Find euclidean distance from a point to rows in pandas dataframe
Translating [(x2 - x1)2 + (y2 - y1)2]1/2 into pandas vectorised operations, you have:
df['distance'] = (df.lat.sub(11.344).pow(2).add(df.long.sub(7.234).pow(2))).pow(.5)
df
lat long distance
id
1 12.654 15.50 8.369161
2 14.364 25.51 18.523838
3 17.636 32.53 26.066777
5 12.334 25.84 18.632320
9 32.224 15.74 22.546096
Alternatively, using arithmetic operators:
(((df.lat - 11.344) ** 2) + (df.long - 7.234) ** 2) ** .5
Use vectorized approach
In [5463]: (df[['lat', 'long']] - np.array(L1)).pow(2).sum(1).pow(0.5)
Out[5463]:
0 8.369161
1 18.523838
2 26.066777
3 18.632320
4 22.546096
dtype: float64
Which can also be
In [5468]: df['distance'] = df[['lat', 'long']].sub(np.array(L1)).pow(2).sum(1).pow(0.5)
In [5469]: df
Out[5469]:
id lat long distance
0 1 12.654 15.50 8.369161
1 2 14.364 25.51 18.523838
2 3 17.636 32.53 26.066777
3 5 12.334 25.84 18.632320
4 9 32.224 15.74 22.546096
Option 2 Use Numpy's built-in np.linalg.norm
vector norm.
In [5473]: np.linalg.norm(df[['lat', 'long']].sub(np.array(L1)), axis=1)
Out[5473]: array([ 8.36916101, 18.52383805, 26.06677732, 18.63231966, 22.5460958 ])
In [5485]: df['distance'] = np.linalg.norm(df[['lat', 'long']].sub(np.array(L1)), axis=1)