Python 确定行中的最低值&；NxN数组的列

我有一个列表，用于存储对象之间的距离

该表如下所示：

+----------+----------+----------+----------+----------+
|          | 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    |
+----------+----------+----------+----------+----------+

这些数字表示行和列标题之间的距离

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)

这大致提供了上表中的信息

我想做的基本上是找到离每个实体最近的对象（反之亦然）。每个对象或实体只能相互匹配，并且应基于接近度

我现在做这件事的方法是简单地按距离大小对嵌套列表排序（在完整的代码中，我有一种方法可以确定哪个对象与每个距离相关联）

因此，在这样做时，我将创建以下关联：

+----------+----------+----------+
| Entity   |  Object  | Distance |
+----------+----------+----------+
| Entity_E | Object_D |     1    |
+----------+----------+----------+
| Entity_F | Object_D |     2    |
+----------+----------+----------+
| Entity_G | Object_B |     1    |
+----------+----------+----------+

这是不正确的，因为它将对象\u D关联两次

协会应：

+----------+----------+----------+
| Entity   |  Object  | Distance |
+----------+----------+----------+
| Entity_E | Object_D |     1    |
+----------+----------+----------+
| Entity_F | Object_A |     3    |
+----------+----------+----------+
| Entity_G | Object_B |     1    |
+----------+----------+----------+

这就是我正在努力的地方——我想不出编写逻辑代码的最佳方法

因为实体_E更接近对象_D，所以它应该得到关联。所以，对于实体F，我需要取第二个最近点

我想做一些事情，记录哪些对象已经被分配，或者尝试做一些事情，首先匹配每列中的最小值，但它们似乎都遇到了问题

我可以用矩阵运算或某种矩阵数学来做这个计算吗

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)

---

不久前，我编写了一些代码来解决类似的问题。我将使用我自己的数组来执行此操作：

[[1,2,3],
 [4,5,6],
 [7,8,9]]

你想做的是得到选择一个特定元素的成本。通过查找选择行的成本，加上选择列的成本，然后减去选择元素本身的成本来实现。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）

。我对行0求和，对列0求和，然后减去元素本身

现在，您有了选择每个元素的成本。找到成本最高的元素。这将是网格中的最低值。选择它并确保行或列中不能选择其他值。重复此操作，直到选择了每行和每列中的一个值

这是我项目的源代码。请注意，它查找的是最高值，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)

---

你能添加完整的代码吗？编辑并添加尽可能多的代码！非常感谢你！这很有道理。然而，我有两个后续问题：1。为什么要减去3*（1）？我假设1是[0][0]处数组的值？2.你是怎么想到“选择元素的成本”的？有什么资料可以让我更熟悉吗？再次感谢！！除了我的经济学教科书，我真的无法引用任何资料。这是一个机会成本的例子。我减去三次有两个原因：第一，我在抓取行和列时加了两次值。第二，当你想要做某事的成本时，收益抵消了成本，所以你减去了价值。