如何在 Python 中将十进制度数转换为度分秒?是否已经编写了公式?
11 回答
这正是divmod
发明的目的:
>>> def decdeg2dms(dd):
... mnt,sec = divmod(dd*3600,60)
... deg,mnt = divmod(mnt,60)
... return deg,mnt,sec
>>> dd = 45 + 30.0/60 + 1.0/3600
>>> print dd
45.5002777778
>>> decdeg2dms(dd)
(45.0, 30.0, 1.0)
这是我基于 Paul McGuire 的更新版本。这个应该正确处理底片。
def decdeg2dms(dd):
is_positive = dd >= 0
dd = abs(dd)
minutes,seconds = divmod(dd*3600,60)
degrees,minutes = divmod(minutes,60)
degrees = degrees if is_positive else -degrees
return (degrees,minutes,seconds)
如果要正确处理负数,则将第一个非零度量设置为负数。将所有度数、分钟数和秒数指定为负数是违反惯例的(维基百科显示 40° 26.7717、-79° 56.93172 作为度数-分钟符号的有效示例,其中度数为负数,分钟数没有符号),如果度数部分为 0,则将度数设置为负数没有任何效果。这是一个充分处理此问题的函数,基于 Paul McGuire 和 baens 的函数:
def decdeg2dms(dd):
negative = dd < 0
dd = abs(dd)
minutes,seconds = divmod(dd*3600,60)
degrees,minutes = divmod(minutes,60)
if negative:
if degrees > 0:
degrees = -degrees
elif minutes > 0:
minutes = -minutes
else:
seconds = -seconds
return (degrees,minutes,seconds)
Just a couple of * 60
multiplications and a couple of int
truncations, i.e.:
>>> decdegrees = 31.125
>>> degrees = int(decdegrees)
>>> temp = 60 * (decdegrees - degrees)
>>> minutes = int(temp)
>>> seconds = 60 * (temp - minutes)
>>> print degrees, minutes, seconds
31 7 30.0
>>>
这是我的 Python 代码:
def DecimaltoDMS(Decimal):
d = int(Decimal)
m = int((Decimal - d) * 60)
s = (Decimal - d - m/60) * 3600.00
z= round(s, 2)
if d >= 0:
print ("N ", abs(d), "º ", abs(m), "' ", abs(z), '" ')
else:
print ("S ", abs(d), "º ", abs(m), "' ", abs(z), '" ')
改进@chqrlie 答案:
def deg_to_dms(deg, type='lat'):
decimals, number = math.modf(deg)
d = int(number)
m = int(decimals * 60)
s = (deg - d - m / 60) * 3600.00
compass = {
'lat': ('N','S'),
'lon': ('E','W')
}
compass_str = compass[type][0 if d >= 0 else 1]
return '{}º{}\'{:.2f}"{}'.format(abs(d), abs(m), abs(s), compass_str)
该标志最好单独返回,以便它可以用于选择('N', 'S')
或('E', 'W')
,例如。
import math
def dd_to_dms(degs):
neg = degs < 0
degs = (-1) ** neg * degs
degs, d_int = math.modf(degs)
mins, m_int = math.modf(60 * degs)
secs = 60 * mins
return neg, d_int, m_int, secs
这是我稍微不同的方法,它与我的 HP Prime 上的正负十进制度相同......
def dms(deg):
f,d = math.modf(deg)
s,m = math.modf(abs(f) * 60)
return (d,m,s * 60)
使用fmod
和舍入来分离度数和分数。将分数乘以 60 并重复以获得分钟和余数。然后将最后一部分再次乘以 60 以获得秒数。
现在我们可以使用 LatLon 库...
https://pypi.org/project/LatLon/
>> palmyra = LatLon(Latitude(5.8833), Longitude(-162.0833)) # Location of Palmyra Atoll in decimal degrees
>> palmyra = LatLon(5.8833, -162.0833) # Same thing but simpler!
>> palmyra = LatLon(Latitude(degree = 5, minute = 52, second = 59.88),
Longitude(degree = -162, minute = -4.998) # or more complicated!
>> print palmyra.to_string('d% %m% %S% %H') # Print coordinates to degree minute second
('5 52 59.88 N', '162 4 59.88 W')`
如果您的数据在 DataFrame 中,您可以使用clean_lat_long()
库DataPrep中的函数。安装 DataPrep 与pip install dataprep
.
from dataprep.clean import clean_lat_long
df = pd.DataFrame({"coord": [(45.5003, -122.4420), (5.8833, -162.0833)]})
df2 = clean_lat_long(df, "coord", output_format="dms")
# print(df2)
coord coord_clean
0 (45.5003, -122.442) 45° 30′ 1.08″ N, 122° 26′ 31.2″ W
1 (5.8833, -162.0833) 5° 52′ 59.88″ N, 162° 4′ 59.88″ W
或者,如果纬度和经度在不同的列中:
df = pd.DataFrame({"latitude": [45.5003, 5.8833], "longitude": [-122.4420, -162.0833]})
df2 = clean_lat_long(df, lat_col="latitude", long_col="longitude", output_format="dms")
# print(df2)
latitude longitude latitude_longitude
0 45.5003 -122.4420 45° 30′ 1.08″ N, 122° 26′ 31.2″ W
1 5.8833 -162.0833 5° 52′ 59.88″ N, 162° 4′ 59.88″ W