OSMxn-如何使用边界框和figsize绘制到特定位置和分辨率

问题描述

我有使用OSMNX(很棒的工具,谢谢GB!)的代码,它试图将OSMNX输出转换为具有特定分辨率和边界框的栅格文件(png图像),因为它需要与现有栅格对齐。我正在使用ox.projection.project_graph()转换为必要的crs(UTM 33N)和ox.plot_graph()来尝试使用所需的边界框和figsize进行绘制,以获得所需的图像分辨率->栅格像元大小。我必须丢失一些东西,因为我始终遇到错误(ValueError:277704x419976像素的图像大小太大。在每个方向上它必须小于2 ^ 16),即使我指定了较小的figsize(3857 x 5833)。边界框以米为单位,projected_to crs也是如此。

任何帮助将不胜感激。

这是这段代码

...
# Fairly standard up to this point.  G is a valid MultiDiGraph and the other variables all have appropriate values.

# This works great: 
fig,ax = ox.plot_graph(G,node_color=nc,node_size=ns,node_zorder=2,edge_color=ec,edge_linewidth=ew)

# So does this:
pG = ox.projection.project_graph(G,to_crs={'init':'epsg:32633'})
fig,ax = ox.plot_graph(pG,edge_linewidth=ew)

# But this gives an error: (ValueError: Image size of 277704x419976 pixels is too large. It must be less than 2^16 in each direction) 
# even though figsize is only 3857x5833.
fig,bBox=(6686683.721299999,6511693.721299999,743864.5602,628154.5602),figsize=(3857,5833),dpi=30,edge_linewidth=ew)

解决方法

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figsize
,

感谢GBoeing的回答以及他在OSMnx方面的出色工作。在尝试提出一种用于创建具有指定范围和像元大小(分辨率)的栅格的标准化过程之后,我进行了整整圆周,并意识到,真正需要做的就是通过除以100来消除dpi。 ,请使用figsize =(58.33,38.57)获得5833x3857图像。这就是我所做的:

...
# (pG is a valid projected MultiDiGraph and the other variables all have appropriate values)
# To create an image that's 5833x3857 using the 100dpi default,# just divide the desired dimensions by 100 and use that for figsize.

fig,ax = ox.plot_graph(pG,figsize=(58.33,38.57),bbox=(6686683.721299999,6511693.721299999,743864.5602,628154.5602),dpi=100,node_color=nc,node_size=ns,node_zorder=2,edge_color=ec,edge_linewidth=ew)

fig.savefig("C:/output_path/output_file.png",pad_inches=0.0)

# This yields an image with the desired bounds,dimensions,and therefore cell size (resolution).