如何使用 Python 底图更快地绘制地理定位的 RGB 数据
How to plot geolocated RGB data faster using Python basemap
我在使用 Python 的带有纬度和经度数据的底图模块绘制 RGB 图像时遇到问题。现在,我可以绘制我想要的图,但问题是它有多慢,因为它能够比 RGB 数据更快地绘制单通道数据,而且一般来说,自己绘制 RGB 图像也是如此快速地。由于我有 lat/lon 数据,这就是事情变得复杂的地方。我已经查看了这个问题的解决方案:
How to plot an irregular spaced RGB image using python and basemap?
这就是我到达现在位置的方式。它本质上归结为以下问题。在底图中使用 pcolormesh
方法时,要绘制 RGB 数据,您必须定义一个 colorTuple 参数,它将逐点映射 RGB 数据。由于数组大小约为 2000x1000,因此这需要一段时间才能完成。下面是我正在谈论的片段(完整的工作代码在下方):
if one_channel:
m.pcolormesh(lons, lats, img[:,:,0], latlon=True)
else:
# This is the part that is slow, but I don't know how to
# accurately plot the data otherwise.
mesh_rgb = img[:, :-1, :]
colorTuple = mesh_rgb.reshape((mesh_rgb.shape[0] * mesh_rgb.shape[1]), 3)
# What you put in for the image doesn't matter because of the color mapping
m.pcolormesh(lons, lats, img[:,:,0], latlon=True,color=colorTuple)
当只绘制一个通道时,它可以在大约10秒左右的时间内制作出地图。绘制 RGB 数据时,可能需要 3-4 分钟。鉴于只有 3 倍多的数据,我觉得一定有更好的方法,特别是当你制作矩形图像时,绘制 RGB 数据可以和单通道数据一样快。
所以,我的问题是:有没有什么方法可以加快计算速度,可以使用其他绘图模块(例如 Bokeh)或以任何方式更改颜色映射?我已经尝试使用 imshow
和精心选择的地图边界,但由于它只是将图像拉伸到地图的整个范围,这对于准确绘制数据来说还不够好。
下面是我的代码的精简版本,适用于具有正确模块的示例:
from pyhdf.SD import SD,SDC
import numpy as np
from scipy.interpolate import interp1d
import matplotlib.pyplot as plt
from mpl_toolkits.basemap import Basemap
def get_hdf_attr(infile,dataset,attr):
f = SD(infile,SDC.READ)
data = f.select(dataset)
index = data.attr(attr).index()
attr_out = data.attr(index).get()
f.end()
return attr_out
def get_hdf_dataset(infile,dataset):
f = SD(infile,SDC.READ)
data = f.select(dataset)[:]
f.end()
return data
class make_rgb:
def __init__(self,file_name):
sds_250 = get_hdf_dataset(file_name, 'EV_250_Aggr1km_RefSB')
scales_250 = get_hdf_attr(file_name, 'EV_250_Aggr1km_RefSB', 'reflectance_scales')
offsets_250 = get_hdf_attr(file_name, 'EV_250_Aggr1km_RefSB', 'reflectance_offsets')
sds_500 = get_hdf_dataset(file_name, 'EV_500_Aggr1km_RefSB')
scales_500 = get_hdf_attr(file_name, 'EV_500_Aggr1km_RefSB', 'reflectance_scales')
offsets_500 = get_hdf_attr(file_name, 'EV_500_Aggr1km_RefSB', 'reflectance_offsets')
data_shape = sds_250.shape
along_track = data_shape[1]
cross_track = data_shape[2]
rgb = np.zeros((along_track, cross_track, 3))
rgb[:, :, 0] = (sds_250[0, :, :] - offsets_250[0]) * scales_250[0]
rgb[:, :, 1] = (sds_500[1, :, :] - offsets_500[1]) * scales_500[1]
rgb[:, :, 2] = (sds_500[0, :, :] - offsets_500[0]) * scales_500[0]
rgb[rgb > 1] = 1.0
rgb[rgb < 0] = 0.0
lin = np.array([0, 30, 60, 120, 190, 255]) / 255.0
nonlin = np.array([0, 110, 160, 210, 240, 255]) / 255.0
scale = interp1d(lin, nonlin, kind='quadratic')
self.img = scale(rgb)
def plot_image(self):
fig = plt.figure(figsize=(10, 10))
ax = fig.add_subplot(111)
ax.set_yticks([])
ax.set_xticks([])
plt.imshow(self.img, interpolation='nearest')
plt.show()
def plot_geo(self,geo_file,one_channel=False):
fig = plt.figure(figsize=(10, 10))
ax = fig.add_subplot(111)
lats = get_hdf_dataset(geo_file, 0)
lons = get_hdf_dataset(geo_file, 1)
lat_0 = np.mean(lats)
lat_range = [np.min(lats), np.max(lats)]
lon_0 = np.mean(lons)
lon_range = [np.min(lons), np.max(lons)]
map_kwargs = dict(projection='cass', resolution='l',
llcrnrlat=lat_range[0], urcrnrlat=lat_range[1],
llcrnrlon=lon_range[0], urcrnrlon=lon_range[1],
lat_0=lat_0, lon_0=lon_0)
m = Basemap(**map_kwargs)
if one_channel:
m.pcolormesh(lons, lats, self.img[:,:,0], latlon=True)
else:
