Question

问题陈述：

一个不愿透露姓名的游客在纽约迷路了。他只有一张 M 的地图地铁站，它显示了车站及其坐标自己的坐标，他在附近的指针上看到了。游客是不确定每个站是否都打开，因此，以防万一，他正在寻找最近的 N 个电台。游客的举动像每个纽约客（{{3}}）一样穿越纽约市。帮助游客找到这些车站。

示例输入

5 2
А 1 2
B 4.5 1.2
C 100500 100500
D 100501 100501
E 100502 100502
1 1

示例输出

A B

我的代码：

import scipy.spatial.distance as d
import math

#finds N nearest metro stations in relation to the tourist
def find_shortest_N(distance_list, name_list, number_of_stations):
    result = []
    for num in range(0, number_of_stations):
        min_val_index = distance_list.index(min(distance_list))
        result.append(name_list[min_val_index])
        distance_list.pop(min_val_index)
        name_list.pop(min_val_index)
    return result

#returns a list with distances between touri and stations
def calculate_nearest(list_of_coords, tourist_coords):
    distances = []
    for metro_coords in list_of_coords:
        distances.append(math.fabs(d.cityblock(metro_coords, tourist_coords)))
    return distances


station_coords = []
station_names = []

input_stations = input("Input a number of stations: ").split()
input_stations = list(map(int, input_stations))

#all station coordinates and their names
station_M = input_stations[0]

#number of stations a tourist wants to visit
stations_wanted_N = input_stations[1]

#distribute the station names in station_names list 
#and the coordinates in station_coords list
for data in range(0, station_M):
    str_input = input()
    list_input = str_input.split()
    station_names.append(list_input[0])
    list_input.pop(0)
    list_input = list(map(float, list_input))
    station_coords.append(list_input)

tourist_coordinates = input("Enter tourist position: ").split()
tourist_coordinates = list(map(float, tourist_coordinates))

distance_values = calculate_nearest(station_coords, tourist_coordinates)

result = find_shortest_N(distance_values, station_names, stations_wanted_N)

for name in result:
    print(name, end=" ")

Answer 1

例如，您还可以直接使用cdist函数：

import numpy as np
from scipy.spatial.distance import cdist

sample_input = '''
5 2
А 1 2
B 4.5 1.2
C 100500 100500
D 100501 100501
E 100502 100502
1 1
'''

# Parsing the input data:
sample_data = [line.split()
                for line in sample_input.strip().split('\n')]

tourist_coords = np.array(sample_data.pop(),  dtype=float) # takes the last line
nbr_stations, nbr_wanted = [int(n) for n in sample_data.pop(0)] # takes the first line

stations_coords = np.array([line[1:] for line in sample_data], dtype=float)
stations_names = [line[0] for line in sample_data]

# Computing the distances:
tourist_coords = tourist_coords.reshape(1, 2)  # have to be a 2D array
distance = cdist(stations_coords, tourist_coords, metric='cityblock')

# Sorting the distances:
sorted_distance = sorted(zip(stations_names, distance), key=lambda x:x[1])

# Result:
result = [name for name, dist in sorted_distance[:nbr_wanted]]
print(result)

Answer 2

使用scipy.spatial.KDTree

mydata <- tibble(ID = 1:4, 
                 Year = c(1999L, 2001L, 2000L, 2001L), 
                 Source_1999 = c("ABC", "ABC", NA, NA), 
                 Source_2000 = c("ABC", "BBB", "ABC", NA), 
                 Source_2001 = c("ABC", "XYZ", "BBB", NA), 
                 Source_2002 = c("ABC", NA, "BBB", NA))

如何使用numpy改进此解决方案以使其更快？

2 个答案: