added the curve code and calculated the required dimensions for the pattern
QUESTION
I have multiple rectangle bounding boxes that I know are part of the same object (parts of a newspaper article), as in the first image. I'm trying to work out a way to merge these into one polygon bounding box, for the whole article, as in the second image.
I have seen lots of solutions based on merging overlapping bounding boxes, but I don't care if they're overlapping or not - I already know they're part of the same article. In some cases the headline is quite far away (for example above a picture), so solutions based on padding don't work either.
I feel like there should be a cv2 function that does this, but if there is, I am missing it. Any suggestions would be super helpful.
Code to create these two images:
...ANSWER
Answered 2021-Nov-14 at 18:00You could draw the convex hull of the contour points (this was drawn by hand):
(wrong convex hull image)
Then keep only the outer contour and try the polygonal approximation. I must admit that I cannot think of a smarter way of getting only vertical and horizontal lines.
As Christoph Rackwitz observed, I was wrong. Convex hull, doesn't work. Maybe an α-shape could solve the problem, but I'm not sure. An alternative approach could be to extract all the equations of the lines which define the bounding boxes, then for each point, compute the segment that connects it to the nearest line. If the line is part of the same bounding box or if the point falls outside of the bounding box, remove the segment.
It was harder than I expected, because my Python OpenCV proficiency is not enough. Nevertheless you can get nearly what you wanted in the question:
QUESTION
I play with CodeWorld and use the "entrypoint" activityOf() to simulate a physical system consisting of tanks. Here we have one tank with an outlet flow qout(h) that depends on the height of volume level. Of course, for serious simulation work one should use other software, but still fun to do here!
...ANSWER
Answered 2021-Oct-21 at 09:43I’ll respond using syntax and types form the “regular” Haskell environment, https://code.world/haskell.
The type of a activityOf is:
QUESTION
I have a list of perhaps hundreds of unsorted 2D line segments at random angles as shown in blue above, and I need to determine the sorted sequence of points a – i from them as shown in red. Here are some constraints and observations:
- The red polyline sequence a - i forms a tight enclosure of the outer points of all the segments. It's like a convex hull, but "sucked" into concavities. A tight enclosure of the entirety of all segments would be a fine solution; I can delete the unused parts.
- An enclosure does exist. The segments are not completely random.
- The y value of one point of each segment is 0, and the other end is a point on the enclosure.
- The enclosure cannot cross any line segment or itself. This also means that all segments are on one side of the enclosure polyline.
- There will be either one or two enclosure segments connected to each enclosure point a - i.
- The points may be spaced widely differently, not as tidily as shown.
- I should be able to determine point a or i if necessary to start off the algorithm.
It seems like starting from a bounding box around all segments and shrinking it to form a tight enclosure would be a reasonable approach, but I can't come up with a decent algorithm. This will eventually be coded in C#, but an algorithm or pseudo-code would be fine.
...ANSWER
Answered 2021-Sep-25 at 23:36History: this answer is now marked community wiki. The original conjecture was proven wrong by @dont talk just code. @Matt Timmermans provided the current conjecture in a comment under the question.
The following conjecture is submitted without proof: To compare the order of two line segments, compare the bottom x, and return the opposite if the segments cross.
If the conjecture holds, a comparator can be written for use with any comparison sort. Pseudocode for the comparator:
QUESTION
I'm using d3.js v6 to create a donut graph
I have created the donut and everything the problem is regarding putting the text in rect as mentioned below
Im hoping to achieve the expected result as shown in the image I've also attached the my current output image below
Problem the only remaining thing that I wanted is the boxes to have the value center aligned
...ANSWER
Answered 2021-Aug-17 at 13:51Fixed a few problems in your code:
QUESTION
I am doing a License/Number plate recognition project and I'm on the stage of completion but there is a small problem, I have successfully recognized the characters, consider the below example:
This is an input image, I got the prediction as 2791 2g rj14
As you can, the ocr did a great job but the arrangement is destroyed (DESTROYING the whole purpose). Sometimes it does outputs in the correct sequence but sometimes it does not, so when it does not output in the correct sequence I'm trying to develop an algorithm which will take the predicted num_plate string as input and rearrange it on the basis of my country (India).
Below are some images which tell us about the format of Indian Number/License Plate.
Also, I have collected all the states but for right now, I just want to do for only the 3 states which are: Delhi (DL), Haryana (HR), UttarPradesh (UP). More info : https://en.wikipedia.org/wiki/List_of_Regional_Transport_Office_districts_in_India
...ANSWER
Answered 2021-Aug-12 at 21:00If you can get the coordinates of each identified text box, then:
- Rotate the coordinates so the boxes are parallel with the X-axis
- Scale the Y-coordinates so they can be rounded to integers, so that boxes that are side-by-side will get the same integer Y-coordinate (like a line number)
- Sort the data by Y, then X coordinate
- Extract the texts in that order
Here is an example of such sequence:
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