如何计算从一个数据帧到另一个数据帧的距离?

2024-05-06 02:41:02 发布

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假设我有一个由点组成的数据帧:

df1型:

x   y  z     label
1.1 2.1 3.1     2
4.1 5.1 6.1     1
7.1 8.1 9.1     0

我还有另一个点的数据帧:

df2型:

x  y  z   label
4  5  6    0
7  8  9    1
1  2  3    2

是否有任何方法可以通过df one查看哪个点最接近df2的内部,并将标签替换为最接近so的点的标签。。你知道吗

结果我想:

x   y    z  label
1.1 2.1 3.1   2
4.1 5.1 6.1   0
7.1 8.1 9.1   1

谢谢你阅读我的问题!你知道吗


Tags: 数据方法dfso标签onelabeldf1
3条回答

我只能从scipy想到distance

from scipy.spatial import distance
df1['label']=df2.label.iloc[distance.cdist(df1.iloc[:,:-1], df2.iloc[:,:-1], metric='euclidean').argmin(1)].values
df1
Out[446]: 
     x    y    z  label
0  1.1  2.1  3.1      2
1  4.1  5.1  6.1      0
2  7.1  8.1  9.1      1

这是一个使用kd-trees的版本,对于大型数据集来说这可能要快得多。你知道吗

import numpy as np
import pandas as pd
from  sklearn.neighbors import KDTree
np.random.seed(0)
#since you have df1 and df2, you will want to convert the dfs to array here with
#X=df1['x'.'y','z'].to_numpy()
#Y=df2['x','y','z'].to_numpy()
X = np.random.random((10, 3))  # 10 points in 3 dimensions
Y = np.random.random((10, 3))
tree = KDTree(Y, leaf_size=2)  


#loop though the x array and find the closest point in y to each x  
#note the you can find as many as k nearest neighbors by this method
#though yours only calls for the k=1 case
dist, ind = tree.query(X, k=1) 

df1=pd.DataFrame(X, columns=['x','y','z']) 

#set the labels to the closest point to each neighbor
df1['label']=ind 

#this is cheesy, but it removes the list brackets 
#get rid of the following line if you want more than k=1 nearest neighbors
df1['label']=df1['label'].str.get(0).str.get(0)  
print(df1)

df1:
          x         y         z
0  0.548814  0.715189  0.602763
1  0.544883  0.423655  0.645894
2  0.437587  0.891773  0.963663
3  0.383442  0.791725  0.528895
4  0.568045  0.925597  0.071036
5  0.087129  0.020218  0.832620
6  0.778157  0.870012  0.978618
7  0.799159  0.461479  0.780529
8  0.118274  0.639921  0.143353
9  0.944669  0.521848  0.414662
df2:
          x         y         z
0  0.264556  0.774234  0.456150
1  0.568434  0.018790  0.617635
2  0.612096  0.616934  0.943748
3  0.681820  0.359508  0.437032
4  0.697631  0.060225  0.666767
5  0.670638  0.210383  0.128926
6  0.315428  0.363711  0.570197
7  0.438602  0.988374  0.102045
8  0.208877  0.161310  0.653108
9  0.253292  0.466311  0.244426

Out:
          x         y         z  label
0  0.548814  0.715189  0.602763      0
1  0.544883  0.423655  0.645894      6
2  0.437587  0.891773  0.963663      2
3  0.383442  0.791725  0.528895      0
4  0.568045  0.925597  0.071036      7
5  0.087129  0.020218  0.832620      8
6  0.778157  0.870012  0.978618      2
7  0.799159  0.461479  0.780529      2
8  0.118274  0.639921  0.143353      9
9  0.944669  0.521848  0.414662      3

这是一张你可以用来检查结果的图片。蓝色的点是x点,橙色的点是y点。 enter image description here

下面是使用matplotlib版本3.0.2的绘图代码

fig = plt.figure()
ax = Axes3D(fig)
ax.scatter(X[:,0],X[:,1],X[:,2])
ax.scatter(Y[:,0],Y[:,1],Y[:,2])
for i in range(len(X)): #plot each point + it's index as text above
    ax.text(X[i,0],X[i,1],X[i,2],  '%s' % (str(i)), size=20, zorder=1, color='blue') 
for i in range(len(Y)): #plot each point + it's index as text above
    ax.text(Y[i,0],Y[i,1],Y[i,2],  '%s' % (str(i)), size=20, zorder=1, color='orange') 
SELECT ABS($df1 - $df2) as nearest, ...
FROM yourtable
ORDER BY nearest ASC
LIMIT 1

按“X”索引对它们排序,然后比较$result数组 这将查找表之间最近的数字。你知道吗

https://www.w3schools.com/sql/func_sqlserver_abs.asp ABS函数返回一个绝对数,所以只要在df2上有整数,它就是一个很好的解决方案。你知道吗

希望有帮助。你知道吗

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