对 Python 中的 x,y 网格执行计算

Question

我想对 X、Y 数据执行计算以生成计算出的 Z。我的代码如下：

injection_wells.csv

的示例数据集

Name	X	Y	Q
MW-1	2517700	996400	5
MW-2	2517770	996420	5

import pandas as pd
import math
import numpy as np
import matplotlib.mlab as mlab
import matplotlib.tri as tri

IW = pd.read_csv (r'Injection_wells.csv')

`Note that - Injection wells is a table of three wells with names, X, Y, and Q (flow rate).`

#pull all the relevant well information by well into their own arrays

MW1 = IW[IW['Name'] == 'MW1']
MW2 = IW[IW['Name'] == 'MW2']
MW3 = IW[IW['Name'] == 'MW3']

#initiate grid
xi = np.linspace(2517675,2517800,625)
yi = np.linspace(996300,996375,375)

#make it so i can apply np.float to an array
vector = np.vectorize(np.float)

X,Y = np.meshgrid(xi,yi)

#perform calculation over every X and Y.
PSI = ((MW1['Q']/(2*math.pi))*(np.arctan(((vector(X[None,:]))-np.float(MW1['X']))/(vector(Y[:,None])-np.float(MW1['Y']))))+
      (MW2['Q']/(2*math.pi))*(np.arctan(((vector(X[None,:])-np.float(MW2['X']))/vector((Y[:,None])-np.float(MW2['Y'])))))+
      (MW3['Q']/(2*math.pi))*(np.arctan(((vector((X[None,:])-np.float(MW3['X']))/vector((Y[:,None])-np.float(MW3['Y'])))))))

我收到错误：

ValueError                                Traceback (most recent call last)
<ipython-input-11-fd6ee058014f> in <module>
     17 X,Y = np.meshgrid(xi,yi)
     18 
---> 19 PSI = ((MW1['Q']/(2*math.pi))*(np.arctan(((vector(X[None,:]))-np.float(MW1['X']))/(vector(Y[:,None])-np.float(MW1['Y']))))+
     20       (MW2['Q']/(2*math.pi))*(np.arctan(((vector(X[None,:])-np.float(MW2['X']))/vector((Y[:,None])-np.float(MW2['Y'])))))+
     21       (MW3['Q']/(2*math.pi))*(np.arctan(((vector((X[None,:])-np.float(MW3['X']))/vector((Y[:,None])-np.float(MW3['Y'])))))))

~\Anaconda3\lib\site-packages\pandas\core\ops\common.py in new_method(self, other)
     63         other = item_from_zerodim(other)
     64 
---> 65         return method(self, other)
     66 
     67     return new_method

~\Anaconda3\lib\site-packages\pandas\core\ops\__init__.py in wrapper(left, right)
    343         result = arithmetic_op(lvalues, rvalues, op)
    344 
--> 345         return left._construct_result(result, name=res_name)
    346 
    347     wrapper.__name__ = op_name

~\Anaconda3\lib\site-packages\pandas\core\series.py in _construct_result(self, result, name)
   2755         # We do not pass dtype to ensure that the Series constructor
   2756         #  does inference in the case where `result` has object-dtype.
-> 2757         out = self._constructor(result, index=self.index)
   2758         out = out.__finalize__(self)
   2759 

~\Anaconda3\lib\site-packages\pandas\core\series.py in __init__(self, data, index, dtype, name, copy, fastpath)
    311                 try:
    312                     if len(index) != len(data):
--> 313                         raise ValueError(
    314                             f"Length of passed values is {len(data)}, "
    315                             f"index implies {len(index)}."

ValueError: Length of passed values is 375, index implies 1.

我知道这与我试图将函数应用于只接受一个值的数组有关。我正在努力克服这个问题，并能够按原样对整个网格化数据集执行此计算。如有任何帮助，我们将不胜感激。

下面是我要计算的等式。请注意，等式中的 theta 是从网格节点到注入井（对于每个网格节点）的距离的反正切值，这就是我要在代码中复制的内容。

谢谢

Answer 1

我想在这里跳枪，因为我想我现在明白了这个问题，在多看了一点之后。

所以你有一个注入井数据的 DataFrame，有四列：

name    x    y    q
str   int  int  int

并且您有一个要计算的函数 f(x, y, q) -> z。我不确定我是否完全按照你的函数在做什么，因为它的格式很难阅读，所以我将使用一个简化的例子：

def func(x, y, q):
    return (q / 2 * np.pi) * np.arctan(y, x)

现在不用将井数据分成不同的数组，只需按行将函数应用于整个数据帧即可：

df["z"] = func(df.x, df.y, df.q)

对 Python 中的 x,y 网格执行计算

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