我有一个数据框：
routeId latitude_value longitude_value
r1 28.210216 22.813209
r2 28.216103 22.496735
r3 28.161786 22.842318
r4 28.093110 22.807081
r5 28.220370 22.503500
r6 28.220370 22.503500
r7 28.220370 22.503500
从此我想生成一个像这样的数据框 df2 ：
routeId nearest
r1 r3 (for example)
r2 ... similarly for all the routes.
我想要实现的逻辑是
对于每条路线，我应该找到所有其他路线的欧氏距离。 并在routeId上迭代它。
有一个计算欧氏距离的功能。
dist = math.hypot(x2 - x1, y2 - y1)
但我很困惑如何构建一个我将传递数据帧的函数，或者使用.apply（）
def get_nearest_route():
.....
return df2
答案 0 :(得分：6)
我建议使用pdist function from scipy.spatial.distance。
matrix = scipy.spatial.distance.pdist(df[['latitude_value', 'longitude_value']], metric='euclidean')
将返回形状（n，）的压缩距离矩阵，并计算所有成对距离。
然后你可以使用squareform来获得方形的成对距离矩阵：
matrix = scipy.spatial.distance.squareform(matrix)
然后，对于每一行matrix[i]，您可以找到索引处的最大值，例如matrix[i][j]你知道，对于第i点，它的最近点是第j点。
答案 1 :(得分：5)
我们可以使用scipy.spatial.distance.cdist或多个for循环然后用路径替换min并找到最接近的，即
mat = scipy.spatial.distance.cdist(df[['latitude_value','longitude_value']],
df[['latitude_value','longitude_value']], metric='euclidean')
# If you dont want scipy, you can use plain python like
# import math
# mat = []
# for i,j in zip(df['latitude_value'],df['longitude_value']):
# k = []
# for l,m in zip(df['latitude_value'],df['longitude_value']):
# k.append(math.hypot(i - l, j - m))
# mat.append(k)
# mat = np.array(mat)
new_df = pd.DataFrame(mat, index=df['routeId'], columns=df['routeId'])
new_df
routeId r1 r2 r3 r4 r5 r6 r7
routeId
r1 0.000000 0.316529 0.056505 0.117266 0.309875 0.309875 0.309875
r2 0.316529 0.000000 0.349826 0.333829 0.007998 0.007998 0.007998
r3 0.056505 0.349826 0.000000 0.077188 0.343845 0.343845 0.343845
r4 0.117266 0.333829 0.077188 0.000000 0.329176 0.329176 0.329176
r5 0.309875 0.007998 0.343845 0.329176 0.000000 0.000000 0.000000
r6 0.309875 0.007998 0.343845 0.329176 0.000000 0.000000 0.000000
r7 0.309875 0.007998 0.343845 0.329176 0.000000 0.000000 0.000000
#Replace minimum distance with column name and not the minimum with `False`.
# new_df[new_df != 0].min(),0). This gives a mask matching minimum other than zero.
closest = np.where(new_df.eq(new_df[new_df != 0].min(),0),new_df.columns,False)
# Remove false from the array and get the column names as list .
df['close'] = [i[i.astype(bool)].tolist() for i in closest]
routeId latitude_value longitude_value close
0 r1 28.210216 22.813209 [r3]
1 r2 28.216103 22.496735 [r5, r6, r7]
2 r3 28.161786 22.842318 [r1]
3 r4 28.093110 22.807081 [r3]
4 r5 28.220370 22.503500 [r2]
5 r6 28.220370 22.503500 [r2]
6 r7 28.220370 22.503500 [r2]
如果您不想忽略零，那么
# Store the array values in a variable
arr = new_df.values
# We dont want to find mimimum to be same point, so replace diagonal by nan
arr[np.diag_indices_from(new_df)] = np.nan
# Replace the non nan min with column name and otherwise with false
new_close = np.where(arr == np.nanmin(arr, axis=1)[:,None],new_df.columns,False)
# Get column names ignoring false.
df['close'] = [i[i.astype(bool)].tolist() for i in new_close]
routeId latitude_value longitude_value close
0 r1 28.210216 22.813209 [r3]
1 r2 28.216103 22.496735 [r5, r6, r7]
2 r3 28.161786 22.842318 [r1]
3 r4 28.093110 22.807081 [r3]
4 r5 28.220370 22.503500 [r6, r7]
5 r6 28.220370 22.503500 [r5, r7]
6 r7 28.220370 22.503500 [r5, r6]
答案 2 :(得分：2)
无法与scipy.spatial.distance.pdist进行比较，但仍可正常使用
from itertools import product
import pandas as pd
import math
df['New']=list(zip(df['latitude_value'],df['longitude_value']))
DF=pd.DataFrame(list(product(df.routeId, df.routeId)), columns=['l1', 'l2'])
New=df[['routeId','New']].merge(DF,left_on='routeId',right_on='l1',how='left').merge(df[['routeId','New']],left_on='l2',right_on='routeId')
New['Cal']=New.apply(lambda x : math.hypot(x.New_x[0] - x.New_y[0] , x.New_x[1] - x.New_y[1] ),axis=1)
New=New.loc[New.l1!=New.l2,:]
New.sort_values('Cal').drop_duplicates(['l1'],keep='first')
Out[386]:
routeId_x New_x l1 l2 routeId_y New_y Cal
47 r6 (28.22037, 22.5035) r6 r7 r7 (28.22037, 22.5035) 0.000000
41 r7 (28.22037, 22.5035) r7 r6 r6 (28.22037, 22.5035) 0.000000
39 r5 (28.22037, 22.5035) r5 r6 r6 (28.22037, 22.5035) 0.000000
43 r2 (28.216103, 22.496735) r2 r7 r7 (28.22037, 22.5035) 0.007998
2 r3 (28.161786, 22.842318) r3 r1 r1 (28.210216, 22.813209) 0.056505
14 r1 (28.210216, 22.813209) r1 r3 r3 (28.161786, 22.842318) 0.056505
17 r4 (28.09311, 22.807081) r4 r3 r3 (28.161786, 22.842318) 0.077188
答案 3 :(得分：0)
您可以使用itertuples来迭代DataFrame
data = {index: ((df['latitude_value'] - lat)**2 + (df['longitude_value'] - long)**2).drop(index).argmin() for index, lat, long in df.itertuples()}
pd.Series(data)
大型数据集可能需要很长时间