Converting pandas data frame with degree minute second (DMS) coordinates to decimal degrees

Question

I have a data frame like the below and would like to convert the Latitude and Longitude columns in Degree, Minute, Second format into decimal degrees - with negative for the correct hemisphere. Is there an easy way to do that?

Parent Company  CPO PKO Latitude    Longitude
Incasi Raya X       0°51'56.29"S    101°26'46.29"E
Incasi Raya X       1°23'39.29"S    101°35'30.45"E
Incasi Raya X       0°19'56.63"N    99°22'56.36"E
Incasi Raya X       0°21'45.91"N    99°37'59.68"E
Incasi Raya X       1°41'6.56"S 102°14'7.68"E
Incasi Raya X       1°15'2.13"S 101°34'30.38"E
Incasi Raya X       2°19'44.26"S    100°59'34.55"E
Musim Mas   X       1°44'55.94"N    101°22'15.94"E

For example 0°51'56.29"S would be converted to -0.8656361

Answer 1

You can use vectorized operations using pd.Series.str.extract. For the latitude, for example:

parts = df.Latitude.str.extract('(\d+)°(\d+)\'([^"]+)"([N|S|E|W])', expand=True)
>>> (parts[0].astype(int) + parts[1].astype(float) / 60 + parts[2].astype(float) / 3600) * parts[3].map({'N':1, 'S':-1, 'E': 1, 'W':-1})
0    101.446192
1    101.591792
2     99.382322
3     99.633244
4    102.235467
5    101.575106
6    100.992931
7    101.371094

Answer 2

Basing my answer on a function from SO you can do it like this:

Interestingly this answer is also 2x as fast as MaxU and Amis answer for a dataset with +500 rows. My bet is that the bottleneck is str.extract(). But something is clearly strange.

import pandas as pd
import re

#https://stackoverflow.com/questions/33997361
def dms2dd(s):
    # example: s = """0°51'56.29"S"""
    degrees, minutes, seconds, direction = re.split('[°\'"]+', s)
    dd = float(degrees) + float(minutes)/60 + float(seconds)/(60*60);
    if direction in ('S','W'):
        dd*= -1
    return dd

df = pd.DataFrame({'CPO': {0: 'Raya', 1: 'Raya'},
 'Latitude': {0: '0°51\'56.29"S', 1: '1°23\'39.29"S'},
 'Longitude': {0: '101°26\'46.29"E', 1: '101°35\'30.45"E'},
 'PKO': {0: 'X', 1: 'X'},
 'ParentCompany': {0: 'Incasi', 1: 'Incasi'}})

df['Latitude'] = df['Latitude'].apply(dms2dd)
df['Longitude'] = df['Longitude'].apply(dms2dd)

printing df returns:

    CPO   Latitude   Longitude PKO ParentCompany
0  Raya  -0.865636  101.446192   X        Incasi
1  Raya  -1.394247  101.591792   X        Incasi

---

Update: To correct your mistake you could do something in the lines of:

m = df['Latitude'].str[-2] != '"'
df.loc[m, 'Latitude'] = df.loc[m, 'Latitude'].str[:-1] + '"' + df.loc[m, 'Latitude'].str[-1]

Full example:

import re

s1 = """0°51'56.29"S"""
s2 = """0°51'56.29S"""

df = pd.Series((s1,s2)).to_frame(name='Latitude')

m = df['Latitude'].str[-2] != '"'
df.loc[m, 'Latitude'] = df.loc[m, 'Latitude'].str[:-1] + '"' + df.loc[m, 'Latitude'].str[-1]

print(df)

Answer 3

Here is a vectorized approach, that also uses matrix * vector ([1, 1./60, 1./3600]) multiplication:

In [233]: %paste
def dms2dec(s):
    x = (s.str.upper()
          .str.split(r'[°\'"]', expand=True)
          .replace(['S','W','N','E'], [-1,-1,1,1])
          .astype('float'))
    return x.iloc[:, :3].dot([1, 1./60, 1./3600]).mul(x.iloc[:, 3])

## -- End pasted text --

In [234]: df[['Latitude','Longitude']] = df[['Latitude','Longitude']].apply(dms2dec)

In [235]: df
Out[235]:
  Parent Company CPO PKO  Latitude   Longitude
0    Incasi Raya       X -0.865636  101.446192
1    Incasi Raya       X -1.394247  101.591792
2    Incasi Raya       X  0.332397   99.382322
3    Incasi Raya       X  0.362753   99.633244
4    Incasi Raya       X -1.685156  102.235467
5    Incasi Raya       X -1.250592  101.575106
6    Incasi Raya       X -2.328961  100.992931
7      Musim Mas       X  1.748872  101.371094

step by step explanation:

In [239]: x = (s.str.upper()
     ...:       .str.split(r'[°\'"]', expand=True)
     ...:       .replace(['S','W','N','E'], [-1,-1,1,1])
     ...:       .astype('float'))

In [240]: x
Out[240]:
     0     1      2    3
0  0.0  51.0  56.29 -1.0
1  1.0  23.0  39.29 -1.0
2  0.0  19.0  56.63  1.0
3  0.0  21.0  45.91  1.0
4  1.0  41.0   6.56 -1.0
5  1.0  15.0   2.13 -1.0
6  2.0  19.0  44.26 -1.0
7  1.0  44.0  55.94  1.0

In [241]: x.iloc[:, :3].dot([1, 1./60, 1./3600])
Out[241]:
0    0.865636
1    1.394247
2    0.332397
3    0.362753
4    1.685156
5    1.250592
6    2.328961
7    1.748872
dtype: float64

In [242]: x.iloc[:, :3].dot([1, 1./60, 1./3600]).mul(x.iloc[:, 3])
Out[242]:
0   -0.865636
1   -1.394247
2    0.332397
3    0.362753
4   -1.685156
5   -1.250592
6   -2.328961
7    1.748872
dtype: float64