使用Python捕获嵌入式谷歌地图图像，无需使用浏览器

Question

我注意到，从 Google 地图页面，您可以获得一个“嵌入”链接，将其放入 iframe 中并在浏览器中加载地图。 （这里没有新闻）

图像尺寸可以调整到非常大，所以我有兴趣将一些大图像作为单个.PNG。

更具体地说，我想从边界框（右上和左下坐标）定义一个矩形区域，并使用适当的缩放系数获取相应的图像。

但我的问题是：如何使用Python来获取该地图的“像素内容”作为图像对象？

（我的理由是：如果浏览器可以获取并渲染这样的图像内容，那么Python也应该能够做到）。

编辑：这是显示我的示例地图的 HTML 文件的内容：

<iframe 
    width="2000"
    height="1500"
    frameborder="0"
    scrolling="yes"
    marginheight="0"
    marginwidth="0"
    src="http://maps.google.com.br/maps?hl=pt-BR&amp;ll=-30.027489,-51.229248&amp;spn=1.783415,2.745209&amp;z=10&amp;output=embed"/>

编辑：我按照 Ned Batchelder 的建议进行操作，并使用 读取 urllib.urlopen() 调用的内容src 地址取自上面的 iframe。结果是大量的 JavaScript 代码，我认为这与 Google Maps JavaScript API 有关。所以，问题仍然存在：我怎样才能从Python中的所有这些东西中做一些有用的事情来获取地图图像？

编辑：此链接似乎包含一些有关 Google 地图如何平铺地图的非常相关的信息： http://www.codeproject.com/KB/scrapbook/googlemap.aspx

还： http://econym.org.uk/gmap/howitworks.htm

Answer 1

我感谢所有的答案。我最终以另一种方式解决了这个问题，使用 Google Maps Static API 和一些公式将坐标空间转换为像素空间，这样我就可以获得很好地“缝合”在一起的精确图像。

对于任何感兴趣的人，这里是代码。如果对某人有帮助，请评论！

=============================

import Image, urllib, StringIO
from math import log, exp, tan, atan, pi, ceil

EARTH_RADIUS = 6378137
EQUATOR_CIRCUMFERENCE = 2 * pi * EARTH_RADIUS
INITIAL_RESOLUTION = EQUATOR_CIRCUMFERENCE / 256.0
ORIGIN_SHIFT = EQUATOR_CIRCUMFERENCE / 2.0

def latlontopixels(lat, lon, zoom):
    mx = (lon * ORIGIN_SHIFT) / 180.0
    my = log(tan((90 + lat) * pi/360.0))/(pi/180.0)
    my = (my * ORIGIN_SHIFT) /180.0
    res = INITIAL_RESOLUTION / (2**zoom)
    px = (mx + ORIGIN_SHIFT) / res
    py = (my + ORIGIN_SHIFT) / res
    return px, py

def pixelstolatlon(px, py, zoom):
    res = INITIAL_RESOLUTION / (2**zoom)
    mx = px * res - ORIGIN_SHIFT
    my = py * res - ORIGIN_SHIFT
    lat = (my / ORIGIN_SHIFT) * 180.0
    lat = 180 / pi * (2*atan(exp(lat*pi/180.0)) - pi/2.0)
    lon = (mx / ORIGIN_SHIFT) * 180.0
    return lat, lon

############################################

# a neighbourhood in Lajeado, Brazil:

upperleft =  '-29.44,-52.0'  
lowerright = '-29.45,-51.98'

zoom = 18   # be careful not to get too many images!

