在beauthulsoup和Python中使用SVG路径

Question

我正在编写一个Python脚本，它将在我市人口普查区块组的各个区域（其中有18个）根据各自家庭收入中值在SVG格式的地图上显示不同的颜色。在

听起来很简单，对吧？嗯，我不知道怎么做，虽然我有点进步。到目前为止，我所做的是根据SVG引用的方式列出每个块组路径，列出家庭收入中值，然后传递给它们着色的代码。然而，这只是。。不管是什么原因，似乎没用。你们中有谁能帮我弄清楚我在哪里失火吗？在

import csv
from bs4 import BeautifulSoup

icbg = []
reader = csv.reader(open('censusdata.csv'),delimiter=",")

#read and get income
for row in reader:
   income = row[6]
   income = int(income)
   icbg.append(income)

svg = open('NM2.svg','r')
soup = BeautifulSoup(svg,"lxml")

#find CBGs and incomes
path1 = soup.find('path')
path2 = path1.find_next('path')
path3 = path2.find_next('path')
path4 = path3.find_next('path')
path5 = path4.find_next('path')
path6 = path5.find_next('path')
path7 = path6.find_next('path')
path8 = path7.find_next('path')
path9 = path8.find_next('path')
path10 = path9.find_next('path')
path11 = path10.find_next('path')
path12 = path11.find_next('path')
path13 = path12.find_next('path')
path14 = path13.find_next('path')
path15 = path14.find_next('path')
path16 = path15.find_next('path')
path17 = path16.find_next('path')
path18 = path17.find_next('path')
incomep1 = icbg[0]
incomep2 = icbg[1]
incomep3 = icbg[2]
incomep4 = icbg[3]
incomep5 = icbg[4]
incomep6 = icbg[5]
incomep7 = icbg[6]
incomep8 = icbg[7]
incomep9 = icbg[8]
incomep10 = icbg[9]
incomep11 = icbg[10]
incomep12 = icbg[11]
incomep13 = icbg[12]
incomep14 = icbg[13]
incomep15 = icbg[14]
incomep16 = icbg[15]
incomep17 = icbg[16]
incomep18 = icbg[17]

paths = (path1, path2, path3, path4, path5, path6, path7, path8, path9, path10,
             path11, path12, path13, path14, path15, path16, path17, path18)
incomes = (incomep1,incomep2,incomep3,incomep4,incomep5,incomep6,incomep7,incomep8,
               incomep9,incomep10,incomep11,incomep12,incomep13,incomep14,incomep15,incomep16,incomep17,incomep18)

#set colors
colors = ['fee5d9','fcae91','fb6a4a','de2d26','a50f15']

for p in paths:
    for i in range(0,17):
        it = incomes[i]
        if it > 20000:
            color_class = 2
        elif it > 25000:
            color_class = 1
        elif it > 30000:
            color_class = 3
        elif it > 35000:
            color_class = 4

        color = colors[color_class]
        path_style = "font-size:12px;fill:#%s;fill-rule:nonzero;stroke:#000000;stroke-opacity:1;stroke-width:0.1;stroke-miterlimit:4;stroke-dasharray:none;stroke-linecap:butt;marker-start:none;stroke-linejoin:bevel" % color
        p['style'] = path_style
    print(soup.prettify())

运行这个命令可以得到这样一个SVG文件：fill:#fb6a4a;fill-rule:nonzero;stroke:#000000;stroke-opacity:1;stroke-width:0.1;stroke-miterlimit:4;stroke-dasharray:none;stroke-linecap:butt;marker-start:none;stroke-linejoin:bevel">出现18次，这意味着对于每个可用的路径，即使这些路径的收入不同。在

问题可能出在我写比较的方式上吗？在

Answer 1

根据我对您正在尝试做什么的理解，您的问题是您有2个for个循环，而不是一个。你应该同时循环路径和收入。你现在这样做的方式是在每一条路径上循环所有的收入。下面的代码只是将路径移到与收入相同的循环中，以便它们同时循环。在

for i in range(0,17):
    it = incomes[i]
    p = paths[i]
    if it > 20000:
        color_class = 2
    elif it > 25000:
        color_class = 1
    elif it > 30000:
        color_class = 3
    elif it > 35000:
        color_class = 4

    color = colors[color_class]
    path_style = "font-size:12px;fill:#%s;fill-rule:nonzero;stroke:#000000;stroke-opacity:1;stroke-width:0.1;stroke-miterlimit:4;stroke-dasharray:none;stroke-linecap:butt;marker-start:none;stroke-linejoin:bevel" % color
    p['style'] = path_style