PyShp让Python从多段线识别点

Question

我对Python还比较陌生，所以我怀疑我遇到的这个问题是因为我太天真了，但是如果有任何帮助，我将不胜感激。在

目前我有一个小型的海岸演化模型。最初，这将使用NumPy在一些约束内沿着定义的x轴随机生成100个点。规范的相关部分如下：

# assign the number of points along the beach       
Num_P_Coast  = 100

# node spacing
dx = 10
x_vec = np.zeros(Num_P_Coast)

for i in range(0,Num_P_Coast):
    x_vec[i] = dx*i

# initialize beach and cliff with random numbers
Bch_Width = np.random.uniform(0, 30, Num_P_Coast)
Cl_Loc = np.random.uniform(30, 60, Num_P_Coast)

然后模型使用这两条初始线进行模型计算。在模型中随机生成一个点来代替所生成的折线。我好像在试密码：

^{pr2}$

当我尝试对代码进行小的更改时，这会不断地给出诸如“TypeError:an integer is required”和“ValueError:sequence too large；must be small than 32”之类的错误。在

有人知道需要做些什么来让代码接受折线上的点，而不是随机生成的点吗？我觉得应该很简单，但是我已经浏览了PyShp的文档和其他问题，似乎找不到该做什么。在

干杯

完整代码：

def toy_beach_model(): #this is the main part of the program, which will call the     evol_equations function to do all the jiggery pokery of outputting beach evolution

#where do you want to put it
#OutputDirectory = str(raw_input("Enter output directory:"))  
OutputDirectory = 'c:/python_results/graphs/'  
OutputModelName =  eg.enterbox(title = 'Model name', msg = 'Enter model ID:')
ModelDesc = eg.enterbox(title = 'Model description', msg = 'Enter short description of model' )

# assign the number of points along the beach    
Num_P_Coast  = 100

# node spacing
dx = 10
x_vec = np.zeros(Num_P_Coast)

# some parameters
Cl_Eros_NoBch = int(eg.enterbox(title = 'Erosion rate', msg = 'Enter erosion rate:'))


Cl_Eros_HRock = Cl_Eros_NoBch * 0.5 #simulating harder rock = headland evolution

Cl_Eros_Efold = 50

Sup_Rate = int(eg.enterbox(title = 'Sediment supply', msg = 'Enter sediment supply rate:'))
K = 0.5


# set up x vector
for i in range(0,Num_P_Coast):
    x_vec[i] = dx*i

BS1 = shp.Reader("Beach2.shp") #calling shapefile of beach front location
CL1 = shp.Reader("Cliff2.shp") #calling shapefile of cliff location

p1 = BS1.shapes()
b = p1[0]
BeachShp1 = b.points

p2 = CL1.shapes()
c = p2[0]
CliffShp1 = c.points


# initialize beach and cliff with random numbers
#Bch_Width = np.random.uniform(0, 30, Num_P_Coast)
#Cl_Loc = np.random.uniform(30, 60, Num_P_Coast)

Bch_Width = np.random.uniform(0, 30, BeachShp1) #Attempt to use polyline of beach to generate initial beach location
Cl_Loc = np.random.uniform(30, 60, CliffShp1)  #Attempt to use polyline of beach to generate initial beach location


#H_Loc = Cl_Loc
#H_Loc = np.random.uniform(0, 8, Num_P_Coast)    
H_Loc = np.random.random_sample(Num_P_Coast)
BcH_Eros = Bch_Width+Cl_Loc

# lets have this evolve through time

# time spacing in years    
dt = 10

# number of time steps
n_timesteps = int(eg.enterbox(title = 'Timestep', msg = 'Enter number of iterations'))

# set up the beach erosion vector
Cl_Eros = np.zeros(Num_P_Coast)

H_Eros = np.zeros(Num_P_Coast)    

# initial headland erosion
H_Eros = dt*Cl_Eros_HRock*np.exp(-Bch_Width/Cl_Eros_Efold)

# initial erosion
Cl_Eros = dt*Cl_Eros_NoBch*np.exp(-Bch_Width/Cl_Eros_Efold)    

Answer 1

错误是因为您试图用Bch_Width = np.random.uniform(0, 30, BeachShp1)等（可能）生成几千维数组

^{}需要一个低、高以及将要生成的数组的形状。在

例如：

In [1]: import numpy as np

In [2]: np.random.uniform(0, 5) # Generate a single random number between 0 and 5
Out[2]: 4.149771995503083

In [3]: np.random.uniform(0, 5, 3) # Generate an array of three numbers
Out[3]: array([ 2.25725653,  0.70070352,  0.62541689])

In [4]: np.random.uniform(0, 5, (2,2)) # Generate a 2x2 array of 4 numbers
Out[4]:
array([[ 0.89355128,  3.30796407],
       [ 1.23816971,  1.12224456]])

跑步时：

^{pr2}$

BeachShp1是x，y坐标的序列。基本上，你想做的是：

np.random.uniform(low, high, [[19.554, 45.998], [20.889, 24.009], ... ])

……这没有任何意义。在

您是否尝试在shapefile表示的线条上添加随机海滩宽度？或者计算悬崖轮廓和海滩轮廓之间的距离？或者完全是别的什么？在