Question

我有一个.dat文件，其中包含数千条记录，每条记录都包含单独的Degree Minute Seconds字段。我试图将其转换为十进制度数，这是大多数地图/ GIS软件的标准。

转换的数学计算非常简单，但我没有得到预期的输出。我假设这是因为数据以string的形式存在，我投放到int和float类型进行计算。

这里的计算如下： latitude = (degs+(mins/60)+(secs/3600))

脚本：

import csv

georgia = r'C:\Users\rhewitt\Desktop\CellTowers\DOF\13-GA.Dat'
output = r'C:\Users\rhewitt\Desktop\CellTowers\CSV\georgia.csv'
header = r'C:\Users\rhewitt\Desktop\CellTowers\header.csv'

dat = open(georgia, 'r')
f = open(output, 'wb')
wr = csv.writer(f, quoting=csv.QUOTE_ALL)

# Create header
header = ["ObstacleNo", "ver_status", "Country", "State", "City", "latdeg", "latmin", "latsec", "N_S", "latitude", "londeg", "lonmin", "lonsec", "E_W", "longitude", "Type", "QTY", "AGL_HT", "AMSL_HT", "Lighting", "horz_acc", "vert_acc", "marking", "FAA_no", "Action", "Julian_dt"]
wr.writerow(header)

i=-1
for line in dat:
    i+=1

    # The top five rows of the .dat file contains header information (skip them)
    if i < 5:
        continue

    # .dat files have fixed-width columns, so I can pull out information like this:
    else:
        ORS     = line[:9]
        v       = line[9:11]
        country = line[11:14]
        state   = line[14:17]
        city    = line[17:35]
        latDeg  = line[35:37]
        latMin  = line[37:40]
        latSec  = line[40:46]
        latHem  = line[46:47]
        lonDeg  = line[47:51]
        lonMin  = line[51:54]
        lonSec  = line[54:60]
        lonHem  = line[60:61]
        obType  = line[61:75]
        uknown  = line[75:76]
        agl     = line[76:82]
        amsl    = line[82:88]
        lt      = line[88:90]
        ah      = line[90:92]
        ccv     = line[92:94]
        MIN     = line[94:96]
        study   = line[96:111]
        action  = line[111:113]
        jDate   = line[113:]

        # Convert Degrees Minutes Seconds to Decimal Degrees for easier mapping
        if latHem == "N":
            latitude = (int(latDeg)+(int(latMin)/60)+(float(latSec)/3600))
        else:
            latitude = -1*(int(latDeg)+(int(latMin)/60)+(float(latSec)/3600))

        if lonHem == "E":
            longitude = (int(lonDeg)+(int(lonMin)/60)+(float(lonSec)/3600))
        else:
            longitude = -1*(int(lonDeg)+(int(lonMin)/60)+(float(lonSec)/3600))

        csvLine = [ORS, v, country, state, city, latDeg, latMin, latSec, latHem, latitude, lonDeg, lonMin, lonSec, lonHem, longitude, obType, uknown, agl, amsl, lt, ah, ccv, MIN, study, action, jDate.rstrip("\n")]
        wr.writerow(csvLine)

f.close()
dat.close()

以下是我脚本的示例输出： (34 + (59/60)+(10.00/3600)) = 34.0027777778，但正确的值实际为34.9861111111

使用Python将度数分钟秒转换为十进制度有问题

0 个答案: