Python 沿直线生成lat long

我需要找到更靠近其他坐标的坐标（lat/long），这两个坐标之间的距离占一定百分比

例如：

location1 = (32.7991094663, -117.234719251)
location2 = (32.7094778234, -117.136413578)

# find a new location closer by 60% from location1 than location2
find_closer_location(location1, location2, 60)

为了计算距离，我使用下面的算法

from math import radians, cos, sin, asin, sqrt


def haversine(lon1, lat1, lon2, lat2):
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    km = 6371 * c
    return km

计算它：

def moveMe(t1,t2,perc):
    x1, y1= t1  # location1 = (32.7991094663, -117.234719251)
    x2, y2 = t2 # location2 = (32.7094778234, -117.136413578)

    dx = (x2-x1) / 100.0 * perc
    dy = (y2-y1) / 100.0 * perc

    return (x1+dx,y1+dy)

location1 = (32.7991094663, -117.234719251)
location2 = (32.7094778234, -117.136413578)
print(moveMe(location1,location2,10))

输出：

(32.808072630590004, -117.2445498183)

---

这将在

perc

100）

t1

和

t2

之间的直线上插入一个新点。如果你想从这条线中获得分数，你就必须更具创造性。

谢谢大家的努力

最后，我想我有了一个解决办法

from math import radians, cos, sin, asin, sqrt, atan2, degrees, log, tan, pi

R = 6378.1  #Radius of the Earth in km


def get_distance(lon1, lat1, lon2, lat2):
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
    dlon = lon2 - lon1
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    km = R * c
    return km


def calculateBearing(lat1, lng1, lat2, lng2):
    '''calculates the azimuth in degrees from start point to end point'''
    startLat = radians(lat1)
    startLong = radians(lng1)
    endLat = radians(lat2)
    endLong = radians(lng2)
    dLong = endLong - startLong
    dPhi = log(tan(endLat/2.0+pi/4.0)/tan(startLat/2.0+pi/4.0))
    if abs(dLong) > pi:
         if dLong > 0.0:
             dLong = -(2.0 * pi - dLong)
         else:
             dLong = (2.0 * pi + dLong)
    bearing = (degrees(atan2(dLong, dPhi)) + 360.0) % 360.0;
    return bearing


def getDestinationLatLong(lat, lng, azimuth, d):
    '''returns the lat an long of destination point 
    given the start lat, long, aziuth, and distance'''
    brng = radians(azimuth) #Bearing is degrees converted to radians.
    lat1 = radians(lat) #Current dd lat point converted to radians
    lon1 = radians(lng) #Current dd long point converted to radians
    lat2 = asin(sin(lat1) * cos(d/R) + cos(lat1)* sin(d/R)* cos(brng))
    lon2 = lon1 + atan2(sin(brng) * sin(d/R)* cos(lat1), cos(d/R)- sin(lat1)* sin(lat2))
    #convert back to degrees
    lat2 = degrees(lat2)
    lon2 = degrees(lon2)
    return[lat2, lon2]


location1 = (32.7991094663, -117.234719251)
location2 = (32.7094778234, -117.136413578)
percentage = 60

distance = get_distance(location1[1], location1[0], location2[1], location2[0])
print distance
new_distance = distance - (distance * percentage / 100)
print new_distance
azimuth = calculateBearing(location1[0], location1[1], location2[0], location2[1])
coords = getDestinationLatLong(location1[0], location1[1], azimuth, new_distance)
print coords

---

参考资料：

什么百分比？Ken Syme，如果我理解正确，两个坐标之间的距离百分比。你得了2分。如果你想把其中一个点移动到离另一个点近60%的地方，你能指定要移动的点吗？地球不是平的！用

（10,10）

和

（15,15）

的方式有20%没有

（9,9）

。。你也不应该鼓励低质量的问题。@sayse修复了它-为什么你认为它是一个低质量的问题？如果你把贴在这里的每一个问题都记下来，去掉那些可以通过思考、谷歌搜索或其他研究来解决的问题，那么几乎没有一个问题会被解决。@KenSyme-嗯，他没有定义距离是在地面上测量的。你认为他想把“在地面上行驶”的距离减半吗？@PatrickArtner-因为op没有用他们自己的任何努力来解决他们的问题。