我需要解密SVG文档中的路径元素以沿着该路径驱动CNC机器。我想知道是否有任何Python库解析SVG并为d属性提供某种pythonic列表，例如：
<path d="M 20 30 L 20 20 20 40 40 40"/>
解析
[["M", 20, 30],
["L", 20, 20],
["L", 20, 40],
["L", 40, 40]]
答案 0 :(得分：3)
这是一个由我和python 2.7.2编写的开始。如果您愿意，只需删除测试和打印语句即可。
Copyright 2012 Christopher L. Ramsey
Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at
http://www.apache.org/licenses/LICENSE-2.0
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
from collections import OrderedDict
from re import match
from re import split
from re import sub
class PathIterator(object):
EOI = 'End of Iteration'
PATH_IDENTIFIERS = r'[MLHVCSQTAmlhvcsqa]'
NUMBERS = r'[0-9.-^A-z]'
SEPERATORS = r'\s|\,'
PATH_END = r'[Zz]'
def __init__(self, path):
self.parseable = path.translate(None, '\t\f')
self.parseable = self.parseable.replace('\n', ' ')
print 'strip_newlines: {}'.format(self.parseable)
self.parseable = sub(r'([A-Za-z])([0-9]|\-)', self.insert, self.parseable)
print 'add_space: {}'.format(self.parseable)
self.parseable = self.parseable.replace(',', ' ')
print 'replace_commas: {}'.format(self.parseable)
self.parseable = sub(r'\s+', ' ', self.parseable) # replace any double space with a single space
print 'strip_extra_space: {}'.format(self.parseable)
self.tokens = split(' ', self.parseable)
self.map = self.produce_map(self.tokens)
print self.map
self.elements = self.process(self.map)
def produce_map(self, tkns):
self.m = OrderedDict()
self.i = 0
while self.i < len(tkns):
if match(self.PATH_IDENTIFIERS, tkns[self.i]):
self.m[self.i] = tkns[self.i]
elif match(self.PATH_END, tkns[self.i]):
self.m[self.i] = tkns[self.i]
else:
pass
self.i += 1
return self.m.items()
def process(self, map):
self.mm = []
self.l = len(map)
for e in range(self.l):
try:
self.element = map[e]
self.future = map[e + 1]
self.ident = self.element[1]
self.start = self.element[0] + 1
self.end = self.future[0]
self.nbrs = self.tokens[self.start:self.end]
except:
self.element = map[e]
self.ident = self.element[1]
self.start = self.element[0] + 1
self.end = len(self.tokens)
self.nbrs = self.tokens[self.start:self.end]
print 'start: {} end {}'.format(self.start, self.end)
finally:
self.numbers = []
for number in self.nbrs:
self.numbers.append(float(number))
self.mm.append((self.ident, self.numbers))
return iter(self.mm)
def next(self):
try:
return self.elements.next()
except:
return self.EOI
def insert(self, match_obj):
self.group = match_obj.group()
return '{} {}'.format(self.group[0], self.group[1])
if __name__ == '__main__':
inkscape_path = "M 12,90 C 8.676,90 6,87.324 6,84 L 6,82 6,14 6,12 c 0,-0.334721 0.04135,-0.6507 0.09375,-0.96875 0.0487,-0.295596 0.09704,-0.596915 0.1875,-0.875 C 6.29113,10.12587 6.302142,10.09265 6.3125,10.0625 6.411365,9.774729 6.5473802,9.515048 6.6875,9.25 6.8320918,8.976493 7.0031161,8.714385 7.1875,8.46875 7.3718839,8.223115 7.5612765,7.995278 7.78125,7.78125 8.221197,7.353194 8.72416,6.966724 9.28125,6.6875 9.559795,6.547888 9.8547231,6.440553 10.15625,6.34375 9.9000482,6.443972 9.6695391,6.580022 9.4375,6.71875 c -0.00741,0.0044 -0.023866,-0.0045 -0.03125,0 -0.031933,0.0193 -0.062293,0.04251 -0.09375,0.0625 -0.120395,0.0767 -0.2310226,0.163513 -0.34375,0.25 -0.1061728,0.0808 -0.2132809,0.161112 -0.3125,0.25 C 8.4783201,7.442683 8.3087904,7.626638 8.15625,7.8125 8.0486711,7.942755 7.9378561,8.077785 7.84375,8.21875 7.818661,8.25713 7.805304,8.30462 7.78125,8.34375 7.716487,8.446782 7.6510225,8.548267 7.59375,8.65625 7.4927417,8.850956 7.3880752,9.071951 7.3125,9.28125 7.30454,9.30306 7.288911,9.3218 7.28125,9.34375 7.2494249,9.4357 7.2454455,9.530581 7.21875,9.625 7.1884177,9.731618 7.1483606,9.828031 7.125,9.9375 7.0521214,10.279012 7,10.635705 7,11 l 0,2 0,68 0,2 c 0,2.781848 2.2181517,5 5,5 l 2,0 68,0 2,0 c 2.781848,0 5,-2.218152 5,-5 l 0,-2 0,-68 0,-2 C 89,10.635705 88.94788,10.279012 88.875,9.9375 88.83085,9.730607 88.78662,9.539842 88.71875,9.34375 88.71105,9.3218 88.69545,9.30306 88.6875,9.28125 88.62476,9.107511 88.549117,8.913801 88.46875,8.75 88.42717,8.6672 88.38971,8.580046 88.34375,8.5 88.28915,8.40279 88.216976,8.31165 88.15625,8.21875 88.06214,8.077785 87.951329,7.942755 87.84375,7.8125 87.700576,7.63805 87.540609,7.465502 87.375,7.3125 87.36383,7.3023 87.35502,7.29135 87.34375,7.28125 87.205364,7.155694 87.058659,7.046814 86.90625,6.9375 86.803679,6.86435 86.701932,6.784136 86.59375,6.71875 c -0.0074,-0.0045 -0.02384,0.0044 -0.03125,0 -0.232039,-0.138728 -0.462548,-0.274778 -0.71875,-0.375 0.301527,0.0968 0.596455,0.204138 0.875,0.34375 0.55709,0.279224 1.060053,0.665694 1.5,1.09375 0.219973,0.214028 0.409366,0.441865 0.59375,0.6875 0.184384,0.245635 0.355408,0.507743 0.5,0.78125 0.14012,0.265048 0.276135,0.524729 0.375,0.8125 0.01041,0.03078 0.02133,0.06274 0.03125,0.09375 0.09046,0.278085 0.1388,0.579404 0.1875,0.875 C 89.95865,11.3493 90,11.665279 90,12 l 0,2 0,68 0,2 c 0,3.324 -2.676,6 -6,6 l -72,0 z"
mdn_path = "M10 80 Q 52.5 10, 95 80 T 180 80"
w3c_path = "M100,200 C100,100 250,100 250,200 S400,300 400,200"
w3c_path_neg = "M-100,200 C100,100 250,100 250,200 S-400,300 400,200"
w3c_path_nl = '''
M600,350 l 50,-25
a25,25 -30 0,1 50,-25 l 50,-25
a25,50 -30 0,1 50,-25 l 50,-25
a25,75 -30 0,1 50,-25 l 50,-25
a25,100 -30 0,1 50,-25 l 50,-25
'''
paths = [inkscape_path, mdn_path, w3c_path, str.strip(w3c_path_nl), w3c_path_neg]
for path in paths:
p = PathIterator(path)
char = ''
while char != PathIterator.EOI:
char = p.next()
print char
答案 1 :(得分：2)
使用svgpathtools获取d-string可以在几行中删除，其余的可以使用正则表达式完成。
from svgpathtools import svg2paths
paths, attributes = svg2paths('some_svg_file.svg')
路径是svgpathtools路径对象的列表（仅包含曲线信息，没有颜色，样式等）。 attributes 是属性的字典对象列表。
假设您感兴趣的路径是SVG中列出的 first （第0个），然后只提取您可以使用的d-string：
d = attributes[0]['d'] # d-string from first path in SVG
# Now for some regular expressions magic
import re
split_by_letters = re.findall('[A-Z|a-z][^A-Z|a-z]*', d)
split_as_you_want = []
for x in split_by_letters:
nums = x[1:].replace(',',' ').split() # list of numbers after letter
for k in range(len(nums) // 2):
split_as_you_want.append([x[0]] + [nums[k]] + [nums[k+1]])
print split_as_you_want
我没有在这里将数字转换为字符串，因为你想要这样做取决于他们是否总是整数以及你是否关心他们保持这种方式。在大多数情况下，这可以通过下面的＆＃34; nums = ...＆＃34;线。
for k, n in enumerate(nums):
try:
nums[k] = int(n)
except ValueError:
nums[k] = float(n)