我有这张地图参考：SC201703
我有以下python转换函数将上面的内容转换为lat / lon：
import numpy as np
import pandas as pd
from math import sqrt, pi, sin, cos, tan, atan2 as arctan2
import re
# From http://all-geo.org/volcan01010/2012/11/change-coordinates-with-pyproj/
# Region codes for 100 km grid squares.
regions=[['HL','HM','HN','HO','HP','JL','JM'],
['HQ','HR','HS','HT','HU','JQ','JR'],
['HV','HW','HX','HY','HZ','JV','JW'],
['NA','NB','NC','ND','NE','OA','OB'],
['NF','NG','NH','NJ','NK','OF','OG'],
['NL','NM','NN','NO','NP','OL','OM'],
['NQ','NR','NS','NT','NU','OQ','OR'],
['NV','NW','NX','NY','NZ','OV','OW'],
['SA','SB','SC','SD','SE','TA','TB'],
['SF','SG','SH','SJ','SK','TF','TG'],
['SL','SM','SN','SO','SP','TL','TM'],
['SQ','SR','SS','ST','SU','TQ','TR'],
['SV','SW','SX','SY','SZ','TV','TW']]
# Reshuffle so indices correspond to offsets
regions=np.array( [ regions[x] for x in range(12,-1,-1) ] )
regions=regions.transpose()
#-------------------------------------------------------------------------------
def to_osgb36(coords):
"""Reformat British National Grid references to OSGB36 numeric coordinates.
Grid references can be 4, 6, 8 or 10 figures. Returns a tuple of x, y.
Examples:
Single value
>>> to_osgb36('NT2755072950')
(327550, 672950)
For multiple values, use the zip function
>>> gridrefs = ['HU431392', 'SJ637560', 'TV374354']
>>> xy=to_osgb36(gridrefs)
>>> x, y = zip(*xy)
>>> x
(443100, 363700, 537400)
>>> y
(1139200, 356000, 35400)
"""
#
# Check for individual coord, or list, tuple or array of coords
#
if type(coords)==list:
return [to_osgb36(c) for c in coords]
elif type(coords)==tuple:
return tuple([to_osgb36(c) for c in coords])
elif type(coords)==type(np.array('string')):
return np.array([ to_osgb36(str(c)) for c in list(coords) ])
#
# Input is grid reference...
#
elif type(coords)==str and re.match(r'^[A-Za-z]{2}(\d{2}|\d{6}|\d{8}|\d{10})$', coords):
region=coords[0:2].upper()
x_box, y_box = np.where(regions==region)
try: # Catch bad region codes
x_offset = 100000 * x_box[0] # Convert index in 'regions' to offset
y_offset = 100000 * y_box[0]
except IndexError:
raise ValueError('Invalid 100km grid square code')
nDigits = (len(coords)-2)//2
factor = 10**(5-nDigits)
x,y = (int(coords[2:2+nDigits])*factor + x_offset,
int(coords[2+nDigits:2+2*nDigits])*factor + y_offset)
return x, y
#
# Catch invalid input
#
else:
raise TypeError('Valid inputs are 6,8 or 10-fig references as strings e.g.'
'"NN123321", or lists/tuples/arrays of strings.')
# From http://hannahfry.co.uk/2012/02/01/converting-british-national-grid-to-latitude-and-longitude-ii/
def OSGB36toWGS84(E,N):
#E, N are the British national grid coordinates - eastings and northings
a, b = 6377563.396, 6356256.909 #The Airy 180 semi-major and semi-minor axes used for OSGB36 (m)
F0 = 0.9996012717 #scale factor on the central meridian
lat0 = 49*pi/180 #Latitude of true origin (radians)
lon0 = -2*pi/180 #Longtitude of true origin and central meridian (radians)
N0, E0 = -100000, 400000 #Northing & easting of true origin (m)
e2 = 1 - (b*b)/(a*a) #eccentricity squared
n = (a-b)/(a+b)
#Initialise the iterative variables
lat,M = lat0, 0
while N-N0-M >= 0.00001: #Accurate to 0.01mm
lat = (N-N0-M)/(a*F0) + lat;
M1 = (1 + n + (5./4)*n**2 + (5./4)*n**3) * (lat-lat0)
M2 = (3*n + 3*n**2 + (21./8)*n**3) * sin(lat-lat0) * cos(lat+lat0)
M3 = ((15./8)*n**2 + (15./8)*n**3) * sin(2*(lat-lat0)) * cos(2*(lat+lat0))
M4 = (35./24)*n**3 * sin(3*(lat-lat0)) * cos(3*(lat+lat0))
#meridional arc
M = b * F0 * (M1 - M2 + M3 - M4)
#transverse radius of curvature
nu = a*F0/sqrt(1-e2*sin(lat)**2)
#meridional radius of curvature
rho = a*F0*(1-e2)*(1-e2*sin(lat)**2)**(-1.5)
eta2 = nu/rho-1
secLat = 1./cos(lat)
VII = tan(lat)/(2*rho*nu)
VIII = tan(lat)/(24*rho*nu**3)*(5+3*tan(lat)**2+eta2-9*tan(lat)**2*eta2)
IX = tan(lat)/(720*rho*nu**5)*(61+90*tan(lat)**2+45*tan(lat)**4)
X = secLat/nu
XI = secLat/(6*nu**3)*(nu/rho+2*tan(lat)**2)
XII = secLat/(120*nu**5)*(5+28*tan(lat)**2+24*tan(lat)**4)
XIIA = secLat/(5040*nu**7)*(61+662*tan(lat)**2+1320*tan(lat)**4+720*tan(lat)**6)
dE = E-E0
#These are on the wrong ellipsoid currently: Airy1830. (Denoted by _1)
lat_1 = lat - VII*dE**2 + VIII*dE**4 - IX*dE**6
lon_1 = lon0 + X*dE - XI*dE**3 + XII*dE**5 - XIIA*dE**7
#Want to convert to the GRS80 ellipsoid.
#First convert to cartesian from spherical polar coordinates
H = 0 #Third spherical coord.
x_1 = (nu/F0 + H)*cos(lat_1)*cos(lon_1)
y_1 = (nu/F0+ H)*cos(lat_1)*sin(lon_1)
z_1 = ((1-e2)*nu/F0 +H)*sin(lat_1)
#Perform Helmut transform (to go between Airy 1830 (_1) and GRS80 (_2))
s = -20.4894*10**-6 #The scale factor -1
tx, ty, tz = 446.448, -125.157, + 542.060 #The translations along x,y,z axes respectively
rxs,rys,rzs = 0.1502, 0.2470, 0.8421 #The rotations along x,y,z respectively, in seconds
rx, ry, rz = rxs*pi/(180*3600.), rys*pi/(180*3600.), rzs*pi/(180*3600.) #In radians
x_2 = tx + (1+s)*x_1 + (-rz)*y_1 + (ry)*z_1
y_2 = ty + (rz)*x_1 + (1+s)*y_1 + (-rx)*z_1
z_2 = tz + (-ry)*x_1 + (rx)*y_1 + (1+s)*z_1
#Back to spherical polar coordinates from cartesian
#Need some of the characteristics of the new ellipsoid
a_2, b_2 =6378137.000, 6356752.3141 #The GSR80 semi-major and semi-minor axes used for WGS84(m)
e2_2 = 1- (b_2*b_2)/(a_2*a_2) #The eccentricity of the GRS80 ellipsoid
p = sqrt(x_2**2 + y_2**2)
#Lat is obtained by an iterative proceedure:
lat = arctan2(z_2,(p*(1-e2_2))) #Initial value
latold = 2*pi
while abs(lat - latold)>10**-16:
lat, latold = latold, lat
nu_2 = a_2/sqrt(1-e2_2*sin(latold)**2)
lat = arctan2(z_2+e2_2*nu_2*sin(latold), p)
#Lon and height are then pretty easy
lon = arctan2(y_2,x_2)
H = p/cos(lat) - nu_2
#Uncomment this line if you want to print the results
#print [(lat-lat_1)*180/pi, (lon - lon_1)*180/pi]
#Convert to degrees
lat = lat*180/pi
lon = lon*180/pi
#Job's a good'n.
return lat, lon
现在我运行以上内容：
OSGB36toWGS84(*to_osgb36('SC201703'))
它返回以下lat / lon元组：
(54.1304582982807, -4.754389466075483)
如果我然后转到任何在线地图，例如：http://gridreferencefinder.com/并输入SC201703并与Lat = 54.1304582982807 lon = -4.754389466075483的输入进行比较，则它们不是同一点。
导致这种不一致的原因是什么？ python代码不正确吗？