# This is the part that is slow, but I don't know how to
# accurately plot the data otherwise.
mesh_rgb = self.img[:, :-1, :]
colorTuple = mesh_rgb.reshape((mesh_rgb.shape[0] * mesh_rgb.shape[1]), 3)
m.pcolormesh(lons, lats, self.img[:,:,0], latlon=True,color=colorTuple)
m.drawcoastlines()
m.drawcountries()
plt.show()
if __name__ == '__main__':
# https://ladsweb.nascom.nasa.gov/archive/allData/6/MOD021KM/2015/183/
data_file = 'MOD021KM.A2015183.1005.006.2015183195350.hdf'
# https://ladsweb.nascom.nasa.gov/archive/allData/6/MOD03/2015/183/
geo_file = 'MOD03.A2015183.1005.006.2015183192656.hdf'
# Very Fast
make_rgb(data_file).plot_image()
# Also Fast, takes about 10 seconds
make_rgb(data_file).plot_geo(geo_file,one_channel=True)
# Much slower, takes several minutes
make_rgb(data_file).plot_geo(geo_file)
我通过将 1.0 添加到 colorTuple 的每个部分的值以将其转换为 RGBA 数组来解决这个问题。我检查了 pcolormesh
函数,发现它调用颜色转换器将 RGB 转换为 RGBA 数组 4 次不同的时间,每次大约需要 50 秒。如果你给它一个 RGBA 数组作为开始,它会绕过它并在合理的时间范围内生成绘图。添加的附加代码行如下所示:
if one_channel:
m.pcolormesh(lons, lats, img[:,:,0], latlon=True)
else:
mesh_rgb = img[:, :-1, :]
colorTuple = mesh_rgb.reshape((mesh_rgb.shape[0] * mesh_rgb.shape[1]), 3)
# ADDED THIS LINE
colorTuple = np.insert(colorTuple,3,1.0,axis=1)
# What you put in for the image doesn't matter because of the color mapping
m.pcolormesh(lons, lats, img[:,:,0], latlon=True,color=colorTuple)
我在使用 Python 的带有纬度和经度数据的底图模块绘制 RGB 图像时遇到问题。现在,我可以绘制我想要的图,但问题是它有多慢,因为它能够比 RGB 数据更快地绘制单通道数据,而且一般来说,自己绘制 RGB 图像也是如此快速地。由于我有 lat/lon 数据,这就是事情变得复杂的地方。我已经查看了这个问题的解决方案:
How to plot an irregular spaced RGB image using python and basemap?
这就是我到达现在位置的方式。它本质上归结为以下问题。在底图中使用 pcolormesh
方法时,要绘制 RGB 数据,您必须定义一个 colorTuple 参数,它将逐点映射 RGB 数据。由于数组大小约为 2000x1000,因此这需要一段时间才能完成。下面是我正在谈论的片段(完整的工作代码在下方):
if one_channel:
m.pcolormesh(lons, lats, img[:,:,0], latlon=True)
else:
