[英]Pandas Dataframe: join items in range based on their geo coordinates (longitude and latitude)
I got a dataframe that contains places with their latitude and longitude. 我有一个包含纬度和经度的地方的数据框。 Imagine for example cities.
想象一下,例如城市。
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. 现在我试图让所有城市都在另一个城市的半径范围内。 Let's say all cities in a distance of 500km from Berlin, 500km from Hamburg and so on.
假设距离柏林500公里,距汉堡500公里等所有城市。 I would do this by duplicating the original dataframe and joining both with a distance-function.
我会通过复制原始数据帧并使用距离函数连接来完成此操作。
The intermediate result would be somewhat like this: 中间结果有点像这样:
Berlin --> Potsdam
Berlin --> Hamburg
Potsdam --> Berlin
Potsdam --> Hamburg
Hamburg --> Potsdam
Hamburg --> Berlin
This final result after grouping (reducing) should be like this. 分组(减少)后的最终结果应该是这样的。 Remark: Would be cool if the list of values includes all columns of the city.
备注:如果值列表包含城市的所有列,那将会很酷。
Berlin --> [Potsdam, Hamburg]
Potsdam --> [Berlin, Hamburg]
Hamburg --> [Berlin, Potsdam]
Or just the count of cities 500km around one city. 或者只是一个城市周围500公里的城市数量。
Berlin --> 2
Potsdam --> 2
Hamburg --> 2
Since I'm quite new to Python, I would appreciate any starting point. 由于我对Python很陌生,所以我会很感激任何起点。 I'm familiar with haversine distance.
我很熟悉长距离。 But not sure if there are useful distance/spatial methods in Scipy or Pandas.
但不确定Scipy或Pandas中是否有有用的距离/空间方法。
Glad if you can give me a starting point. 很高兴,如果你能给我一个起点。 Up to now I tried following this post .
到目前为止,我尝试过这篇文章 。
Update: The original idea behind this question comes from the Two Sigma Connect Rental Listing Kaggle Competition . 更新:这个问题背后的原始想法来自两西格玛连接租赁列表Kaggle比赛 。 The idea is to get those listing 100m around another listing.
我们的想法是让那些在另一个列表中上市100米。 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.
其中a)表示密度,因此表示热门区域; b)如果比较地址,您可以查看是否存在交叉,因此是否存在噪声区域。 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.
因此,您不需要完整的项目与项目关系,因为您不仅需要比较距离,还需要比较地址和其他元数据。 PS: I'm not uploading a solution to Kaggle.
PS:我没有向Kaggle上传解决方案。 I just want to learn.
我只是想学习。
You can use: 您可以使用:
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
: 首先需要使用
merge
交叉连接,通过boolean indexing
删除city_x
和city_y
具有相同值的行:
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: 然后应用hasrsine函数:
df['dist'] = df.apply(lambda row: haversine(row['lng_x'],
row['lat_x'],
row['lng_y'],
row['lat_y']), axis=1)
Filter distance: 过滤距离:
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
: 最后创建
list
或使用groupby
获取size
:
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
: 也可以使用
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: 更新:我建议首先尝试距离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): 有效地产生成对距离DF(没有重复):
In [106]: x
Out[106]:
dist
Berlin Potsdam 27.198616
Hamburg 255.063541
Potsdam Hamburg 242.311890
Old answer: 老答案:
Here is a bit optimized version, which uses scipy.spatial.distance.pdist method: 这是一个有点优化的版本,它使用scipy.spatial.distance.pdist方法:
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: 这给了我们:
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 : 让我们计算距离大于30的城市数量:
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
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