I got a dataframe that contains places with their latitude and longitude. Imagine for example cities.
df = pd.DataFrame([{'city':"Berlin", 'lat':52.5243700, 'lng':13.4105300},
{'city':"Potsdam", 'lat':52.3988600, 'lng':13.0656600},
{'city':"Hamburg", 'lat':53.5753200, 'lng':10.0153400}]);
Now I'm trying to get all cities in a radius around another. Let's say all cities in a distance of 500km from Berlin, 500km from Hamburg and so on. I would do this by duplicating the original dataframe and joining both with a distance-function.
The intermediate result would be somewhat like this:
Berlin --> Potsdam
Berlin --> Hamburg
Potsdam --> Berlin
Potsdam --> Hamburg
Hamburg --> Potsdam
Hamburg --> Berlin
This final result after grouping (reducing) should be like this. Remark: Would be cool if the list of values includes all columns of the city.
Berlin --> [Potsdam, Hamburg]
Potsdam --> [Berlin, Hamburg]
Hamburg --> [Berlin, Potsdam]
Or just the count of cities 500km around one city.
Berlin --> 2
Potsdam --> 2
Hamburg --> 2
Since I'm quite new to Python, I would appreciate any starting point. I'm familiar with haversine distance. But not sure if there are useful distance/spatial methods in Scipy or Pandas.
Glad if you can give me a starting point. Up to now I tried following this post.
Update: The original idea behind this question comes from the Two Sigma Connect Rental Listing Kaggle Competition. The idea is to get those listing 100m around another listing. Which a) indicates a density and therefore a popular area and b) if the addresses are compares, you can find out if there is a crossing and therefore a noisy area. Therefore you not need the full item to item relation since you need to compare not only the distance but also the address and other meta-data. PS: I'm not uploading a solution to Kaggle. I just want to learn.
You can use:
from math import radians, cos, sin, asin, sqrt
def haversine(lon1, lat1, lon2, lat2):
lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
# haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * asin(sqrt(a))
r = 6371 # Radius of earth in kilometers. Use 3956 for miles
return c * r
First need cross join with merge, remove row with same values in city_x and city_y by boolean indexing:
df['tmp'] = 1
df = pd.merge(df,df,on='tmp')
df = df[df.city_x != df.city_y]
print (df)
city_x lat_x lng_x tmp city_y lat_y lng_y
1 Berlin 52.52437 13.41053 1 Potsdam 52.39886 13.06566
2 Berlin 52.52437 13.41053 1 Hamburg 53.57532 10.01534
3 Potsdam 52.39886 13.06566 1 Berlin 52.52437 13.41053
5 Potsdam 52.39886 13.06566 1 Hamburg 53.57532 10.01534
6 Hamburg 53.57532 10.01534 1 Berlin 52.52437 13.41053
7 Hamburg 53.57532 10.01534 1 Potsdam 52.39886 13.06566
Then apply haversine function:
df['dist'] = df.apply(lambda row: haversine(row['lng_x'],
row['lat_x'],
row['lng_y'],
row['lat_y']), axis=1)
Filter distance:
df = df[df.dist < 500]
print (df)
city_x lat_x lng_x tmp city_y lat_y lng_y dist
1 Berlin 52.52437 13.41053 1 Potsdam 52.39886 13.06566 27.215704
2 Berlin 52.52437 13.41053 1 Hamburg 53.57532 10.01534 255.223782
3 Potsdam 52.39886 13.06566 1 Berlin 52.52437 13.41053 27.215704
5 Potsdam 52.39886 13.06566 1 Hamburg 53.57532 10.01534 242.464120
6 Hamburg 53.57532 10.01534 1 Berlin 52.52437 13.41053 255.223782
7 Hamburg 53.57532 10.01534 1 Potsdam 52.39886 13.06566 242.464120
And last create list or get size with groupby:
df1 = df.groupby('city_x')['city_y'].apply(list)
print (df1)
city_x
Berlin [Potsdam, Hamburg]
Hamburg [Berlin, Potsdam]
Potsdam [Berlin, Hamburg]
Name: city_y, dtype: object
df2 = df.groupby('city_x')['city_y'].size()
print (df2)
city_x
Berlin 2
Hamburg 2
Potsdam 2
dtype: int64
Also is possible use numpy haversine solution:
def haversine_np(lon1, lat1, lon2, lat2):
"""
Calculate the great circle distance between two points
on the earth (specified in decimal degrees)
All args must be of equal length.
"""
lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])
dlon = lon2 - lon1
dlat = lat2 - lat1
a = np.sin(dlat/2.0)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2.0)**2
c = 2 * np.arcsin(np.sqrt(a))
km = 6367 * c
return km
df['tmp'] = 1
df = pd.merge(df,df,on='tmp')
df = df[df.city_x != df.city_y]
#print (df)
df['dist'] = haversine_np(df['lng_x'],df['lat_x'],df['lng_y'],df['lat_y'])
city_x lat_x lng_x tmp city_y lat_y lng_y dist
1 Berlin 52.52437 13.41053 1 Potsdam 52.39886 13.06566 27.198616
2 Berlin 52.52437 13.41053 1 Hamburg 53.57532 10.01534 255.063541
3 Potsdam 52.39886 13.06566 1 Berlin 52.52437 13.41053 27.198616
5 Potsdam 52.39886 13.06566 1 Hamburg 53.57532 10.01534 242.311890
6 Hamburg 53.57532 10.01534 1 Berlin 52.52437 13.41053 255.063541
7 Hamburg 53.57532 10.01534 1 Potsdam 52.39886 13.06566 242.311890
UPDATE: I would suggest first to buiuld a distance DataFrame:
from scipy.spatial.distance import squareform, pdist
from itertools import combinations
# see definition of "haversine_np()" below
x = pd.DataFrame({'dist':pdist(df[['lat','lng']], haversine_np)},
index=pd.MultiIndex.from_tuples(tuple(combinations(df['city'], 2))))
efficiently produces pairwise distance DF (without duplicates):
In [106]: x
Out[106]:
dist
Berlin Potsdam 27.198616
Hamburg 255.063541
Potsdam Hamburg 242.311890
Old answer:
Here is a bit optimized version, which uses scipy.spatial.distance.pdist method:
from scipy.spatial.distance import squareform, pdist
# slightly modified version: of http://stackoverflow.com/a/29546836/2901002
def haversine_np(p1, p2):
"""
Calculate the great circle distance between two points
on the earth (specified in decimal degrees)
All args must be of equal length.
"""
lat1, lon1, lat2, lon2 = np.radians([p1[0], p1[1],
p2[0], p2[1]])
dlon = lon2 - lon1
dlat = lat2 - lat1
a = np.sin(dlat/2.0)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2.0)**2
c = 2 * np.arcsin(np.sqrt(a))
km = 6367 * c
return km
x = pd.DataFrame(squareform(pdist(df[['lat','lng']], haversine_np)),
columns=df.city.unique(),
index=df.city.unique())
this gives us:
In [78]: x
Out[78]:
Berlin Potsdam Hamburg
Berlin 0.000000 27.198616 255.063541
Potsdam 27.198616 0.000000 242.311890
Hamburg 255.063541 242.311890 0.000000
let's count number of cities where the distance is greater than 30:
In [81]: x.groupby(level=0, as_index=False) \
...: .apply(lambda c: c[c>30].notnull().sum(1)) \
...: .reset_index(level=0, drop=True)
Out[81]:
Berlin 1
Hamburg 2
Potsdam 1
dtype: int64
来源：https://stackoverflow.com/questions/42877802/pandas-dataframe-join-items-in-range-based-on-their-geo-coordinates-longitude