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我有一些多边形（加拿大省份），使用 读取GeoPandas，并希望使用这些创建一个掩码，以应用于二维纬度-经度网格上的网格数据（使用从netcdf文件读取iris）。最终目标是只保留给定省份的数据，其余数据被屏蔽掉。因此，对于省内的网格框，掩码将为 1，而对于省外的网格框，掩码将为 0 或 NaN。

可以从此处的 shapefile 获取多边形： https ://www.dropbox.com/s/o5elu01fetwnobx/CAN_adm1.shp?dl=0

我使用的 netcdf 文件可以在这里下载： https ://www.dropbox.com/s/kxb2v2rq17m7lp7/t2m.20090815.nc?dl=0

我想这里有两种方法，但我都在苦苦挣扎：

1）使用多边形在经纬度网格上创建一个掩码，以便可以将其应用于python之外的大量数据文件（首选）

2）使用多边形对已读入的数据进行屏蔽，只提取感兴趣省内的数据，进行交互处理。

到目前为止我的代码：

import iris
import geopandas as gpd

#read the shapefile and extract the polygon for a single province
#(province names stored as variable 'NAME_1')
Canada=gpd.read_file('CAN_adm1.shp')
BritishColumbia=Canada[Canada['NAME_1'] == 'British Columbia']

#get the latitude-longitude grid from netcdf file
cubelist=iris.load('t2m.20090815.nc')
cube=cubelist[0]
lats=cube.coord('latitude').points
lons=cube.coord('longitude').points

#create 2d grid from lats and lons (may not be necessary?)
[lon2d,lat2d]=np.meshgrid(lons,lats)

#HELP!

非常感谢任何帮助或建议。

更新：按照下面@DPeterK 的出色解决方案，我的原始数据可以被屏蔽，给出以下内容：

使用加拿大各省的 shapefile 掩码的 netcdf 文件中的温度数据

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2 回答 2

8

看起来你已经开始了！从 shapefile 加载的几何图形公开了各种地理空间比较方法，在这种情况下，您需要该contains方法。您可以使用它来测试立方体水平网格中的每个点是否包含在不列颠哥伦比亚省几何体中。（请注意，这不是一个快速操作！）您可以使用此比较来构建一个 2D 掩码数组，该数组可以应用于您的多维数据集的数据或以其他方式使用。

我编写了一个 Python 函数来执行上述操作——它需要一个立方体和一个几何图形，并为立方体的（指定）水平坐标生成一个掩码，并将掩码应用于立方体的数据。功能如下：

def geom_to_masked_cube(cube, geometry, x_coord, y_coord,
                        mask_excludes=False):
    """
    Convert a shapefile geometry into a mask for a cube's data.

    Args:

    * cube:
        The cube to mask.
    * geometry:
        A geometry from a shapefile to define a mask.
    * x_coord: (str or coord)
        A reference to a coord describing the cube's x-axis.
    * y_coord: (str or coord)
        A reference to a coord describing the cube's y-axis.

    Kwargs:

    * mask_excludes: (bool, default False)
        If False, the mask will exclude the area of the geometry from the
        cube's data. If True, the mask will include *only* the area of the
        geometry in the cube's data.

    .. note::
        This function does *not* preserve lazy cube data.

    """
    # Get horizontal coords for masking purposes.
    lats = cube.coord(y_coord).points
    lons = cube.coord(x_coord).points
    lon2d, lat2d = np.meshgrid(lons,lats)

    # Reshape to 1D for easier iteration.
    lon2 = lon2d.reshape(-1)
    lat2 = lat2d.reshape(-1)

    mask = []
    # Iterate through all horizontal points in cube, and
    # check for containment within the specified geometry.
    for lat, lon in zip(lat2, lon2):
        this_point = gpd.geoseries.Point(lon, lat)
        res = geometry.contains(this_point)
        mask.append(res.values[0])

    mask = np.array(mask).reshape(lon2d.shape)
    if mask_excludes:
        # Invert the mask if we want to include the geometry's area.
        mask = ~mask
    # Make sure the mask is the same shape as the cube.
    dim_map = (cube.coord_dims(y_coord)[0],
               cube.coord_dims(x_coord)[0])
    cube_mask = iris.util.broadcast_to_shape(mask, cube.shape, dim_map)

    # Apply the mask to the cube's data.
    data = cube.data
    masked_data = np.ma.masked_array(data, cube_mask)
    cube.data = masked_data
    return cube

如果您只需要 2D 蒙版，则可以在上述函数将其应用于立方体之前将其返回。

要在原始代码中使用此函数，请在代码末尾添加以下内容：

geometry = BritishColumbia.geometry
masked_cube = geom_to_masked_cube(cube, geometry,
                                  'longitude', 'latitude',
                                  mask_excludes=True)

如果这没有掩盖任何东西，则很可能意味着您的立方体和几何图形是在不同范围内定义的。也就是说，您的立方体的经度坐标从 0°–360° 运行，如果几何的经度值从 -180°–180° 运行，则包含测试将永远不会返回True。您可以通过使用以下方法更改多维数据集的范围来解决此问题：

cube = cube.intersection(longitude=(-180, 180))
于 2017-12-14T12:17:11.850 回答
4

我找到了上面@DPeterK 发布的优秀解决方案的替代解决方案，它产生了相同的结果。它用于matplotlib.path测试点是否包含在从形状文件加载的几何图形描述的外部坐标内。我发布这个是因为这种方法比@DPeterK 给出的方法快约 10 倍（2:23 分钟 vs 25:56 分钟）。我不确定什么是可取的：优雅的解决方案，或快速的蛮力解决方案。也许一个人可以两者兼得？！

