我有一個資料檔案，其中包含一堆帶有“代碼”名稱和“組”字母的位置坐標：
code, x, y, group, n, importance
x1, 4.5692, 2.747, P, 1, 3
x2, 2.2551, 16.7944, H, 2, 1
x3, 15.761, 15.1464, R, 3, 4
x4, 15.6516, 5.1249, C, 4, 5
x5, 6.0939, 9.8601, S, 5, 2
...
我正在嘗試撰寫一個函式，該函式輸出與輸入位置距離最小的位置字典：
def search(inputFile,LocationCode,radius):
from math import sin, cos, sqrt, atan2, radians
data = dict()
with open(inputFile) as file:
next(file)
readin = file.readlines()
for lines in readin:
line = lines.strip('\n')
(code, x, y, group,n,importance) = line.split(",")
data[code] = x,y,group,n,importance
float_rad = float(radius)
to_search = []
for i in LocationCode:
find = data.get(i)
to_search.append(find)
search_x = []
search_y = []
for j in range(len(to_search)):
sx = float(to_search[j][0])
sy = float(to_search[j][1])
search_x.append(sx)
search_y.append(sy)
search_coord = list(zip(search_x,search_y))
search_result = []
distance = []
for k in range(len(search_coord)):
result_in_loop = []
for i in data:
all_x = float(data[i][0])
all_y = float(data[i][1])
if ((all_x - search_coord[k][0])**2 (all_y - search_coord[k][1])**2) < (float_rad**2):
result_in_loop.append(data[i])
# Here starts to find distance & minimum...
distance_in_loop = []
for o in range(len(result_in_loop)):
dx = float(result_in_loop[o][0])
dy = float(result_in_loop[o][1])
if dx != search_coord[k][0] or dy != search_coord[k][1]:
distance_in_loop.append(sqrt((dx - search_coord[k][0])**2 (dy - search_coord[k][1])**2))
distance.append(distance_in_loop)
testing = []
for check in range(len(distance)):
lmin = min(distance[check])
testing.append(lmin)
search_result.append(result_in_loop)
#Test_run
search('sample.csv',["x26", "x52"],3.5)
基本上，在函式中首先讀入inputFile并以“代碼”作為鍵的字典形式設定它。
然后 dict.get()locationCode使用 for 回圈輸入的關聯值，然后將它們附加到to_search串列中。
接下來獲得x和y協調來自值to_search，并將它們添加到串列中search_x，并search_y隨后在最后拉鏈他們到一個名為清單search_coord（格式為類似于此：[(4.5692,2.747),(2.2551, 16.7944)]）
Then it starts to search locations that are within the input radius with coordinates of the input LocationCode (which are in search_coord) being the centers, using firstly a for loop on search_coord then a nested loop on the original data dictionary that runs the calculation to find locations within the radius. The data locations that are within the radius will then put into result_in_loop.
Later I made it to calculate distances of locations from the central input coordinates, append to distance_in_loop then finally put lists of values to distance outside the for k in range(len(search_coord)) loop which did return me a list such as [[dist_1,dist_2,dist_3...],[dist_1,dist_2,dist_3...]].
But now I am not sure how to create dictionaries that find minimum distance in distance then use data's group to group the locations values and in forms of something such as this:
>> [{'H': ('x7', 2.3034), 'R': ('x81', 0.7736), 'C': ('x99', 2.0607), 'S': ('x65', 1.556)}, {'P': ('x46', 2.4717), 'H': ('x22', 1.4374), 'R': ('x88', 2.5338), 'S': ('x30', 2.0482)}]
換句話說，試圖使函式回傳包含以下形式的字典的串列：
{ "group" : ("code" , the_distance) }
而且，已經嘗試過使用min()上distance卻是迷茫于如何使一個關系到他們的group（以使distance相關值group和x,y值在原來的data字典）...
uj5u.com熱心網友回復：
如果您可以將檔案中的資料打包成如下結構，這里有相對簡單的代碼來解決您的問題：
{'group_name':('item_name',x,y)} -> 以 x,y 為坐標
然后您可以使用以下功能：
def find_min_distances(coordinates,data_dict):
dx,dy=coordinates
result={}
for k, v in data_dict.items():
for i in v:
d=((i[1]-dx)**2 (i[2]-dy)**2)**0.5
print(d,i[0],k) # just to control the result print can be removed later on!
if k in result:
if result[k][1]>d: # >= may make also sense depending what you want
result[k]=(i[0],d)
else:
result[k]=(i[0],d)
return result
我測驗了：
b={'H':[('x2',3.4,8.9),('x3',4.9,18.9)],'B':[('x4',1.4,1.9),('x5',14.9,8.9)],'C':[('x6',1.4,8.9),('x7',4.9,18.9)]}
print(find_min_distances((1,2),b))
交付：
{'C': ('x6', 6.911584478250989), 'B': ('x4', 0.41231056256176596), 'H': ('x2', 7.305477397131552)}
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