用减号解析字符串,保留减号

Question

我有以下代码:

latlonginput = 'Latitude=28.0358214, Longitude=-82.59970229999999, Accuracy=4232m'

import re

try:
    #latlonginput = input['latlong']
    geo_lat = format(re.split(r'(Latitude=|,)', latlonginput)[2])
    geo_long = format(re.split(r'(Longitude=|,)', latlonginput)[2])
    accuracy = format(re.split(r'(Accuracy=)', latlonginput)[2])
    geo_coord = geo_lat + ',' + geo_long
    latlong = 'Detected Lat/Long: '+ geo_coord + '\n'
    latlong = latlong + 'Accuracy: ' + accuracy + '\n'
    latlong = latlong + 'Maps: placeholder'
except:
    latlong = None

return {'lat_long': latlong }

在这种情况下,经度为负值,省略负​​数.例如,结果如下:

Detected Lat/Long: 28.0358214,
Accuracy: 4232m
Maps: placeholder

我需要允许纬度或经度的负数.我该如何调整呢？

谢谢,Noob

Answer 1

您可以使用带有可选匹配的正则表达式来检查所需数字前面的负号.

此外,如果这是您的字符串格式,re.split则不需要使用,您可以简单地匹配.

如果这些数字是字符串中唯一的数字,您只需使用:

In [11]: re.findall(r'-?\d+\.?\d+m?', s)
Out[11]: ['28.0358214', '-82.59970229999999', '4232m']

如果你想让它更健壮,你可以匹配整个格式:

import re

rgx = r'Latitude=(-?\d+\.\d+).*Longitude=(-?\d+\.\d+).*Accuracy=-?(\d+m)'
s = 'Latitude=28.0358214, Longitude=-82.59970229999999, Accuracy=4232m'

lat, long, accuracy = re.findall(rgx, s)[0]
print('Lat/Long: {}, {}\nAccuracy: {}'.format(lat, long, accuracy))

输出:

Lat/Long: 28.0358214, -82.59970229999999
Accuracy: 4232m

Answer 2

正则表达式对此有些过分; ", "首先拆分字符串,然后拆分每个元素"=":

>>> dict(x.split('=') for x in latlonginput.split(', '))
{'Latitude': '28.0358214', 'Longitude': '-82.59970229999999', 'Accuracy': '4232m'}

然后根据需要使用字典.