xarray 选择最近的纬度/经度与多维坐标
bla*_*kbk 9 python python-xarray
我有一个具有不规则间隔纬度和经度坐标的 xarray 数据集。我的目标是在最接近某个纬度/经度的点处找到变量的值。
由于x和y尺寸不是纬度/经度值，因此ds.sel()在这种情况下似乎不能单独使用该方法。是否有一种以 xarray 为中心的方法来通过参考多维纬度/经度维度来定位最接近所需纬度/经度的点？例如，我想提取最接近lat=21.2和的 SPEED 值lon=-122.68。
下面是一个示例数据集...
lats = np.array([[21.138 , 21.14499, 21.15197, 21.15894, 21.16591],
[21.16287, 21.16986, 21.17684, 21.18382, 21.19079],
[21.18775, 21.19474, 21.20172, 21.2087 , 21.21568],
[21.21262, 21.21962, 21.22661, 21.23359, 21.24056],
[21.2375 , 21.2445 , 21.25149, 21.25848, 21.26545]])
lons = np.array([[-122.72 , -122.69333, -122.66666, -122.63999, -122.61331],
[-122.7275 , -122.70082, -122.67415, -122.64746, -122.62078],
[-122.735 , -122.70832, -122.68163, -122.65494, -122.62825],
[-122.7425 , -122.71582, -122.68912, -122.66243, -122.63573],
[-122.75001, -122.72332, -122.69662, -122.66992, -122.64321]])
speed = np.array([[10.934007, 10.941321, 10.991583, 11.063932, 11.159435],
[10.98778 , 10.975482, 10.990983, 11.042522, 11.131154],
[11.013505, 11.001573, 10.997754, 11.03566 , 11.123781],
[11.011163, 11.000227, 11.010223, 11.049 , 11.1449 ],
[11.015698, 11.026604, 11.030653, 11.076904, 11.201464]])
ds = xarray.Dataset({'SPEED':(('x', 'y'),speed)},
coords = {'latitude': (('x', 'y'), lats),
'longitude': (('x', 'y'), lons)},
attrs={'variable':'Wind Speed'})
的价值ds：
<xarray.Dataset>
Dimensions: (x: 5, y: 5)
Coordinates:
latitude (x, y) float64 21.14 21.14 21.15 21.16 ... 21.25 21.26 21.27
longitude (x, y) float64 -122.7 -122.7 -122.7 ... -122.7 -122.7 -122.6
Dimensions without coordinates: x, y
Data variables:
SPEED (x, y) float64 10.93 10.94 10.99 11.06 ... 11.03 11.03 11.08 11.2
Attributes:
variable: Wind Speed
同样，ds.sel(latitude=21.2, longitude=-122.68)不起作用，因为纬度和经度不是数据集维度。
我想出了一个不纯粹使用 xarray 的方法。我首先手动找到最近邻居的索引，然后使用该索引访问 xarray 维度。
# A 2D plot of the SPEED variable, assigning the coordinate values,
# and plot the verticies of each point
ds.SPEED.plot(x='longitude', y='latitude')
plt.scatter(ds.longitude, ds.latitude)
# I want to find the speed at a certain lat/lon point.
lat = 21.22
lon = -122.68
# First, find the index of the grid point nearest a specific lat/lon.
abslat = np.abs(ds.latitude-lat)
abslon = np.abs(ds.longitude-lon)
c = np.maximum(abslon, abslat)
([xloc], [yloc]) = np.where(c == np.min(c))
# Now I can use that index location to get the values at the x/y diminsion
point_ds = ds.sel(x=xloc, y=yloc)
# Plot requested lat/lon point blue
plt.scatter(lon, lat, color='b')
plt.text(lon, lat, 'requested')
# Plot nearest point in the array red
plt.scatter(point_ds.longitude, point_ds.latitude, color='r')
plt.text(point_ds.longitude, point_ds.latitude, 'nearest')
plt.title('speed at nearest point: %s' % point_ds.SPEED.data)
另一个潜在的解决方案（同样，不是 xarray）是使用 scipy 的 KDTree
- 稍后再讲，但如果你转到上面的笔记本，你会得到一个很好的可视化结果，显示这个答案正在取 l-无穷范数的最小值 (https://math.stackexchange.com/questions/2842541/about- l-2-和-l-infty-范数）。执行 `np.min(np.maximum(abslon, abslat))` 是 l-无穷范数，执行 `np.min(abslon**2 + abslat**2)` 是执行 L2 范数最小值。两人似乎都会在这里工作。 (2认同)
我认为您需要以不同的方式创建数据集，以确保latitude并longitude具有可解释的维度（请参阅文章Xarray 的基本数据结构）。
例如：
import numpy as np
import pandas as pd
import xarray
import matplotlib.pyplot as plt
from scipy.interpolate import griddata
lats = np.array([21.138, 21.14499, 21.15197, 21.15894, 21.16591,
21.16287, 21.16986, 21.17684, 21.18382, 21.19079,
21.18775, 21.19474, 21.20172, 21.2087, 21.21568,
21.21262, 21.21962, 21.22661, 21.23359, 21.24056,
21.2375, 21.2445, 21.25149, 21.25848, 21.26545])
lons = np.array([-122.72, -122.69333, -122.66666, -122.63999, -122.61331,
-122.7275, -122.70082, -122.67415, -122.64746, -122.62078,
-122.735, -122.70832, -122.68163, -122.65494, -122.62825,
-122.7425, -122.71582, -122.68912, -122.66243, -122.63573,
-122.75001, -122.72332, -122.69662, -122.66992, -122.64321])
speed = np.array([10.934007, 10.941321, 10.991583, 11.063932, 11.159435,
