matplotlib.mlab.griddata非常慢，并且在输入有效数据时返回nan数组 [英] matplotlib.mlab.griddata very slow and returns array of nan when valid data is input

问题描述

我正在尝试将不规则网格的数据集(原始卫星数据)与相关的纬度和经度映射到由basemap.makegrid()给出的规则网格化的经度和纬度集.我在安装mpl_toolkits.natgrid的情况下使用matplotlib.mlab.griddata.以下是ipython中whos用作输出的变量的列表以及该变量的一些统计信息:

I am trying to map an irregularly gridded dataset (raw satellite data) with associated latitudes and longitudes to a regularly gridded set of latitudes and longitudes given by basemap.makegrid(). I am using matplotlib.mlab.griddata with mpl_toolkits.natgrid installed. Below is a list of the variables being used as output by whos in ipython and some stats on the variables:

Variable   Type       Data/Info
-------------------------------
datalat    ndarray    666x1081: 719946 elems, type `float32`, 2879784 bytes (2 Mb)
datalon    ndarray    666x1081: 719946 elems, type `float32`, 2879784 bytes (2 Mb)
gridlat    ndarray    1200x1000: 1200000 elems, type `float64`, 9600000 bytes (9 Mb)
gridlon    ndarray    1200x1000: 1200000 elems, type `float64`, 9600000 bytes (9 Mb)
var        ndarray    666x1081: 719946 elems, type `float32`, 2879784 bytes (2 Mb)

In [11]: var.min()
Out[11]: -30.0

In [12]: var.max()
Out[12]: 30.0

In [13]: datalat.min()
Out[13]: 27.339874

In [14]: datalat.max()
Out[14]: 47.05302

In [15]: datalon.min()
Out[15]: -137.55658

In [16]: datalon.max()
Out[16]: -108.41629

In [17]: gridlat.min()
Out[17]: 30.394031556984299

In [18]: gridlat.max()
Out[18]: 44.237140350357713

In [19]: gridlon.min()
Out[19]: -136.17646180595321

In [20]: gridlon.max()
Out[20]: -113.82353819404671

datalat和datalon是原始数据坐标

gridlat和gridlon是要插值到的坐标

var包含实际数据

使用这些变量，当我调用griddata(datalon, datalat, var, gridlon, gridlat)时，它花了长达20分钟的时间才能完成，并返回nan的数组.通过查看数据，经度和纬度似乎是正确的，原始坐标与新区域的一部分重叠，而一些数据点位于新区域之外.有没有人有什么建议? nan值表明我在做一些愚蠢的事情...

Using these variables, when I call griddata(datalon, datalat, var, gridlon, gridlat) it has taken as long as 20 minutes to complete and returns an array of nan. From looking at the data, the latitudes and longitudes appear to be correct with the original coordinates overlapping a portion of the new area and a few data points lying outside of the new area. Does anyone have any suggestions? The nan values suggest that I'm doing something stupid...

推荐答案

看来mlab.griddata例程可能会对输出数据引入其他不必要的约束.尽管输入位置可以是任何位置，但输出位置必须是常规网格-由于您的示例位于纬度/经度空间中，因此您选择的地图投影可能会违反此规定(即，x/y中的常规网格不是纬度/经度中的常规网格).

It looks like the mlab.griddata routine may introduce additional constraints on your output data that may not be necessary. While the input locations may be anything, the output locations must be a regular grid - since your example is in lat/lon space, your choice of map projection may violate this (i.e. regular grid in x/y is not a regular grid in lat/lon).

您可以从interpolate.griddata例程="nofollow"> SciPy 作为替代方案-但是，由于调用签名不同，因此您需要将位置变量合并为一个数组:类似

You might try the interpolate.griddata routine from SciPy as an alternative - you'll need to combine your location variables into a single array, though, since the call signature is different: something like

import scipy.interpolate
data_locations = np.vstack(datalon.ravel(), datalat.ravel()).T
grid_locations = np.vstack(gridlon.ravel(), gridlat.ravel()).T
grid_data      = scipy.interpolate.griddata(data_locations, val.ravel(),
                                            grid_locations, method='nearest')

用于最近邻插值.这会将位置放入2列对应于2个维度的数组中.您可能还想在地图投影的变换空间中执行插值.

for nearest-neighbor interpolation. This gets the locations into an array with 2 columns corresponding to your 2 dimensions. You may also want to perform the interpolation in the transformed space of your map projection.

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