Python 熊猫-根据其他行中的相对值计算新列

数据如下

data = """
Class,Location,Long,Lat
A,ABC11,139.6295542,35.61144069
A,ABC20,139.630596,35.61045559
A,ABC03,139.6300307,35.61327781
B,ABC54,139.7787818,35.68847945
B,ABC05,139.7814447,35.6816882
B,ABC06,139.7788191,35.681865
B,ABC24,139.7790396,35.67781697
"""
df = pd.read_csv(StringIO(data))

每行包含与位置相关的数据。对于每个位置，需要按如下方式查找到其他位置（行）的距离（为方便简化）

如果是在熊猫之外做的，我会做如下

import math

rows = df.to_dict('records')

# distance of each location w.r.t other locations excluding self
results = {}
for row in rows:
    loc = row['Location']
    results[loc] = {}
    # get a new list excl the curr row
    nrows = [row for row in rows if row['Location'] != loc]
    for nrow in nrows:
        dist = math.sqrt((row["Long"] - nrow["Long"])**2 + (row["Lat"] - nrow["Lat"])**2)
        results[loc][nrow["Location"]] = dist

# find the location with min distance 
fin_results = {}
for k, v in results.items():
    fin_results[k] = {}
    minValKey = min(v, key = v.get)
    fin_results[k]["location"] = minValKey 
    fin_results[k]["dist"] = v[minValKey]

这将给出一个如下的输出，对于每个位置，该输出将给出距离该位置最近的位置和距离

{'ABC11': {'location': 'ABC20', 'dist': 0.001433795400325211}, 'ABC20': {'location': 'ABC11', 'dist': 0.001433795400325211}, 'ABC03': {'location': 'ABC11', 'dist': 0.001897909941062068}, 'ABC54': {'location': 'ABC06', 'dist': 0.006614555169662396}, 'ABC05': {'location': 'ABC06', 'dist': 0.002631545857463665}, 'ABC06': {'location': 'ABC05', 'dist': 0.002631545857463665}, 'ABC24': {'location': 'ABC06', 'dist': 0.004054030973106164}}

虽然这在功能上可行，但我想知道这样做的

pandas

方法是什么

期望输出

+----------+-------------------+----------------------------+
| location |  nearest_location |  nearest_location_distance |
+----------+-------------------+----------------------------+
| 'ABC11'  | 'ABC20'           | 0.001433795400325211       |
| 'ABC20'  | 'ABC11'           | 0.001433795400325211       |
| 'ABC03'  | 'ABC11'           | 0.001897909941062068       |
| 'ABC54'  | 'ABC06'           | 0.006614555169662396       |
| 'ABC05'  | 'ABC06'           | 0.002631545857463665       |
| 'ABC06'  | 'ABC05'           | 0.002631545857463665       |
| 'ABC24'  | 'ABC06'           | 0.004054030973106164       |
+----------+-------------------+----------------------------+

---

您可以使用

numpy

广播

long_ = df.Long.to_numpy()
lat   = df.Lat.to_numpy() 

distances = np.sqrt((long_ - long_[:, None]) ** 2 + (lat - lat[:,None]) **2)

dist_df = pd.DataFrame(distances, index=df.Location, columns=df.Location)

输出数据帧类似于

        nearest_location  nearest_location_distance
Location                                            
ABC11               ABC20                   0.001434
ABC20               ABC11                   0.001434
ABC03               ABC11                   0.001898
ABC54               ABC06                   0.006615
ABC05               ABC06                   0.002632
ABC06               ABC05                   0.002632
ABC24               ABC06                   0.004054

---

---

这将找到从一行到所有其他行的距离。我就是这样解释这个问题的，不知道你的目标是不是。

你可以使用

numpy

广播

long_ = df.Long.to_numpy()
lat   = df.Lat.to_numpy() 

distances = np.sqrt((long_ - long_[:, None]) ** 2 + (lat - lat[:,None]) **2)

dist_df = pd.DataFrame(distances, index=df.Location, columns=df.Location)

输出数据帧类似于

        nearest_location  nearest_location_distance
Location                                            
ABC11               ABC20                   0.001434
ABC20               ABC11                   0.001434
ABC03               ABC11                   0.001898
ABC54               ABC06                   0.006615
ABC05               ABC06                   0.002632
ABC06               ABC05                   0.002632
ABC24               ABC06                   0.004054

---

---

这将找到从一行到所有其他行的距离。这就是我对这个问题的解释，我不确定你的目标是不是。

当安瑟夫提出同样的解决方案时，我完成了一点

import pandas as pd 
from io import StringIO

df = pd.read_csv(StringIO(data))
df['result']= (df['Lat'].diff(-1).pow(2)+df['Long'].diff(-1).pow(2)).pow(1/2)

---

随着ansev提出同样的解决方案，这项工作已经完成了一点

import pandas as pd 
from io import StringIO

df = pd.read_csv(StringIO(data))
df['result']= (df['Lat'].diff(-1).pow(2)+df['Long'].diff(-1).pow(2)).pow(1/2)

