我有一个列表，用于存储对象之间的距离。
表格如下所示：
+----------+----------+----------+----------+----------+
| | Object_A | Object_B | Object_C | Object_D |
+----------+----------+----------+----------+----------+
| Entity_E | 2 | 3 | 6 | 1 |
+----------+----------+----------+----------+----------+
| Entity_F | 3 | 4 | 7 | 2 |
+----------+----------+----------+----------+----------+
| Entity_G | 9 | 1 | 2 | 3 |
+----------+----------+----------+----------+----------+
数字表示行与行之间的距离。列标题。
这大致计算如下：
entites = [Entity_E, Entity_F, Entity_G]
objects = [Object_A, Object_B, Object_C, Obhect_D]
distances = []
for object in objects:
distanceEntry = []
for entity in entities:
distance = getDistance(entity, object)
distanceEntry.append(distance)
distances.append(distanceEntry)
这给了我大致上表中的信息。
我要做的是基本上找到每个实体最接近的对象（反之亦然）。每个对象或实体只能彼此匹配，并且应该基于邻近度。
我现在这样做的方法是简单地按距离按大小排序嵌套列表（在完整代码中我有办法确定哪个对象与每个距离相关联）。
因此，在这样做时，我会创建以下关联：
+----------+----------+----------+
| Entity | Object | Distance |
+----------+----------+----------+
| Entity_E | Object_D | 1 |
+----------+----------+----------+
| Entity_F | Object_D | 2 |
+----------+----------+----------+
| Entity_G | Object_B | 1 |
+----------+----------+----------+
这是不正确的，因为它将Object_D关联两次。
该协会应该是：
+----------+----------+----------+
| Entity | Object | Distance |
+----------+----------+----------+
| Entity_E | Object_D | 1 |
+----------+----------+----------+
| Entity_F | Object_A | 3 |
+----------+----------+----------+
| Entity_G | Object_B | 1 |
+----------+----------+----------+
这是我正在努力的地方 - 我无法找到为此编码逻辑的最佳方法。
由于Entity_E更接近Object_D，它应该获得关联。所以，对于Entity_F，我需要选择第二个。
我正在考虑做一些事情，我记录哪些对象已经被分配，或者尝试做一些事情，我首先匹配每列中的最小值，但它们似乎都遇到了问题。
我可以使用矩阵运算或某种矩阵运算来进行此计算吗？
任何建议都将不胜感激！
编辑 - 添加完整代码：
# Create an array that stores the distances between each label and symbol. Only calculate the distance for label that
# are "in front" of the symbol.
# Example Table:
# +---------------+--------------+--------------+--------------+--------------+
# | | Label_1 | Label_2 | Label_3 | Label_4 |
# +---------------+--------------+--------------+--------------+--------------+
# | Measurement_1 | 2 | 3 | Not_in_front | 1 |
# +---------------+--------------+--------------+--------------+--------------+
# | Measurement_2 | 3 | 4 | 1 | Not_in_front |
# +---------------+--------------+--------------+--------------+--------------+
# | Measurement_3 | Not_in_front | Not_in_front | 2 | 1 |
# +---------------+--------------+--------------+--------------+--------------+
# Data Structures:
# measurementsDictionary = {['Type', 'Handle', 'X-Coord', 'Y-Coord', 'Z-Coord', 'Rotation', 'True Strike']}
# dipsDictionary = {['Handle', 'Text', 'Unit', 'X-Coord', 'Y-Coord', 'Z-Coord']}
#The two functions below grab the information from a csv-like file.
measurementsDictionary = getMeasurementsInformationFromFile()
dipsDictionary = getDipsInformationFromFile()
exportHeaders = [" "]
exportArray = []
for measurementSymbol in measurementsDictionary:
measurementEntry = measurementsDictionary[measurementSymbol]
measurementCoord = [measurementEntry[2], measurementEntry[3]]
measurementDistance = []
measurementDistance.append(measurementEntry[1])
measurementCartesianAngle = getCartesianAngle(measurementEntry[6])
measurementLineEquation = generateLineEquation(measurementCoord,measurementCartesianAngle)
for dip in dipsDictionary:
dipEntry = dipsDictionary[dip]
dipCoord = [dipEntry[3],dipEntry[4]]
isPointInFrontOfLineTest = isPointInFrontOfLine(measurementCartesianAngle, measurementLineEquation, dipCoord)
if isPointInFrontOfLineTest == 1:
measurementSymbolDistance = calculateDistance(measurementCoord, dipCoord)
# string = dipEntry[0] +":" + str(measurementSymbolDistance)
# measurementDistance.append(string)
measurementDistance.append(measurementSymbolDistance)
elif isPointInFrontOfLineTest == 0:
string = ""
measurementDistance.append(string)
exportArray.append(measurementDistance)
for dip in dipsDictionary:
dipEntry = dipsDictionary[dip]
exportHeaders.append(dipEntry[0])
exportedArray = [exportHeaders] + exportArray
export = np.array(exportedArray)
np.savetxt("exportArray2.csv", export, fmt='%5s', delimiter=",")
print(exportHeaders)
答案 0 :(得分：1)
我为PE345做了一些代码来解决类似的问题。我将使用自己的数组：
[[1,2,3],
[4,5,6],
[7,8,9]]
您想要做的是获得选择特定元素的成本。通过查找选择行的成本，添加选择列的成本，然后减去选择元素本身的成本来做到这一点。因此，在我的数组中，选择元素[0][0]的成本是(1+2+3)+(1+4+7)-3*(1)。我将第0行求和，将第0列相加，然后减去元素本身。
现在您需要选择每个元素。找到成本最高的元素。这将是网格中的最低值。选择它并确保无法选择行或列中的其他值。重复，直到选择了每行和每列中的值。
这是我项目的源代码。请注意，它找到最高值，而不是最低值。
from random import randint
from itertools import permutations
def make_a_grid(radius=5,min_val=0,max_val=999):
"""Return a radius x radius grid of numbers ranging from min_val to max_val.
Syntax: make_a_grid(radius=5,min_val=0,max_val=999
"""
return [[randint(min_val,max_val) for i in xrange(radius)] for j in xrange(radius)]
def print_grid(grid,rjustify = 4):
"""Print a n x m grid of numbers, justified to a column.
Syntax: print_grid(grid,rjustify = 4)
"""
for i in grid:
outstr = ''
for j in i:
outstr += str(j).rjust(rjustify)
print outstr
def brute_search(grid):
"""Brute force a grid to find the maximum sum in the grid."""
mx_sum = 0
for r in permutations('01234',5):
mx_sum = max(sum([grid[i][int(r[i])] for i in xrange(5)]),mx_sum)
return mx_sum
def choose(initial_grid,row,col):
"""Return a grid with a given row and column zeroed except where they intersect."""
grid = [sub[:] for sub in initial_grid] #We don't actually want to alter initial_grid
Special_value = grid[row][col] #This is where they intersect.
grid[row] = [0]*len(grid[row]) #zeroes the row.
for i in xrange(len(grid)):
grid[i][col] = 0
grid[row][col] = Special_value
print_grid(grid)
return grid
def smart_solve(grid):
"""Solve the grid intelligently.
"""
rowsum = [sum(row) for row in grid] #This is the cost for any element in a given row.
print rowsum
colsum = [sum(row) for row in [[grid[i][j] for i in xrange(len(grid))] for j in xrange(len(grid[0]))]] #Cost for any element in a given column
print colsum,"\n"
total_cost = [map(lambda x: x+i,rowsum) for i in colsum] #This adds rowsum and colsum together, yielding the cost at a given index.
print_grid(total_cost,5)
print "\n"
#Total_cost has a flaw: It counts the value of the cell twice. It needs to count it -1 times, subtracting its value from the cost.
for i in xrange(len(grid)): #For each row
for j in xrange(len(grid[0])): #For each column
total_cost[i][j] -= 3*(grid[i][j]) #Remove the double addition and subtract once.
###Can I turn ^^that into a list comprehension? Maybe use zip or something?###
print_grid(total_cost,5)
return total_cost
def min_value(grid):
"""return the minimum value (And it's index) such that value>0.
Output is: (value,col,row)"""
min_value,row,col = grid[0][0],0,0
for row_index,check_row in enumerate(grid):
for col_index,check_val in enumerate(check_row):
#print "Min_value: %d\n Check_Val: %d\n Location: (%d,%d)"%(min_value,check_val,row_index,col_index)
if min_value>check_val>0:
min_value = check_val
row = row_index
col = col_index
return (min_value,row,col)