0 
"POLYGON ((12 13,22 23,16 17,22 24))",101,Something,100000
"POLYGON ((10 12,40 42,46 34,16 24,88 22,33 24,18 20 ))",102,another,200000

如何在 csv 文件中获得如下内容：

UID（如 101,102 等）表示每个多边形的唯一标识符。

UID#1,County,population,Point#1_Lat,Point#1_Long
UID#1,County,population,Point#2_Lat,Point#2_Long
UID#1,County,population,Point#3_Lat,Point#3_Long
UID#1,County,population,Point#n_Lat,Point#n_Long

UID#2,County,population,Point#1_Lat,Point#1_Long
UID#2,County,population,Point#2_Lat,Point#2_Long
UID#2,County,population,Point#3_Lat,Point#3_Long
UID#2,County,population,Point#n_Lat,Point#n_Long
4

2 回答 2

2

这是一个使用pyparsing. 如果这对您不起作用，请告诉我 - 想出只使用标准库（例如re等）的东西应该不会太难，但它肯定会更丑陋。

import csv
from pyparsing import Group, Literal, OneOrMore, Optional, Word
from pyparsing import delimitedList
from pyparsing import alphas, nums

data = """
"POLYGON ((12 13,22 23,16 17,22 24))",101,Something,100000
"POLYGON ((10 12,40 42,46 34,16 24,88 22,33 24,18 20 ))",102,another,200000
"""

def parse_line(line):
    latitude = Word(nums)
    longitude = Word(nums)
    point = Group(latitude + longitude)
    point_sequence = delimitedList(point, delim=',')

    name = Word("POLYGON").suppress()
    paren_left = Literal("((").suppress()
    paren_right = Literal("))").suppress()
    quote = Literal('"').suppress()
    polygon = Group(quote + name + paren_left + point_sequence + paren_right + quote)

    uid = Word(nums)
    county = Word(alphas)
    population = Word(nums)
    sep = Literal(",").suppress()
    parser = polygon + sep + uid + sep + county + sep + population

    result = parser.parseString(line)
    return result

def parse_lines(data, outfile):
    with open(outfile, 'w') as f:
        writer = csv.writer(f, lineterminator='\n')
        lines = data.split('\n')
        for line in lines:
            if not line:
                continue
            points, uid, county, population = parse_line(line)
            for lat, long in points:
                writer.writerow([uid, county, population, lat, long])
            writer.writerow('')

parse_lines(data, r'd:\out.txt') # change the path to wherever you want output

结果：

101,Something,100000,12,13
101,Something,100000,22,23
101,Something,100000,16,17
101,Something,100000,22,24

102,another,200000,10,12
102,another,200000,40,42
102,another,200000,46,34
102,another,200000,16,24
102,another,200000,88,22
102,another,200000,33,24
102,another,200000,18,20
于 2014-01-16T18:34:34.003 回答
1

谢谢 senshin 的解决方案。这是我第一次尝试 Python。我尝试了你的建议。确实尝试找出替代方案并取得了良好的结果。

几何，区域，类型，UID
"POLYGON(x1 y1,x2 y2,x3 y3,x4 y4)",name1,abc,100
"POLYGON(x1 y1,x2 y2,x3 y3,x4 y4,x5 y5,x6 y6) ",name2,pqr,101

import csv
import re
import sys


l_InputFileName ='D:/Example1.txt'             # make changes here..
l_OutputFileName ='D:/Example1_o.txt'          # make changes here..

fo = open(l_OutputFileName, "a+")
with open( l_InputFileName, 'r') as csvfile:

csvR = csv.reader( csvfile, delimiter=',', quotechar='"')
#Get first row which has column names
header = csvR.next()
#Convert first row which is a list into a string
print ','.join(header[0:])
#Write the header to output csv file,index starts at 1 here(splitting point into lat and        long)
fo.write('longitude,latitude,')
fo.write(','.join(header[1:])+'\n')

for row in csvR:

    #Remove the string POLYGON and brackets from first column
    coodlist=row[0].strip('POLYGON').strip('()').split(',')
    #Get the number of columns
    #print "\nNo of Columns="+str(len(row))
    #Get remaining columns 
    strRemainingCols =','.join(row[1:])
    #Print each lat,long in a seperate row..reamining columns will not change
    #Seperate latitude and longitude using comma
    for i in range(len(coodlist)):
       print coodlist[i].replace(' ',',')+','+strRemainingCols
       fo.write(coodlist[i].replace(' ',',')+','+strRemainingCols+'\n')

csvfile.close()
fo.close()
于 2014-01-19T12:04:27.760 回答