您不能独立于坐标系来评估纬度的“接近程度”-您必须评估坐标对与输入坐标的接近程度。
Lat/Lon实际上只是球坐标。给定两个点（lat 1，lon 1）（lat 2，lon 2），接近度（用大圆表示）由这两个点之间的球向量之间的Angular 给出（近似地球为球体）。
你可以通过构造这两个点的笛卡尔向量，并取点积来计算，也就是a * b * cos（theta），其中theta是你想要的。

import numpy as np

def lat_lon_cartesian(lats,lons):

    lats = np.ravel(lats) #make both inputs 1-dimensional
    lons = np.ravel(lons)

    x = np.cos(np.radians(lons))*np.cos(np.radians(lats))
    y = np.sin(np.radians(lons))*np.cos(np.radians(lats))
    z = np.sin(np.radians(lats))
    return np.c_[x,y,z]

def closest(in_lats,in_lons,data_lats,data_lons):
    in_vecs = lat_lon_cartesian(in_lats,in_lons)
    data_vecs = lat_lon_cartesian(data_lats,data_lons)
    indices = []
    for in_vec in in_vecs: # if input lats/lons is small list then doing a for loop is ok
        # otherwise can be vectorized with some array gymnastics
        dot_product = np.sum(in_vec*data_vecs,axis=1)
        angles = np.arccos(dot_product) # all are unit vectors so a=b=1
        indices.append(np.argmin(angles))
    return indices

def get_local_value(data, in_lats, in_lons, data_lats, data_lons):
    raveled_data = np.ravel(data)
    raveled_lats = np.ravel(data_lats)
    raveled_lons = np.ravel(data_lons)
    inds = closest(in_lats,in_lons,raveled_lats,raveled_lons)
    dtypes = []
    closest_lat_lons = []
    
    for ind in inds:
                
        #if data is 2-d with same shape as the lat and lon meshgrids, then
        #it should be raveled as well and indexed by the same index
        dtype = raveled_data[ind]
        dtypes.append(dtype)

        closest_lat_lons.append((raveled_lats[ind],raveled_lons[ind]))
        #can return the closes matching lat lon data in the grib if you want
    return dtypes

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编辑：或者使用插值。

import numpyp as np
from scipy.interpolate import RegularGridInterpolator

#assuming a grb object from pygrib
#see https://jswhit.github.io/pygrib/api.html#example-usage

lats, lons = grb.latlons()
#source code for pygrib looks like it calls lons,lats = np.meshgrid(...)
#so the following should give the unique lat/lon sequences
lat_values = lats[:,0]
lon_values = lons[0,:]

grb_values = grb.values

#create interpolator
grb_interp = RegularGridInterpolator((lat_values,lon_values),grb_values)

#in_lats, in_lons = desired input points (1-d each)
interpolated_values = grb_interp(np.c_[in_lats,in_lons])

#the result should be the linear interpolation between the four closest lat/lon points in the data set around each of your input lat/lon points.

型
虚拟数据插值示例：

>>> import numpy as np
>>> lats = np.array([1,2,3])
>>> lons = np.array([4,5,6,7])
>>> lon_mesh,lat_mesh = np.meshgrid(lons,lats)
>>> lon_mesh
array([[4, 5, 6, 7],
       [4, 5, 6, 7],
       [4, 5, 6, 7]])
>>> lat_mesh
array([[1, 1, 1, 1],
       [2, 2, 2, 2],
       [3, 3, 3, 3]])
>>> z = lon_mesh + lat_mesh #some example function of lat/lon (simple sum)
>>> z
array([[ 5,  6,  7,  8],
       [ 6,  7,  8,  9],
       [ 7,  8,  9, 10]])
>>> from scipy.interpolate import RegularGridInterpolator
>>> lon_mesh[0,:] #should produce lons
array([4, 5, 6, 7])
>>> lat_mesh[:,0] #should produce lats
array([1, 2, 3])
>>> interpolator = RegularGridInterpolator((lats,lons),z)
>>> input_lats = np.array([1.5,2.5])
>>> input_lons = np.array([5.5,7])
>>> input_points = np.c_[input_lats,input_lons]
>>> input_points
array([[1.5, 5.5],
       [2.5, 7. ]])
>>> interpolator(input_points)
array([7. , 9.5])
>>> #7 = 1.5+5.5 : correct
... #9.5 = 2.5+7 : correct
...
>>>

型

我知道这是一个老问题，但我想把它作为一个简单得多的答案，通过最小化x和y的距离，给予你所需要的：

coordinates = np.unravel_index((np.abs(lats - lat_to_find) + np.abs(lons - lon_to_find)).argmin(), lats.shape)

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lats和隆恩应该是lats和隆恩的网格，lat_to_find和lon_to_find是要在数组中查找的lat/lon对。然后你有你的索引，你可以快速索引任何你需要的变量。