10.98778, 10.975482, 10.990983, 11.042522, 11.131154,
11.013505, 11.001573, 10.997754, 11.03566, 11.123781,
11.011163, 11.000227, 11.010223, 11.049, 11.1449,
11.015698, 11.026604, 11.030653, 11.076904, 11.201464])
fig, (ax1, ax2) = plt.subplots(ncols=2, figsize=(12, 5))
idx = pd.MultiIndex.from_arrays(arrays=[lons, lats], names=["lon", "lat"])
s = pd.Series(data=speed, index=idx)
da = xarray.DataArray.from_series(s)
print(da)
da.plot(ax=ax1)
print('-'*80)
print(da.sel(lat=21.2, lon=-122.68, method='nearest'))
# define grid.
num_points = 100
lats_i = np.linspace(np.min(lats), np.max(lats), num_points)
lons_i = np.linspace(np.min(lons), np.max(lons), num_points)
# grid the data.
speed_i = griddata((lats, lons), speed,
(lats_i[None, :], lons_i[:, None]), method='cubic')
# contour the gridded data
ax2.contour(lats_i, lons_i, speed_i, 15, linewidths=0.5, colors='k')
contour = ax2.contourf(lats_i, lons_i, speed_i, 15, cmap=plt.cm.jet)
plt.colorbar(contour, ax=ax2)
# plot data points.
for i, (lat, lon) in enumerate(zip(lats, lons)):
label = f'{speed[i]:0.2f}'
ax2.annotate(label, (lat, lon))
ax2.scatter(lats, lons, marker='o', c='b', s=5)
ax2.set_title(f'griddata test {num_points} points')
plt.subplots_adjust(wspace=0.2)
plt.show()
结果
<xarray.DataArray (lat: 25, lon: 25)>
array([[ nan, nan, nan, nan, nan, 10.934007,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, 10.941321, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, 10.991583, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, 11.063932, nan, nan, nan,
nan],
[ nan, nan, nan, 10.98778 , nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
11.159435],
[ nan, nan, nan, nan, nan, nan,
nan, nan, 10.975482, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, 10.990983, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
11.042522, nan, nan, nan, nan, nan,
nan],
[ nan, nan, 11.013505, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, 11.131154,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, 11.001573, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
10.997754, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, 11.03566 ,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, 11.011163, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, 11.123781, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
11.000227, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, 11.010223,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, 11.049 , nan,
nan, nan, nan, nan, nan, nan,
nan],
[11.015698, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, 11.1449 , nan, nan,
nan],
[ nan, nan, nan, nan, 11.026604, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, 11.030653, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, 11.076904, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan],
[ nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, nan, nan, nan, nan, nan,
nan, 11.201464, nan, nan, nan, nan,
nan]])
Coordinates:
* lat (lat) float64 21.14 21.14 21.15 21.16 ... 21.24 21.25 21.26 21.27
* lon (lon) float64 -122.8 -122.7 -122.7 -122.7 ... -122.6 -122.6 -122.6
--------------------------------------------------------------------------------
<xarray.DataArray ()>
array(10.997754)
Coordinates:
lat float64 21.2
lon float64 -122.7
以及一个包含网格的绘图，只是为了好玩
在这里参加聚会有点晚了，但我已经多次回到这个问题。如果您的 x 和 y 坐标在地理空间坐标系中，您可以使用 cartopy 将纬度/经度点转换为该坐标系。如果您查看来自 netcdf 的元数据，构建 cartopy 投影通常很简单。
import cartopy.crs as ccrs
# Example - your x and y coordinates are in a Lambert Conformal projection
data_crs = ccrs.LambertConformal(central_longitude=-100)
# Transform the point - src_crs is always Plate Carree for lat/lon grid
x, y = data_crs.transform_point(-122.68, 21.2, src_crs=ccrs.PlateCarree())
# Now you can select data
ds.sel(x=x, y=y)
- 这是最好的答案。请注意，您还可以使用 `.sel(..., method="nearest")` 来选择 `x, y` 中最近的点。请注意，如果您的投影使空间充分扭曲，这可能不是某些点中最近的点。 (4认同)