你如何在Python中将十进制度转换为度数分钟秒?有没有已经写过的公式?
这正是divmod
的发明:
>>> 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。这个应该正确处理否定。
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。与通常的做法相反,将所有度,分和秒指定为负(Wikipedia显示40°26.7717,-79°56.93172作为度 - 分表示法的有效示例,其中度为负且分钟无符号),如果度数部分为0,则将度数设置为负数没有任何影响。这是一个根据Paul McGuire和baens的函数充分处理这个问题的函数:
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)
只需几个* 60
乘法和几个int
截断,即:
>>> 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), '" ')
标志最好单独返回,以便它可以用来选择('N', 'S')
或('E', 'W')
。
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
改善@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)
使用fmod
和舍入来获得度数和分数。将分数乘以60并重复以获得分钟和余数。然后再将该最后一部分乘以60以获得秒数。