############################################

ullat, ullon = map(float, upperleft.split(','))
lrlat, lrlon = map(float, lowerright.split(','))

# Set some important parameters
scale = 1
maxsize = 640

# convert all these coordinates to pixels
ulx, uly = latlontopixels(ullat, ullon, zoom)
lrx, lry = latlontopixels(lrlat, lrlon, zoom)

# calculate total pixel dimensions of final image
dx, dy = lrx - ulx, uly - lry

# calculate rows and columns
cols, rows = int(ceil(dx/maxsize)), int(ceil(dy/maxsize))

# calculate pixel dimensions of each small image
bottom = 120
largura = int(ceil(dx/cols))
altura = int(ceil(dy/rows))
alturaplus = altura + bottom


final = Image.new("RGB", (int(dx), int(dy)))
for x in range(cols):
    for y in range(rows):
        dxn = largura * (0.5 + x)
        dyn = altura * (0.5 + y)
        latn, lonn = pixelstolatlon(ulx + dxn, uly - dyn - bottom/2, zoom)
        position = ','.join((str(latn), str(lonn)))
        print x, y, position
        urlparams = urllib.urlencode({'center': position,
                                      'zoom': str(zoom),
                                      'size': '%dx%d' % (largura, alturaplus),
                                      'maptype': 'satellite',
                                      'sensor': 'false',
                                      'scale': scale})
        url = 'http://maps.google.com/maps/api/staticmap?' + urlparams
        f=urllib.urlopen(url)
        im=Image.open(StringIO.StringIO(f.read()))
        final.paste(im, (int(x*largura), int(y*altura)))
final.show()

Answer 2

您应该直接访问 Google API 以获取静态图形形式的图像，而不是尝试使用嵌入链接。这是 Google 地图静态图像 API 的链接 - 您似乎可以只需在 URL 中传入 long/lat 参数，就像处理普通的可嵌入参数一样。例如：

http://maps.googleapis.com/maps/api/staticmap?center=-30.027489,-51.229248&size=600x600&zoom=14&sensor=false

为您提供以上面给出的坐标为中心的 600x600 街道级别概览，该坐标似乎是巴西的阿雷格里港。现在您可以按照 Ned 的建议使用 urlopen 和 PIL：

from cStringIO import StringIO
import Image
import urllib

url = "http://maps.googleapis.com/maps/api/staticmap?center=-30.027489,-51.229248&size=800x800&zoom=14&sensor=false"
buffer = StringIO(urllib.urlopen(url).read())
image = Image.open(buffer)

Answer 3

编辑：
此答案中的代码已得到改进和简化，此处：https://stackoverflow.com/a/50536888/5859283

---

基于 heltonbiker 的出色回答以及 BenElgar 的更改，下面是 Python 3 的一些更新代码并添加了 API 密钥访问，希望它对某人有用：

"""
Stitch together Google Maps images from lat, long coordinates
Based on work by heltonbiker and BenElgar
Changes: 
  * updated for Python 3
  * added Google Cloud Static Maps API key field (now required for access)
  * handle http request exceptions
"""

import requests
from io import BytesIO
from math import log, exp, tan, atan, pi, ceil
from PIL import Image
import sys

EARTH_RADIUS = 6378137
EQUATOR_CIRCUMFERENCE = 2 * pi * EARTH_RADIUS
INITIAL_RESOLUTION = EQUATOR_CIRCUMFERENCE / 256.0
ORIGIN_SHIFT = EQUATOR_CIRCUMFERENCE / 2.0
GOOGLE_MAPS_API_KEY = 'change this to your API key'

def latlontopixels(lat, lon, zoom):
    mx = (lon * ORIGIN_SHIFT) / 180.0
    my = log(tan((90 + lat) * pi/360.0))/(pi/180.0)
    my = (my * ORIGIN_SHIFT) /180.0
    res = INITIAL_RESOLUTION / (2**zoom)
    px = (mx + ORIGIN_SHIFT) / res
    py = (my + ORIGIN_SHIFT) / res
    return px, py

def pixelstolatlon(px, py, zoom):
    res = INITIAL_RESOLUTION / (2**zoom)
    mx = px * res - ORIGIN_SHIFT
    my = py * res - ORIGIN_SHIFT
    lat = (my / ORIGIN_SHIFT) * 180.0
    lat = 180 / pi * (2*atan(exp(lat*pi/180.0)) - pi/2.0)
    lon = (mx / ORIGIN_SHIFT) * 180.0
    return lat, lon


def get_maps_image(NW_lat_long, SE_lat_long, zoom=18):
  
  ullat, ullon = NW_lat_long
  lrlat, lrlon = SE_lat_long
  
  # Set some important parameters
  scale = 1
  maxsize = 640
  
  # convert all these coordinates to pixels
  ulx, uly = latlontopixels(ullat, ullon, zoom)
  lrx, lry = latlontopixels(lrlat, lrlon, zoom)
  
