# How to calculate distance for every row in a pandas dataframe from a single point efficiently?

You can compute vectorized Euclidean distance (L2 norm) using the formula

sqrt((a1 - b1)2 + (a2 - b2)2 + ...)

df.sub(point, axis=1).pow(2).sum(axis=1).pow(.5)

0    0.474690
1    0.257080
2    0.703857
3    0.503596
4    0.461151
dtype: float64


Which gives the same output as your current code.

Or, using linalg.norm:

np.linalg.norm(df.to_numpy() - point, axis=1)
# array([0.47468985, 0.25707985, 0.70385676, 0.5035961 , 0.46115096])


Another option is use cdist which is a bit faster:

from scipy.spatial.distance import cdist
cdist(point[None,], df.values)


Output:

array([[0.47468985, 0.25707985, 0.70385676, 0.5035961 , 0.46115096]])


Some comparison with 100k rows:

%%timeit -n 10
cdist([point], df.values)
645 µs ± 36.4 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%%timeit -n 10
np.linalg.norm(df.to_numpy() - point, axis=1)
5.16 ms ± 227 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%%timeit -n 10
df.sub(point, axis=1).pow(2).sum(axis=1).pow(.5)
16.8 ms ± 444 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)


Let us do scipy

from scipy.spatial import distance
ary = distance.cdist(df.values, np.array([point]), metric='euclidean')
ary
Out[57]:
array([[0.47468985],
[0.25707985],
[0.70385676],
[0.5035961 ],
[0.46115096]])