问题描述
我试图从中心纬度位置计算出特定半径内的所有值.我使用的代码如下:
I am trying to calculate all the values contained within a particular radius from a central lat lon position.The code which I am using is as given:
import numpy as np
import matplotlib.pylab as pl
import netCDF4 as nc
import haversine
f = nc.Dataset('air_temp.nc')
def haversine(lon1, lat1, lon2, lat2):
# convert decimal degrees to radians
lon1 = np.deg2rad(lon1)
lon2 = np.deg2rad(lon2)
lat1 = np.deg2rad(lat1)
lat2 = np.deg2rad(lat2)
# haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = np.sin(dlat/2)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2)**2
c = 2 * np.arcsin(np.sqrt(a))
r = 6371
return c * r
# Latitude / longitude grid
#lat = np.linspace(50,54,16)
lat = f.variables['lat'][:]
#lon = np.linspace(6,9,12)
lon = f.variables['lon'][:]
clat = 19.7
clon = 69.7
max_dist = 750 # max distance in km
# Calculate distance between center and all other lat/lon pairs
distance = haversine(lon[:,np.newaxis], lat, clon, clat)
# Mask distance array where distance > max_dist
distance_m = np.ma.masked_greater(distance, max_dist)
# Dummy data
air = f.variables['air'][0,:,:,:]
data = np.squeeze(air)
data = np.transpose(data)
#data = np.random.random(size=[lon.size, lat.size])
data_m = np.ma.masked_where(distance >max_dist, data)
# Test: set a value outside the max_dist circle to a large value:
#data[0,0] = 10
#avg = np.nanmean(data_m)-273
我已使用Haversine函数查找距离.现在我面临的问题是我需要距离中心点 2.5 度半径内的值，但我得到的都是公里.因此，如果有人能说出我做错了什么或在正确的过程中如何做对我有帮助，将不胜感激
I have used the haversine function for finding the distance. Now what I am facing the problem is I need values within a radius of 2.5 degrees from the center point, but I am getting all in kilometers. So if anyone can help me by saying what I am doing wrong or how to it in the right procedure it will be highly appreciated
推荐答案
就直线距离(或更短的电弧)而言，1度始终是111公里(假设地球是一个完美的球体(*编辑，而不是正方形").
In terms of straight-line (or rather shortest-arc) distance, 1 degree is always 111km (assuming the earth is a perfect sphere (*edited, not "square")).
地球上任意两点之间的最短弧的中心始终是地球的中心.1度=2π/360弧度，因此距离为R(2π/360)= 6371(2π/360)= 111.19.
The center of the shortest arc between any two points on a globe is always the center of the globe. 1 degree = 2π/360 radian, so the distance is R(2π/360) = 6371(2π/360) = 111.19.
更新:
您错过的不是半正弦计算或度-公里转换，而是对 NetCDF 的元数据格式和 NumPy 的网格网格的理解.f.variables['lat'] 为您提供 37 个纬度值，f.variables['lon'] 为您提供 144 个经度值，因此如果您想进行暴力搜索所有这些，你需要使用 np.meshgrid 生成一个 37*144=5328 点的网格.
What you missed is not the haversine calculation or the degree-km conversion, it's the understanding of NetCDF's metadata format and NumPy's meshgrid. f.variables['lat'] gives you 37 latitude values and f.variables['lon'] gives you 144 longitude values, so if you want to brute force search all of them, you need to use np.meshgrid to generate a grid of 37*144=5328 points.
功能代码如下:
import numpy as np
def haversine(lon1, lat1, lon2, lat2):
# convert decimal degrees to radians
lon1 = np.deg2rad(lon1)
lon2 = np.deg2rad(lon2)
lat1 = np.deg2rad(lat1)
lat2 = np.deg2rad(lat2)
# haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = np.sin(dlat/2)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2)**2
c = 2 * np.arcsin(np.sqrt(a))
r = 6371
return c * r
# center point
ctr_lon, ctr_lat = 69.7, 19.7
# the lon/lat grids
lon = np.arange(0, 360, 2.5)
lat = np.arange(-45, 46, 2.5)
# get coordinates of all points on the grid
grid_lon, grid_lat = np.meshgrid(lon, lat)
dists_in_km = haversine(grid_lon, grid_lat, ctr_lon, ctr_lat)
dists_in_deg = dists_in_km / 111
# find nearby points
thr = 2.5
for i in range(grid_lon.shape[0]):
for j in range(grid_lon.shape[1]):
this_lon = grid_lon[i, j]
this_lat = grid_lat[i, j]
dist = dists_in_deg[i, j]
if dist <= thr:
print('lon=%.1f lat=%.1f dist=%.2fdeg' % (this_lon, this_lat, dist))
输出:
lon=70.0 lat=17.5 dist=2.22deg
lon=67.5 lat=20.0 dist=2.09deg
lon=70.0 lat=20.0 dist=0.41deg
这很有意义.
这篇关于如何在经度和纬度值的中心位置找到半径范围内的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！