7

I am generating a Meshgrid with Numpy and it's taking a lot of memory and quite a bit of time as well. 

xi, yi = np.meshgrid(xi, yi)

I am generating a meshgrid the same resolution as the underlying sitemap image, sometimes 3000px dimensions. It uses several gigs of memory sometimes and takes 10-15 seconds or more while when it's writing it to the page file.

My question is; can I speed this up without upgrading the server? Here is a full copy of my application source code. 

def generateContours(date_collected, substance_name, well_arr, site_id, sitemap_id, image, title_wildcard='', label_over_well=False, crop_contours=False, groundwater_contours=False, flow_lines=False, site_image_alpha=1, status_token=""):
    #create empty arrays to fill up!
    x_values = []
    y_values = []
    z_values = []

    #iterate over wells and fill the arrays with well data
    for well in well_arr:
        x_values.append(well['xpos'])
        y_values.append(well['ypos'])
        z_values.append(well['value'])

    #initialize numpy array as required for interpolation functions
    x = np.array(x_values, dtype=np.float)
    y = np.array(y_values, dtype=np.float)
    z = np.array(z_values, dtype=np.float)

    #create a list of x, y coordinate tuples
    points = zip(x, y)

    #create a grid on which to interpolate data
    start_time = time.time()
    xi, yi = np.linspace(0, image['width'], image['width']), np.linspace(0, image['height'], image['height'])

    xi, yi = np.meshgrid(xi, yi)

    #interpolate the data with the matlab griddata function (http://matplotlib.org/api/mlab_api.html#matplotlib.mlab.griddata)
    zi = griddata(x, y, z, xi, yi, interp='nn')

    #create a matplotlib figure and adjust the width and heights to output contours to a resolution very close to the original sitemap
    fig = plt.figure(figsize=(image['width']/72, image['height']/72))

    #create a single subplot, just takes over the whole figure if only one is specified
    ax = fig.add_subplot(111, frameon=False, xticks=[], yticks=[])

    #read the database image and save to a temporary variable
    im = Image.open(image['tmpfile'])

    #place the sitemap image on top of the figure
    ax.imshow(im, origin='upper', alpha=site_image_alpha)

    #figure out a good linewidth
    if image['width'] > 2000:
        linewidth = 3
    else:
        linewidth = 2

    #create the contours (options here http://cl.ly/2X0c311V2y01)
    kwargs = {}
    if groundwater_contours:
        kwargs['colors'] = 'b'

    CS = plt.contour(xi, yi, zi, linewidths=linewidth, **kwargs)
    for key, value in enumerate(CS.levels):
        if value == 0:
            CS.collections[key].remove()

    #add a streamplot
    if flow_lines:
        dy, dx = np.gradient(zi)
        plt.streamplot(xi, yi, dx, dy, color='c', density=1, arrowsize=3, arrowstyle='<-')

    #add labels to well locations
    label_kwargs = {}
    if label_over_well is True:
        label_kwargs['manual'] = points

    plt.clabel(CS, CS.levels[1::1], inline=5, fontsize=math.floor(image['width']/100), fmt="%.1f", **label_kwargs)

    #add scatterplot to show where well data was read
    scatter_size = math.floor(image['width']/20)
    plt.scatter(x, y, s=scatter_size, c='k', facecolors='none', marker=(5, 1))

    try:
        site_name = db_session.query(Sites).filter_by(site_id=site_id).first().title
    except:
        site_name = "Site Map #%i" % site_id

    sitemap = SiteMaps.query.get(sitemap_id)
    if sitemap.title != 'Sitemap':
        sitemap_wildcard = " - " + sitemap.title
    else:
        sitemap_wildcard = ""

    if title_wildcard != '':
        filename_wildcard = "-" + slugify(title_wildcard)
        title_wildcard = " - " + title_wildcard
    else:
        filename_wildcard = ""
        title_wildcard = ""

    #add descriptive title to the top of the contours
    title_font_size = math.floor(image['width']/72)
    plt.title(parseDate(date_collected) + " - " + site_name + " " + substance_name + " Contour" + sitemap_wildcard + title_wildcard, fontsize=title_font_size)

    #generate a unique filename and save to a temp directory
    filename = slugify(site_name) + str(int(time.time())) + filename_wildcard + ".pdf"
    temp_dir = tempfile.gettempdir()
    tempFileObj = temp_dir + "/" + filename
    savefig(tempFileObj)  # bbox_inches='tight' tightens the white border

    #clears the matplotlib memory
    clf()

    #send the temporary file to the user
    resp = make_response(send_file(tempFileObj, mimetype='application/pdf', as_attachment=True, attachment_filename=filename))

    #set the users status token for javascript workaround to check if file is done being generated
    resp.set_cookie('status_token', status_token)

    return resp
4

3 回答 3

7

如果meshgrid是什么让你慢下来，不要称之为......根据griddata文档：

xi 和 yi 必须描述一个规则网格，可以是 1D 或 2D，但必须是单调递增的。

因此griddata，如果您跳过呼叫并执行以下操作，您的呼叫应该同样有效meshgrid：

xi = np.linspace(0, image['width'], image['width'])
yi = np.linspace(0, image['height'], image['height'])
zi = griddata(x, y, z, xi, yi, interp='nn')

这就是说，如果你的x和y向量很大，实际的插值，即调用griddata可能需要相当长的时间，因为德劳内三角剖分是一项计算密集型操作。你确定你的性能问题来自于meshgrid，而不是来自于griddata吗？

于 2013-08-13T14:34:52.230 回答
2

怎么样xi, yi = np.meshgrid(xi, yi, copy=False)。这样，它只返回原始数组的视图，而不是复制所有数据。

于 2013-08-13T04:16:58.187 回答
1

看起来你可能不需要xi经过yi。meshgrid检查您使用的函数的文档字符串xi和yi. 许多人接受（甚至期望）一维数组。

例如：

In [33]: x
Out[33]: array([0, 0, 0, 1, 1, 1, 2, 2, 2])

In [34]: y
Out[34]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])

In [35]: z
Out[35]: array([0, 1, 4, 1, 2, 5, 2, 3, 6])

In [36]: xi
Out[36]: array([ 0. ,  0.5,  1. ,  1.5,  2. ])

In [37]: yi
Out[37]: 
array([ 0.        ,  0.33333333,  0.66666667,  1.        ,  1.33333333,
        1.66666667,  2.        ])

In [38]: zi = griddata(x, y, z, xi, yi)

In [39]: zi
Out[39]: 
array([[ 0.        ,  0.5       ,  1.        ,  1.5       ,  2.        ],
       [ 0.33333333,  0.83333333,  1.33333333,  1.83333333,  2.33333333],
       [ 0.66666667,  1.16666667,  1.66666667,  2.16666667,  2.66666667],
       [ 1.        ,  1.61111111,  2.        ,  2.61111111,  3.        ],
       [ 2.        ,  2.5       ,  3.        ,  3.5       ,  4.        ],
       [ 3.        ,  3.5       ,  4.        ,  4.5       ,  5.        ],
       [ 4.        ,  4.5       ,  5.        ,  5.5       ,  6.        ]])


In [40]: plt.contour(xi, yi, zi)
Out[40]: <matplotlib.contour.QuadContourSet instance at 0x3ba03b0>
于 2013-08-13T14:34:09.053 回答