---

您可以使用

scipy

的

距离矩阵

，这实际上就是@rafaelc编码的：

from scipy.spatial import distance_matrix

dist_mat = distance_matrix(df[['Long','Lat']],df[['Long','Lat']])

# assign distance matrix with appropriate name
dist_mat = pd.DataFrame(dist_mat, 
                        index=df.Location, 
                        columns=df.Location)

# convert the data frame to dict
(dist_mat.where(dist_mat>0)
     .agg(('idxmin', 'min'))
     .to_dict()
)

输出：

{'ABC11': {'idxmin': 'ABC20', 'min': 0.001433795400325211},
 'ABC20': {'idxmin': 'ABC11', 'min': 0.001433795400325211},
 'ABC03': {'idxmin': 'ABC11', 'min': 0.001897909941062068},
 'ABC54': {'idxmin': 'ABC06', 'min': 0.006614555169662396},
 'ABC05': {'idxmin': 'ABC06', 'min': 0.002631545857463665},
 'ABC06': {'idxmin': 'ABC05', 'min': 0.002631545857463665},
 'ABC24': {'idxmin': 'ABC06', 'min': 0.004054030973106164}}    ​

      idxmin         min
ABC11  ABC20   0.0014338
ABC20  ABC11   0.0014338
ABC03  ABC11  0.00189791
ABC54  ABC06  0.00661456
ABC05  ABC06  0.00263155
ABC06  ABC05  0.00263155
ABC24  ABC06  0.00405403

如果只需要数据帧：

(dist_mat.where(dist_mat>0)
     .agg(('idxmin', 'min'))
     .T
)

输出：

{'ABC11': {'idxmin': 'ABC20', 'min': 0.001433795400325211},
 'ABC20': {'idxmin': 'ABC11', 'min': 0.001433795400325211},
 'ABC03': {'idxmin': 'ABC11', 'min': 0.001897909941062068},
 'ABC54': {'idxmin': 'ABC06', 'min': 0.006614555169662396},
 'ABC05': {'idxmin': 'ABC06', 'min': 0.002631545857463665},
 'ABC06': {'idxmin': 'ABC05', 'min': 0.002631545857463665},
 'ABC24': {'idxmin': 'ABC06', 'min': 0.004054030973106164}}    ​

      idxmin         min
ABC11  ABC20   0.0014338
ABC20  ABC11   0.0014338
ABC03  ABC11  0.00189791
ABC54  ABC06  0.00661456
ABC05  ABC06  0.00263155
ABC06  ABC05  0.00263155
ABC24  ABC06  0.00405403

---

您可以使用

scipy

的

距离矩阵

，这实际上就是@rafaelc编码的：

from scipy.spatial import distance_matrix

dist_mat = distance_matrix(df[['Long','Lat']],df[['Long','Lat']])

# assign distance matrix with appropriate name
dist_mat = pd.DataFrame(dist_mat, 
                        index=df.Location, 
                        columns=df.Location)

# convert the data frame to dict
(dist_mat.where(dist_mat>0)
     .agg(('idxmin', 'min'))
     .to_dict()
)

输出：

{'ABC11': {'idxmin': 'ABC20', 'min': 0.001433795400325211},
 'ABC20': {'idxmin': 'ABC11', 'min': 0.001433795400325211},
 'ABC03': {'idxmin': 'ABC11', 'min': 0.001897909941062068},
 'ABC54': {'idxmin': 'ABC06', 'min': 0.006614555169662396},
 'ABC05': {'idxmin': 'ABC06', 'min': 0.002631545857463665},
 'ABC06': {'idxmin': 'ABC05', 'min': 0.002631545857463665},
 'ABC24': {'idxmin': 'ABC06', 'min': 0.004054030973106164}}    ​

      idxmin         min
ABC11  ABC20   0.0014338
ABC20  ABC11   0.0014338
ABC03  ABC11  0.00189791
ABC54  ABC06  0.00661456
ABC05  ABC06  0.00263155
ABC06  ABC05  0.00263155
ABC24  ABC06  0.00405403

如果只需要数据帧：

(dist_mat.where(dist_mat>0)
     .agg(('idxmin', 'min'))
     .T
)

输出：

{'ABC11': {'idxmin': 'ABC20', 'min': 0.001433795400325211},
 'ABC20': {'idxmin': 'ABC11', 'min': 0.001433795400325211},
 'ABC03': {'idxmin': 'ABC11', 'min': 0.001897909941062068},
 'ABC54': {'idxmin': 'ABC06', 'min': 0.006614555169662396},
 'ABC05': {'idxmin': 'ABC06', 'min': 0.002631545857463665},
 'ABC06': {'idxmin': 'ABC05', 'min': 0.002631545857463665},
 'ABC24': {'idxmin': 'ABC06', 'min': 0.004054030973106164}}    ​

      idxmin         min
ABC11  ABC20   0.0014338
ABC20  ABC11   0.0014338
ABC03  ABC11  0.00189791
ABC54  ABC06  0.00661456
ABC05  ABC06  0.00263155
ABC06  ABC05  0.00263155
ABC24  ABC06  0.00405403

您还可以使用：

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您还可以使用：

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