【问题标题】:PyShp - getting Python to recognise points from a polylinePyShp - 让 Python 识别折线中的点
【发布时间】:2014-01-26 19:06:14
【问题描述】:

我对 Python 比较陌生,所以我怀疑我遇到的这个问题是由于幼稚造成的,但我们将不胜感激。

目前我有一个小型海岸演化模型。最初,这使用 NumPy 在某些约束内沿定义的 x 轴随机生成 100 个点。代码的相关部分是这样的:

# 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。我正在尝试做的是使用 PyShp 读取在 ArcMap 中绘制的折线以替换这些随机生成的点,然后将其用于模型计算。我正在尝试的代码是这样的:

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

Bch_Width = np.random.uniform(0, 30, BS1) #Attempt to use polyline of beach to generate    initial beach location
Cl_Loc = np.random.uniform(30, 60, CL1)  #Attempt to use polyline of beach to generate initial beach location

当我尝试对代码进行小的更改时,这会不断给出错误,例如“TypeError:需要一个整数”和“ValueError:序列太大;必须小于 32”。

有谁知道需要做什么才能让代码接受折线上的点,而不是随机生成的点?我觉得它应该很简单,但是我在这里浏览了 PyShp 文档和其他问题,似乎找不到该怎么做。

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)    

【问题讨论】:

    标签: python numpy shapefile polyline


    【解决方案1】:

    错误是因为您正在尝试使用Bch_Width = np.random.uniform(0, 30, BeachShp1) 等创建一个(可能)数千维数组。

    numpy.random.uniform 需要一个低、高以及将要生成的数组的形状。

    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]])
    

    当你运行时:

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

    BeachShp1 是一个 x,y 坐标序列。基本上,您正在尝试执行以下操作:

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

    ...这没有任何意义。

    您是否尝试将随机海滩宽度添加到 shapefile 表示的线?或者计算悬崖轮廓和海滩轮廓之间的距离?还是完全不同的东西?

    【讨论】:

    • 为答案干杯!我想我知道如何尝试结合 random.uniform 和 BeachShp 是行不通的。我希望 shapefile 中的折线将代表海滩的起始“形状”和位置(以及悬崖的相应 shapefile)。该模型的目的是随着时间的推移通过海滩/悬崖侵蚀。然而,模型方程依赖于生成的点作为海滩/悬崖的起始位置。因此,我希望我能以某种方式将这些点替换为实际绘制的海滩轮廓,以增加模型的真实感。
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