python – Pandas Dataframe：根据地理坐标(经度和纬度)连接范围内的项目
作者：互联网
我有一个包含纬度和经度的地方的数据框.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}]);
现在我试图让所有城市都在另一个城市的半径范围内.假设距离柏林500公里,距汉堡500公里等所有城市.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.
中间结果有点像这样：
Berlin --> Potsdam
Berlin --> Hamburg
Potsdam --> Berlin
Potsdam --> Hamburg
Hamburg --> Potsdam
Hamburg --> Berlin
分组(减少)后的最终结果应该是这样的.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]
或者只是一个城市周围500公里的城市数量.
Berlin --> 2
Potsdam --> 2
Hamburg --> 2
由于我对Python很陌生,所以我会很感激任何起点.我很熟悉长距离.但不确定Scipy或Pandas中是否有有用的距离/空间方法.
很高兴,如果你能给我一个起点.到目前为止,我尝试过跟随this post.
更新：这个问题背后的原始想法来自Two Sigma Connect Rental Listing Kaggle Competition.这个想法是让那些列在另一个列表周围的100米.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.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：我没有向Kaggle上传解决方案.我只是想学习.
解决方法:
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
首先需要与merge交叉连接,在boolean indexing中删除city_x和city_y中具有相同值的行：
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
然后应用haversine功能：
df['dist'] = df.apply(lambda row: haversine(row['lng_x'],
row['lat_x'],
row['lng_y'],
row['lat_y']), axis=1)
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
最后创建列表或使用groupby获取大小：
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
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
标签：python,pandas,latitude-longitude,haversine 来源： https://codeday.me/bug/20191006/1858391.html