# This is the part that is slow, but I don't know how to
# accurately plot the data otherwise.
mesh_rgb = img[:, :-1, :]
colorTuple = mesh_rgb.reshape((mesh_rgb.shape[0] * mesh_rgb.shape[1]), 3)
# What you put in for the image doesn't matter because of the color mapping
m.pcolormesh(lons, lats, img[:,:,0], latlon=True,color=colorTuple)
当只绘制一个通道时,它可以在大约10秒左右的时间内制作出地图。绘制 RGB 数据时,可能需要 3-4 分钟。鉴于只有 3 倍多的数据,我觉得一定有更好的方法,特别是当你制作矩形图像时,绘制 RGB 数据可以和单通道数据一样快。
所以,我的问题是:有没有什么方法可以加快计算速度,可以使用其他绘图模块(例如 Bokeh)或以任何方式更改颜色映射?我已经尝试使用 imshow
和精心选择的地图边界,但由于它只是将图像拉伸到地图的整个范围,这对于准确绘制数据来说还不够好。
下面是我的代码的精简版本,适用于具有正确模块的示例:
from pyhdf.SD import SD,SDC
import numpy as np
from scipy.interpolate import interp1d
import matplotlib.pyplot as plt
from mpl_toolkits.basemap import Basemap
def get_hdf_attr(infile,dataset,attr):
f = SD(infile,SDC.READ)
data = f.select(dataset)
index = data.attr(attr).index()
attr_out = data.attr(index).get()
f.end()
return attr_out
def get_hdf_dataset(infile,dataset):
f = SD(infile,SDC.READ)
data = f.select(dataset)[:]
f.end()
return data
class make_rgb:
def __init__(self,file_name):
sds_250 = get_hdf_dataset(file_name, 'EV_250_Aggr1km_RefSB')
scales_250 = get_hdf_attr(file_name, 'EV_250_Aggr1km_RefSB', 'reflectance_scales')
offsets_250 = get_hdf_attr(file_name, 'EV_250_Aggr1km_RefSB', 'reflectance_offsets')
sds_500 = get_hdf_dataset(file_name, 'EV_500_Aggr1km_RefSB')
scales_500 = get_hdf_attr(file_name, 'EV_500_Aggr1km_RefSB', 'reflectance_scales')
offsets_500 = get_hdf_attr(file_name, 'EV_500_Aggr1km_RefSB', 'reflectance_offsets')
data_shape = sds_250.shape
along_track = data_shape[1]
cross_track = data_shape[2]
rgb = np.zeros((along_track, cross_track, 3))
rgb[:, :, 0] = (sds_250[0, :, :] - offsets_250[0]) * scales_250[0]
rgb[:, :, 1] = (sds_500[1, :, :] - offsets_500[1]) * scales_500[1]
rgb[:, :, 2] = (sds_500[0, :, :] - offsets_500[0]) * scales_500[0]
rgb[rgb > 1] = 1.0
rgb[rgb < 0] = 0.0
lin = np.array([0, 30, 60, 120, 190, 255]) / 255.0
nonlin = np.array([0, 110, 160, 210, 240, 255]) / 255.0
scale = interp1d(lin, nonlin, kind='quadratic')
self.img = scale(rgb)
def plot_image(self):
fig = plt.figure(figsize=(10, 10))
ax = fig.add_subplot(111)
ax.set_yticks([])
ax.set_xticks([])
plt.imshow(self.img, interpolation='nearest')
plt.show()
def plot_geo(self,geo_file,one_channel=False):
fig = plt.figure(figsize=(10, 10))
ax = fig.add_subplot(111)
lats = get_hdf_dataset(geo_file, 0)
lons = get_hdf_dataset(geo_file, 1)
lat_0 = np.mean(lats)
lat_range = [np.min(lats), np.max(lats)]
lon_0 = np.mean(lons)
lon_range = [np.min(lons), np.max(lons)]
map_kwargs = dict(projection='cass', resolution='l',
llcrnrlat=lat_range[0], urcrnrlat=lat_range[1],
llcrnrlon=lon_range[0], urcrnrlon=lon_range[1],
lat_0=lat_0, lon_0=lon_0)
m = Basemap(**map_kwargs)
if one_channel:
m.pcolormesh(lons, lats, self.img[:,:,0], latlon=True)
else:
# This is the part that is slow, but I don't know how to
# accurately plot the data otherwise.
mesh_rgb = self.img[:, :-1, :]
colorTuple = mesh_rgb.reshape((mesh_rgb.shape[0] * mesh_rgb.shape[1]), 3)
m.pcolormesh(lons, lats, self.img[:,:,0], latlon=True,color=colorTuple)
m.drawcoastlines()
m.drawcountries()
plt.show()
if __name__ == '__main__':
# https://ladsweb.nascom.nasa.gov/archive/allData/6/MOD021KM/2015/183/
data_file = 'MOD021KM.A2015183.1005.006.2015183195350.hdf'
# https://ladsweb.nascom.nasa.gov/archive/allData/6/MOD03/2015/183/
geo_file = 'MOD03.A2015183.1005.006.2015183192656.hdf'
# Very Fast
make_rgb(data_file).plot_image()
# Also Fast, takes about 10 seconds
make_rgb(data_file).plot_geo(geo_file,one_channel=True)
# Much slower, takes several minutes
make_rgb(data_file).plot_geo(geo_file)
我通过将 1.0 添加到 colorTuple 的每个部分的值以将其转换为 RGBA 数组来解决这个问题。我检查了 pcolormesh
函数,发现它调用颜色转换器将 RGB 转换为 RGBA 数组 4 次不同的时间,每次大约需要 50 秒。如果你给它一个 RGBA 数组作为开始,它会绕过它并在合理的时间范围内生成绘图。添加的附加代码行如下所示:
if one_channel:
m.pcolormesh(lons, lats, img[:,:,0], latlon=True)
else:
mesh_rgb = img[:, :-1, :]
colorTuple = mesh_rgb.reshape((mesh_rgb.shape[0] * mesh_rgb.shape[1]), 3)
# ADDED THIS LINE
colorTuple = np.insert(colorTuple,3,1.0,axis=1)
# What you put in for the image doesn't matter because of the color mapping
m.pcolormesh(lons, lats, img[:,:,0], latlon=True,color=colorTuple)