这种方法的一个复杂之处是一些几何图形是多多边形- 即形状由几个较小的多边形组成（在这种情况下，不列颠哥伦比亚省包括西海岸的岛屿，这些岛屿无法用大陆的坐标来描述不列颠哥伦比亚多边形）。MultiPolygon 没有外部坐标，但单个多边形有，因此每个多边形都需要单独处理。我发现最巧妙的解决方案是使用从 GitHub ( https://gist.github.com/mhweber/cf36bb4e09df9deee5eb54dc6be74d26 ) 复制的函数，该函数将 MultiPolygons“分解”为单个多边形列表，然后可以单独处理。

下面概述了工作代码，以及我的文档。抱歉，它不是最优雅的代码 - 我对 Python 比较陌生，我确信有很多不必要的循环/更整洁的方法来做事！

import numpy as np
import iris
import geopandas as gpd
from shapely.geometry import Point
import matplotlib.path as mpltPath
from shapely.geometry.polygon import Polygon
from shapely.geometry.multipolygon import MultiPolygon


#-----


#FIRST, read in the target data and latitude-longitude grid from netcdf file
cubelist=iris.load('t2m.20090815.minus180_180.nc')
cube=cubelist[0]
lats=cube.coord('latitude').points
lons=cube.coord('longitude').points

#create 2d grid from lats and lons
[lon2d,lat2d]=np.meshgrid(lons,lats)

#create a list of coordinates of all points within grid
points=[]

for latit in range(0,241):
    for lonit in range(0,480):
        point=(lon2d[latit,lonit],lat2d[latit,lonit])
        points.append(point)

#turn into np array for later
points=np.array(points)

#get the cube data - useful for later
fld=np.squeeze(cube.data)

#create a mask array of zeros, same shape as fld, to be modified by
#the code below
mask=np.zeros_like(fld)


#NOW, read the shapefile and extract the polygon for a single province
#(province names stored as variable 'NAME_1')
Canada=gpd.read_file('/Users/ianashpole/Computing/getting_province_outlines/CAN_adm_shp/CAN_adm1.shp')
BritishColumbia=Canada[Canada['NAME_1'] == 'British Columbia']


#BritishColumbia.geometry.type reveals this to be a 'MultiPolygon'
#i.e. several (in this case, thousands...) if individual polygons.
#I ultimately want to get the exterior coordinates of the BritishColumbia
#polygon, but a MultiPolygon is a list of polygons and therefore has no
#exterior coordinates. There are probably many ways to progress from here,
#but the method I have stumbled upon is to 'explode' the multipolygon into
#it's individual polygons and treat each individually. The function below
#to 'explode' the MultiPolygon was found here:
#https://gist.github.com/mhweber/cf36bb4e09df9deee5eb54dc6be74d26


#---define function to explode MultiPolygons

def explode_polygon(indata):
    indf = indata
    outdf = gpd.GeoDataFrame(columns=indf.columns)
    for idx, row in indf.iterrows():
        if type(row.geometry) == Polygon:
            #note: now redundant, but function originally worked on
            #a shapefile which could have combinations of individual polygons
            #and MultiPolygons
            outdf = outdf.append(row,ignore_index=True)
        if type(row.geometry) == MultiPolygon:
            multdf = gpd.GeoDataFrame(columns=indf.columns)
            recs = len(row.geometry)
            multdf = multdf.append([row]*recs,ignore_index=True)
            for geom in range(recs):
                multdf.loc[geom,'geometry'] = row.geometry[geom]
            outdf = outdf.append(multdf,ignore_index=True)
    return outdf

#-------


#Explode the BritishColumbia MultiPolygon into its constituents
EBritishColumbia=explode_polygon(BritishColumbia)


#Loop over each individual polygon and get external coordinates
for index,row in EBritishColumbia.iterrows():

    print 'working on polygon', index
    mypolygon=[]
    for pt in list(row['geometry'].exterior.coords):
        print index,', ',pt
        mypolygon.append(pt)


    #See if any of the original grid points read from the netcdf file earlier
    #lie within the exterior coordinates of this polygon
    #pth.contains_points returns a boolean array (true/false), in the
    #shape of 'points'
    path=mpltPath.Path(mypolygon)
    inside=path.contains_points(points)


    #find the results in the array that were inside the polygon ('True')
    #and set them to missing. First, must reshape the result of the search
    #('points') so that it matches the mask & original data
    #reshape the result to the main grid array
    inside=np.array(inside).reshape(lon2d.shape)
    i=np.where(inside == True)
    mask[i]=1


print 'fininshed checking for points inside all polygons'


#mask now contains 0's for points that are not within British Columbia, and
#1's for points that are. FINALLY, use this to mask the original data
#(stored as 'fld')
i=np.where(mask == 0)
fld[i]=np.nan

#Done.
于 2017-12-15T15:43:26.903 回答