  # calculate total pixel dimensions of final image
  dx, dy = lrx - ulx, uly - lry
  
  # calculate rows and columns
  cols, rows = int(ceil(dx/maxsize)), int(ceil(dy/maxsize))
  
  # calculate pixel dimensions of each small image
  bottom = 120
  largura = int(ceil(dx/cols))
  altura = int(ceil(dy/rows))
  alturaplus = altura + bottom
  
  # assemble the image from stitched
  final = Image.new("RGB", (int(dx), int(dy)))
  for x in range(cols):
      for y in range(rows):
          dxn = largura * (0.5 + x)
          dyn = altura * (0.5 + y)
          latn, lonn = pixelstolatlon(ulx + dxn, uly - dyn - bottom/2, zoom)
          position = ','.join((str(latn), str(lonn)))
          print(x, y, position)
          urlparams = {'center': position,
                        'zoom': str(zoom),
                        'size': '%dx%d' % (largura, alturaplus),
                        'maptype': 'satellite',
                        'sensor': 'false',
                        'scale': scale}
          if GOOGLE_MAPS_API_KEY is not None:
            urlparams['key'] = GOOGLE_MAPS_API_KEY
            
          url = 'http://maps.google.com/maps/api/staticmap'
          try:                  
            response = requests.get(url, params=urlparams)
            response.raise_for_status()
          except requests.exceptions.RequestException as e:
            print(e)
            sys.exit(1)
            
          im = Image.open(BytesIO(response.content))                  
          final.paste(im, (int(x*largura), int(y*altura)))
          
  return final

############################################

if __name__ == '__main__':
  
  # a neighbourhood in Lajeado, Brazil:
  NW_lat_long =  (-29.44,-52.0)
  SE_lat_long = (-29.45,-51.98)
  
  zoom = 18   # be careful not to get too many images!
  
  result = get_maps_image(NW_lat_long, SE_lat_long, zoom=18)
  result.show()

Answer 4

@4Oh4 的答案是正确的，但数学比他们需要的更复杂。度数和弧度之间的转换发生得比需要的要频繁。无缘无故地调用地球半径——它在所有计算中都被抵消。无缘无故地向像素坐标添加了偏移量。标志的截面积比需要的要大。还有其他一些零碎的东西，这些都已写在更改中。这是我的版本：

#!/usr/bin/env python
"""
Stitch together Google Maps images from lat, long coordinates
Based on work by heltonbiker and BenElgar
Changes: 
* updated for Python 3
* added Google Maps API key (compliance with T&C, although can set to None)
* handle http request exceptions

With contributions from Eric Toombs.
Changes:
* Dramatically simplified the maths.
* Set a more reasonable default logo cutoff.
* Added global constants for logo cutoff and max image size.
* Translated a couple presumably Portuguese variable names to English.
"""

import requests
from io import BytesIO
from math import log, exp, tan, atan, ceil
from PIL import Image
import sys

# circumference/radius
tau = 6.283185307179586
# One degree in radians, i.e. in the units the machine uses to store angle,
# which is always radians. For converting to and from degrees. See code for
# usage demonstration.
DEGREE = tau/360

ZOOM_OFFSET = 8
GOOGLE_MAPS_API_KEY = None  # set to 'your_API_key'

# Max width or height of a single image grabbed from Google.
MAXSIZE = 640
# For cutting off the logos at the bottom of each of the grabbed images.  The
# logo height in pixels is assumed to be less than this amount.
LOGO_CUTOFF = 32


def latlon2pixels(lat, lon, zoom):
    mx = lon
    my = log(tan((lat + tau/4)/2))
    res = 2**(zoom + ZOOM_OFFSET) / tau
    px = mx*res
    py = my*res
    return px, py

def pixels2latlon(px, py, zoom):
    res = 2**(zoom + ZOOM_OFFSET) / tau
    mx = px/res
    my = py/res
    lon = mx
    lat = 2*atan(exp(my)) - tau/4
    return lat, lon


def get_maps_image(NW_lat_long, SE_lat_long, zoom=18):

    ullat, ullon = NW_lat_long
    lrlat, lrlon = SE_lat_long

    # convert all these coordinates to pixels
    ulx, uly = latlon2pixels(ullat, ullon, zoom)
    lrx, lry = latlon2pixels(lrlat, lrlon, zoom)

    # calculate total pixel dimensions of final image
    dx, dy = lrx - ulx, uly - lry

    # calculate rows and columns
    cols, rows = ceil(dx/MAXSIZE), ceil(dy/MAXSIZE)

    # calculate pixel dimensions of each small image
    width = ceil(dx/cols)
    height = ceil(dy/rows)
    heightplus = height + LOGO_CUTOFF

    # assemble the image from stitched
    final = Image.new('RGB', (int(dx), int(dy)))
    for x in range(cols):
        for y in range(rows):
            dxn = width * (0.5 + x)
            dyn = height * (0.5 + y)
            latn, lonn = pixels2latlon(
                    ulx + dxn, uly - dyn - LOGO_CUTOFF/2, zoom)
            position = ','.join((str(latn/DEGREE), str(lonn/DEGREE)))
            print(x, y, position)
            urlparams = {
                    'center': position,
                    'zoom': str(zoom),
                    'size': '%dx%d' % (width, heightplus),
                    'maptype': 'satellite',
                    'sensor': 'false',
                    'scale': 1
                }
            if GOOGLE_MAPS_API_KEY is not None:
                urlparams['key'] = GOOGLE_MAPS_API_KEY

            url = 'http://maps.google.com/maps/api/staticmap'
            try:                  
                response = requests.get(url, params=urlparams)
                response.raise_for_status()
            except requests.exceptions.RequestException as e:
                print(e)
                sys.exit(1)

            im = Image.open(BytesIO(response.content))                  
            final.paste(im, (int(x*width), int(y*height)))

    return final

############################################

if __name__ == '__main__':
    # a neighbourhood in Lajeado, Brazil:
    NW_lat_long =  (-29.44*DEGREE, -52.0*DEGREE)
    SE_lat_long = (-29.45*DEGREE, -51.98*DEGREE)

    result = get_maps_image(
        NW_lat_long,
        SE_lat_long,
        zoom=18 # be careful not to get too many images!
    )
    result.show()

Answer 5

这是 Daniel Roseman 为使用 python 3.x 的人提供的答案：

• 假设您已经 Google 地图静态图像 API

Python 3.x 代码：

from io import BytesIO
from PIL import Image
from urllib import request
import matplotlib.pyplot as plt # this is if you want to plot the map using pyplot

url = "http://maps.googleapis.com/maps/api/staticmap?center=-30.027489,-51.229248&size=800x800&zoom=14&sensor=false"

buffer = BytesIO(request.urlopen(url).read())
image = Image.open(buffer)

# Show Using PIL
image.show()

# Or using pyplot
plt.imshow(image)
plt.show()

Answer 6

一个更简洁的Python 2.x兼容方法是

from io import BytesIO
import Image
import urllib

url = "http://maps.googleapis.com/maps/api/staticmap?center=52.50058,13.31316&size=800x800&zoom=14"
buffer = BytesIO(urllib.urlopen(url).read())
image = Image.open(buffer)
image.save("map.png")

Answer 7

捕获/保存 Google 静态地图图像（作为 png）的最简单方法：

import requests

img = open('tmp.png','wb')
img.write(requests.get('https://maps.googleapis.com/maps/api/staticmap?center=33.0456,131.3009&zoom=12&size=320x385&key=YOUR_API_KEY').content)
img.close()

Answer 8

urllib.urlopen 将打开一个 URL，结果将有一个 .read() 方法，您可以使用它来获取图像字节。 cStringIO 有一个基于内存中字符串的类文件对象。 PIL 有一个 Image.open 函数，可以打开一个类似文件的东西来给你一个图像对象。可以询问图像对象的像素值。