Pandas：基于多列对数据表的行运行计算并将输出存储在新列中

Question

我正在尝试计算与 2 个位置的距离，并且我已经获得了两个目的地的经度和纬度。在我的 CSV 中，我有 4 列（lat1、lon1、lat2、lon2），如何应用下面的代码，以便创建名为“距离”的第 5 列，并使用下面的代码计算距离？

import math
from math import sin, cos, sqrt, atan2, radians

# approximate radius of earth in km
R = 6373.0

#Test
lat1 = radians(25.2296756)
lon1 = radians(36.0122287)
lat2 = radians(51.406374)
lon2 = radians(20.9251681)

dlon = lon2 - lon1
dlat = lat2 - lat1

a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))

distance = R * c

print("Result:", distance)
print("Should be:", 3181.11, "km")

数据框：

df = pd.DataFrame({'Normalised': {(0, 'London,', 'United', 'Kingdom'): '-',
  (1, 'Johannesburg,', 'South', 'Africa'): '-',
  (2, 'London,', 'United', 'Kingdom'): '-',
  (3, 'Johannesburg,', 'South', 'Africa'): '-',
  (4, 'London,', 'United', 'Kingdom'): '-'},
 'City': {(0, 'London,', 'United', 'Kingdom'): 'New',
  (1, 'Johannesburg,', 'South', 'Africa'): 'London,',
  (2, 'London,', 'United', 'Kingdom'): 'New',
  (3, 'Johannesburg,', 'South', 'Africa'): 'London,',
  (4, 'London,', 'United', 'Kingdom'): 'Singapore,'},
 'Pair': {(0, 'London,', 'United', 'Kingdom'): 'York,',
  (1, 'Johannesburg,', 'South', 'Africa'): 'United',
  (2, 'London,', 'United', 'Kingdom'): 'York,',
  (3, 'Johannesburg,', 'South', 'Africa'): 'United',
  (4, 'London,', 'United', 'Kingdom'): 'Singapore'},
 'Departure': {(0, 'London,', 'United', 'Kingdom'): 'United',
  (1, 'Johannesburg,', 'South', 'Africa'): 'Ki...',
  (2, 'London,', 'United', 'Kingdom'): 'United',
  (3, 'Johannesburg,', 'South', 'Africa'): 'Ki...',
  (4, 'London,', 'United', 'Kingdom'): 'SIN'},
 'Code': {(0, 'London,', 'United', 'Kingdom'): 'Stat.',
  (1, 'Johannesburg,', 'South', 'Africa'): 'JNB',
  (2, 'London,', 'United', 'Kingdom'): 'Stat',
  (3, 'Johannesburg,', 'South', 'Africa'): 'JNB',
  (4, 'London,', 'United', 'Kingdom'): 'LHR'},
 'Arrival': {(0, 'London,', 'United', 'Kingdom'): 'LHR',
  (1, 'Johannesburg,', 'South', 'Africa'): 'LHR',
  (2, 'London,', 'United', 'Kingdom'): 'LHR',
  (3, 'Johannesburg,', 'South', 'Africa'): 'LHR',
  (4, 'London,', 'United', 'Kingdom'): '1.3'},
 'Code.1': {(0, 'London,', 'United', 'Kingdom'): 'JFK',
  (1, 'Johannesburg,', 'South', 'Africa'): '-26.1',
  (2, 'London,', 'United', 'Kingdom'): 'JFK',
  (3, 'Johannesburg,', 'South', 'Africa'): '-26.1',
  (4, 'London,', 'United', 'Kingdom'): '103.98'},
 'Departure_lat': {(0, 'London,', 'United', 'Kingdom'): 51.5,
  (1, 'Johannesburg,', 'South', 'Africa'): 28.23,
  (2, 'London,', 'United', 'Kingdom'): 51.5,
  (3, 'Johannesburg,', 'South', 'Africa'): 28.23,
  (4, 'London,', 'United', 'Kingdom'): 51.47},
 'Departure_lon': {(0, 'London,', 'United', 'Kingdom'): -0.45,
  (1, 'Johannesburg,', 'South', 'Africa'): 51.47,
  (2, 'London,', 'United', 'Kingdom'): -0.45,
  (3, 'Johannesburg,', 'South', 'Africa'): 51.47,
  (4, 'London,', 'United', 'Kingdom'): -0.45},
 'Arrival_lat': {(0, 'London,', 'United', 'Kingdom'): 40.64,
  (1, 'Johannesburg,', 'South', 'Africa'): -0.45,
  (2, 'London,', 'United', 'Kingdom'): 40.64,
  (3, 'Johannesburg,', 'South', 'Africa'): -0.45,
  (4, 'London,', 'United', 'Kingdom'): np.nan},
 'Arrival_lon': {(0, 'London,', 'United', 'Kingdom'): -73.79,
  (1, 'Johannesburg,', 'South', 'Africa'): np.nan,
  (2, 'London,', 'United', 'Kingdom'): -73.79,
  (3, 'Johannesburg,', 'South', 'Africa'): np.nan,
  (4, 'London,', 'United', 'Kingdom'): np.nan}})

Answer 1

您可以为距离计算定义自定义函数。然后，使用.apply()调用每一行的函数，得到每一行的距离。

1.定义一个自定义函数进行距离计算，如下：

import math
from math import sin, cos, sqrt, atan2, radians

def get_distance(in_lat1, in_lon1, in_lat2, in_lon2):
    # approximate radius of earth in km
    R = 6373.0

    lat1 = radians(in_lat1)
    lon1 = radians(in_lon1)
    lat2 = radians(in_lat2)
    lon2 = radians(in_lon2)

