Question

我需要的是Code i应该替换它：ZeroMQ with。

我有一个在表格中创建单元格的类。我将这些单元格存储在单元格列表中。

inproc:

我想知道该列表中是否存在单元格。我正在做以下事情：

<-- code (exist) -->

细胞分类

List<Cell> locations = new List<Cell>();

Answer 1

使用IEqualityComparer<Cell>的推动力让您的生活变得轻松。我使用这个解决方案，因为你可能需要在某个地方使用那个seme逻辑。

public class CellComparer : IEqualityComparer<Cell>
{
    public bool Equals(Cell x, Cell y)
    {
        if (x == null && y == null) return true;

        if (x == null || y == null) return false;

        if (x.Column == y.Column && x.Row == y.Row) return true;

        return false;
    }

    public int GetHashCode(Cell cell)
    {
        int hCode = cell.Column ^ cell.Row;
        return hCode.GetHashCode();
    }
}

要使用它只是简单的，如果你检查列表内容，你会看到它包含两个元素，因为第一个和最后一个添加的单元格是相同的。

var list = new HashSet<Cell>(new CellComparer());
list.Add(new Cell(0, 1));
list.Add(new Cell(1, 2));
list.Add(new Cell(0, 1));

感谢HashSet，CellComparer将使用您的Cell来避免public class Cell { public Cell(int col, int row) { Column = col; Row = row; } public int Row { get; private set; } public int Column { get; private set; } public int[] Position { get { return new[] { Column, Row }; } } }重复。

因此，使用auto-property而不是method来返回字段的值。您的单元格类必须如下所示：

from ivisual import *
from numpy.linalg import solve
import numpy as np
import math as m

scene = display(x=0,y=0,width=500,height=500,
                title='String and masses configuration')
tempe = curve(x=range(0,500),color=color.black)

n = 9
eps = 1*10**(-6)
deriv = np.zeros( (n, n), float)
f = np.zeros( (n), float)
x = np.array([0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1., 1., 1.])

def plotconfig():
    for obj in scene.objects:
        obj.visible=0           # to erase the previous configuration
    L1 = 3.0
    L2 = 4.0
    L3 = 4.0
    xa = L1*x[3]                # L1*cos(th1)
    ya = L1*x[0]                # L1 sin(th1)
    xb = xa+L2*x[4]             # L1*cos(th1)+L2*cos(th2)
    yb = ya+L2*x[1]             # L1*sin(th1)+L2*sen(th2)
    xc = xb+L3*x[5]             # L1*cos(th1)+L2*cos(th2)+L3*cos(th3)
    yc = yb-L3*x[2]             # L1*sin(th1)+L2*sen(th2)-L3*sin(th3)
    mx = 100.0                  # for linear coordinate transformation
    bx = -500.0                 # from 0=< x =<10
    my = -100.0                 # to    -500 =<x_window=>500
    by = 400.0                  # same transformation for y
    xap = mx*xa+bx              # to keep aspect ratio
    yap = my*ya+by
    ball1 = sphere(pos=(xap,yap), color=color.cyan,radius=15) 
    xbp = mx*xb+bx
    ybp = my*yb+by
    ball2 = sphere(pos=(xbp,ybp), color=color.cyan,radius=25) 
    xcp = mx*xc+bx
    ycp = my*yc+by
    x0 = mx*0+bx
    y0 = my*0+by
    line1 = curve(pos=[(x0,y0),(xap,yap)], color=color.yellow,radius=4)
    line2 = curve(pos=[(xap,yap),(xbp,ybp)], color=color.yellow,radius=4)
    line3 = curve(pos=[(xbp,ybp),(xcp,ycp)], color=color.yellow,radius=4)
    topline = curve(pos=[(x0,y0),(xcp,ycp)], color=color.red,radius=4)

def F(x, f):                    # Define F function
    f[0] = 3*x[3]  +  4*x[4]  +  4*x[5]  -  8.0
    f[1] = 3*x[0]  +  4*x[1]  -  4*x[2]
    f[2] = x[6]*x[0]  -  x[7]*x[1]  -  10.0
    f[3] = x[6]*x[3]  -  x[7]*x[4]
    f[4] = x[7]*x[1]  +  x[8]*x[2]  -  20.0
    f[5] = x[7]*x[4]  -  x[8]*x[5]
    f[6] = pow(x[0], 2)  +  pow(x[3], 2)  -  1.0
    f[7] = pow(x[1], 2)  +  pow(x[4], 2)  -  1.0
    f[8] = pow(x[2], 2)  +  pow(x[5], 2)  -  1.0

def dFi_dXj(x, deriv, n):       # Define derivative function
    h = 1*10**(-4)                                             
    for j in range(0, n):
         temp = x[j]
         x[j] = x[j] +  h/2.
         F(x, f)                                                 
         for i in range(0, n):  deriv[i, j] = f[i] 
         x[j] = temp

    for j in range(0, n):
         temp = x[j]
         x[j] = x[j] - h/2.
         F(x, f)
         for i in range(0, n): deriv[i, j] = (deriv[i, j] - f[i])/h
         x[j] = temp

for it in range(1, 100):
      rate(1)                   # 1 second between graphs
      F(x, f)                              
      dFi_dXj(x, deriv, n)   
      B = np.array([[-f[0]], [-f[1]], [-f[2]], [-f[3]], [-f[4]], [-f[5]],[-f[6]], [-f[7]], [-f[8]]])      
      sol = solve(deriv, B)
      dx = np.take(sol, (0, ), 1)      # take the first column of matrix sol
      for i in range(0, n):
        x[i]  = x[i]  +  dx[i]
      plotconfig()
      errX = errF = errXi = 0.0

      for i in range(0, n):
        if ( x[i] !=  0.): errXi = abs(dx[i]/x[i])
        else:  errXi = abs(dx[i])
        if ( errXi > errX): errX = errXi                            
        if ( abs(f[i]) > errF ):  errF = abs(f[i])        
        if ( (errX <=  eps) and (errF <=  eps) ): break

print('Number of iterations = ', it)
print('Solution:')
for i in range(0, n):
        print('x[', i, '] =  ', x[i])

Answer 2

您可以将IEquatable接口添加到Cell类：

public class Cell : IEquatable<Cell>

在课程中需要bool方法Equals：

public bool Equals(Cell otherCell)

，在this单元格等于otherCell（例如，this.getRow() == otherCell.getRow()等）时，您可以提供代码告诉。

如果Cell为IEquatable，则允许您在单元格列表中使用Contains方法。因此，在您的主要代码中，您可以使用<-- code (exist) -->而不是locations.Contains(cell)。

你应该考虑使用属性重写你的Cell课程。

Answer 3

你可以使用linq来做那个

public bool cell_in_array(Cell cell)
{
    return locations.Any(c => c.getColumn() == cell.getColumn() && 
                              c.getRow() == cell.getRow())         
}

Any将确定是否有任何元素符合条件。

Answer 4

您可以使用Any来确定序列中的任何元素是否满足条件：

locations.Any(c => c.getColumn() == cell.getColumn() && c.getRow() == cell.getRow())

检查对象列表中是否存在对象

4 个答案: