我正在开发一个如下所示的数据框架:
lat lon
id_zone
0 40.0795 4.338600
1 45.9990 4.829600
2 45.2729 2.882000
3 45.7336 4.850478
4 45.6981 5.043200我在试着做一个Haverisne距离矩阵。基本上,对于每个区域,我想计算它和数据中所有其他区域之间的距离。所以对角线上应该只有0。这是我使用的Haversine函数,但是我不能建立我的矩阵。
def haversine(x):
x.lon, x.lat, x.lon2, x.lat2 = map(radians, [x.lon, x.lat, x.lon2, x.lat2])
# formule de Haversine
dlon = x.lon2 - x.lon
dlat = x.lat2 - x.lat
a = sin(dlat / 2) ** 2 + cos(x.lat) * cos(x.lat2) * sin(dlon / 2) ** 2
c = 2 * atan2(sqrt(a), sqrt(1 - a))
km = 6367 * c
return km发布于 2022-04-01 08:52:24
您可以使用这个答案的解决方案Pandas - Creating Difference Matrix from Data Frame
或者在您的具体案例中,您有一个类似于本例的DataFrame:
lat lon
id_zone
0 40.0795 4.338600
1 45.9990 4.829600
2 45.2729 2.882000
3 45.7336 4.850478
4 45.6981 5.043200您的功能被定义为:
def haversine(first, second):
# convert decimal degrees to radians
lat, lon, lat2, lon2 = map(np.radians, [first[0], first[1], second[0], second[1]])
# haversine formula
dlon = lon2 - lon
dlat = lat2 - lat
a = np.sin(dlat/2)**2 + np.cos(lat) * np.cos(lat2) * np.sin(dlon/2)**2
c = 2 * np.arcsin(np.sqrt(a))
r = 6371 # Radius of earth in kilometers. Use 3956 for miles
return c * r其中传递lat和lon的first位置和second位置。
然后,您可以使用Numpy创建一个距离矩阵,然后用haversine函数的距离结果替换零:
# create a matrix for the distances between each pair of zones
distances = np.zeros((len(df), len(df)))
for i in range(len(df)):
for j in range(len(df)):
distances[i, j] = haversine(df.iloc[i], df.iloc[j])
pd.DataFrame(distances, index=df.index, columns=df.index)您的输出应该类似于以下内容:
id_zone 0 1 2 3 4
id_zone
0 0.000000 659.422944 589.599339 630.083979 627.383858
1 659.422944 0.000000 171.597296 29.555376 37.325316
2 589.599339 171.597296 0.000000 161.731366 174.983855
3 630.083979 29.555376 161.731366 0.000000 15.474533
4 627.383858 37.325316 174.983855 15.474533 0.000000https://stackoverflow.com/questions/71703229
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