    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
    c = 2 * atan2(sqrt(a), sqrt(1 - a))

    distance = R * c

    return distance

2。使用.apply()在每一行调用并应用函数，得到每一行的距离，如下：

df['Distance'] = df.apply(lambda x: get_distance(x['Departure_lat'], x['Departure_lon'], x['Arrival_lat'], x['Arrival_lon']), axis=1)

演示

输入数据框

        City  Departure_lat  Departure_lon  Arrival_lat  Arrival_lon
0  CityName1      25.229676      36.012229    51.406374    20.925168

输出

        City  Departure_lat  Departure_lon  Arrival_lat  Arrival_lon    Distance
0  CityName1      25.229676      36.012229    51.406374    20.925168  3181.11039

Answer 2

您可以将 dlon、dlat、a 和 c 都设置为一些临时列，然后从那里计算（或者将它们全部放在一个难以阅读的行中）。

类似：

df['dlon'] = df['Arrival_lon'] - df['Departure_lon']
df['dlat'] = df['Arrival_lat'] - df['Departure_lat']

df['a'] = sin(df['dlat'] / 2)**2 + cos(df['Departure_lat']) * cos(df['Arrival_lat']) * sin(df['dlon'] / 2)**2
df['c'] = 2 * atan2(sqrt(df['a']), sqrt(1 - df['a']))

df['distance'] = R * df['c']

然后，如果需要，您可以 .drop() 所有这些额外的列，但这应该创建 df['distance'] 作为为每一行计算的新列。

如果我在该代码中出现拼写错误，我不会感到惊讶，但希望您能理解。每一行 df[xxx] = 都会构成新列。

Answer 3

你没有提供数据，所以我根据你的问题自己编了；只需在您的列上使用这些函数的numpy 版本。

import pandas as pd
import numpy as np

row = pd.Series({
    "lat1": 25.2296756,
    "lon1": 36.0122287,
    "lat2": 51.406374,
    "lon2": 20.9251681
})
df = pd.concat([row]*5, axis=1).T.apply(np.radians)

df["dlon"] = df.lon2 - df.lon1
df["dlat"] = df.lat2 - df.lat1

R = 6373
a = np.sin(df.dlat / 2)**2 + np.cos(df.lat1) * np.cos(df.lat2) * np.sin(df.dlon / 2)**2
c = 2 * np.arctan2(np.sqrt(a), np.sqrt(1 - a))
df["distance"] = R*c

生成的数据框如下所示：

       lat1      lon1     lat2      lon2      dlon     dlat    distance
0  0.440341  0.628532  0.89721  0.365213 -0.263319  0.45687  3181.11039
1  0.440341  0.628532  0.89721  0.365213 -0.263319  0.45687  3181.11039
2  0.440341  0.628532  0.89721  0.365213 -0.263319  0.45687  3181.11039
3  0.440341  0.628532  0.89721  0.365213 -0.263319  0.45687  3181.11039
4  0.440341  0.628532  0.89721  0.365213 -0.263319  0.45687  3181.11039

Answer 4

你可以把你的计算代码放在一个函数中：

def calculate_distance(lat1,lon1,lat2,lon2):
  # approximate radius of earth in km
  R = 6373.0

  lat1 = radians(lat1)
  lon1 = radians(lon1)
  lat2 = radians(lat2)
  lon2 = radians(lon2)

  dlon = lon2 - lon1
  dlat = lat2 - lat1

  a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
  c = 2 * atan2(sqrt(a), sqrt(1 - a))

  distance = R * c

  return distance

然后用列表理解将其应用到每一行：

df['distance'] = [calculate_distance(row.lat1, row.lon1, row.lat2, row.lon2) for row in df.itertuples() ]

Answer 5

根据您的数据 csv 文件的格式，可以使用类似于以下内容的内容。

本质上，您需要将计算转换为可调用函数，然后在数据文件中的每一行上调用它，然后使用 csv 库将其导入 python。

import math
import csv # Added import for importing csv into python.
from math import sin, cos, sqrt, atan2, radians

# Import the data from the csv file.
with open('data.csv', newline='') as csvfile:
    data = list(csv.reader(csvfile))
    
# Approximate radius of earth in km.
R = 6373.0

# Create a distance calculation function.
def calculate_distance(lat1_d, lon1_d, lat2_d, lon2_d):
    
    # Convert from degrees to radians.
    lat1 = radians(lat1_d)
    lon1 = radians(lon1_d)
    lat2 = radians(lat2_d)
    lon2 = radians(lon2_d)

    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
    c = 2 * atan2(sqrt(a), sqrt(1 - a))
    
    distance = R * c
    return distance
    
# Use list comprehension to run function on every data row.
distances = [calculate_distance(row[0],row[1],row[2],row[3]) for row in data]

# Append distance column to original array to create output.
output = [row + [distance[index]] for index,row in enumerate(data)]

请注意，row[0],row[1],row[2],row[3] 指的是数据数组/csv 文件中列的顺序。这些可能需要根据需要重新排序，以符合函数声明的输入顺序，即：lat1_d, lon1_d, lat2_d, lon2_d。

# Import the data from the csv file.
with open('data.csv', newline='') as csvfile:
    data = list(csv.reader(csvfile))

还需要调整这些导入参数以考虑 csv 文件的格式和名称。