我有两个要遍历的dataFrame，应用Haversine函数，并在新数组中构造结果。我想获取da_store中第一家餐厅的经度，纬度坐标，对所有的经纬度的纬度，经度应用Haversine函数，存储结果并获取最小值。最终，对于每家商店，从所有da_univ坐标计算出欧几里得距离，并找到最近的大学和学院。
到目前为止，这是我在嵌套的for循环中所做的尝试，我正在努力以正确的格式保存结果并查找最小值。
for index_store, row_store in da_store.iterrows():
store_lat = row_store['lat']
store_lon = row_store['lon']
store_list = []
for index_univ, row_univ in da_univ.iterrows():
univ_lat = row_univ['LATITUDE']
univ_lon = row_univ['LONGITUDE']
distance = haversine_np(store_lon, store_lat, univ_lon, univ_lat)
print(distance)
数据框1：da_store
In [203]: da_store.head()
Out[203]:
Restaurant # Restaurant Name Address City State Zip Code lat lon
0 3006 Weymouth Dual Riverway Plaza, Weymouth, MA Weymouth MA 2191 42.244559 -70.936438
1 3009 Somerset Dual Somerset Plaza, Rt. 6, Somerset, MA Somerset MA 2725 41.734643 -71.152320
2 3502 Westboro Mass Pike West Mile Post 105; Mass Turnpike W., Westboro, MA Westboro MA 1581 42.253973 -71.663506
3 3503 Charlton Mass Pike East Mile Post 81; Mass Turnpike E., Charlton, MA Charlton MA 1507 42.101589 -72.018530
4 3504 Charlton Mass Pike West Mile Post 89; Mass Turnpike W., Charlton City, MA Charlton City MA 1508 42.101497 -72.018247
数据框2：da_univ
In [204]: da_univ.head()
Out[204]:
INSTNM ZIP CITY STABBR LATITUDE LONGITUDE
0 Hult International Business School 02141-1805 Cambridge MA 42.369968 -71.070645
1 New England College of Business and Finance 2110 Boston MA 42.353619 -71.056671
2 Assumption College 01609-1296 Worcester MA 42.294226 -71.828991
3 Bancroft School of Massage Therapy 1604 Worcester MA 42.268973 -71.778113
4 Bay State College 2116 Boston MA 42.351760 -71.076991
Haversine函数：haversine_np
from math import radians, cos, sin, asin, sqrt
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
我需要练习我的基本编程，谢谢您的帮助！
最佳答案
如果对pandas和numpy使用loop，则很可能您做错了。学习并应用这些库提供的矢量化功能：
# Build an index that contain every pairing of Store - University
idx = pd.MultiIndex.from_product([da_store.index, da_univ.index], names=['Store', 'Univ'])
# Pull the coordinates of the store and the universities together
# We don't need their name here
df = pd.DataFrame(index=idx) \
.join(da_store[['lat', 'lon']], on='Store') \
.join(da_univ[['LATITUDE', 'LONGITUDE']], on='Univ')
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['Distance'] = haversine_np(*df[['lat', 'lon', 'LATITUDE', 'LONGITUDE']].values.T)
# The closest university to each store
min_distance = df.loc[df.groupby('Store')['Distance'].idxmin(), 'Distance']
# Pulling everything together
min_distance.to_frame().join(da_store, on='Store').join(da_univ, on='Univ') \
[['Restaurant Name', 'INSTNM', 'Distance']]
结果：
Restaurant Name INSTNM Distance
Store Univ
0 1 Weymouth Dual New England College of Business and Finance 15.651923
1 4 Somerset Dual Bay State College 68.921108
2 3 Westboro Mass Pike West Bancroft School of Massage Therapy 9.580468
3 2 Charlton Mass Pike East Assumption College 26.514269
4 2 Charlton Mass Pike West Assumption College 26.508821