I want to create equations using numpy array multiplication ie I want to keep all variables in an array and its coefficients in other array and multiply both with each other to produce an expression so that I can use m.Equation() method of GEKKO. I tried the mentioned code but failed, please let me know how I can achieve my goal.
By "it failed" I meant that it just gave an error and did not let me use x*y==1 as equation in m.Equation() method available in GEKKO. My target is that I want to keep variables in one array and their coefficients in the other array and I multiply them to get mathematical equations to be used as input in m.Equation() method.
import numpy as np
from gekko import GEKKO
X = np.array([x,y,z])
y = np.array([4,5,6])
m = GEKKO(remote=False)
m.Equation(x*y==1)
# I wanted to get a result like 4x+5y+6z=1
The error I get is below
Traceback (most recent call last):
File "C:\Users\kk\AppData\Local\Programs\Python\Python37\MY WORK FILES\numpy practise.py", line 5, in <module>
X = np.array([x,y,z])
NameError: name 'x' is not defined
You need to define variables and make the coefficients into a Gekko object. You can use an array to make the variables and a parameter for the coefficients:
from gekko import GEKKO
m = GEKKO(remote=False)
X = m.Array(m.Var, 3)
y = m.Param([4, 5, 6])
eq = m.Equation(X.dot(y) == 1)
print(eq.value)
Output:
((((v1)*(4))+((v2)*(5)))+((v3)*(6)))=1
Related
I am using python3, Scipy
I have a 3d points (x,y,z]
From them I make s apline using scipy.interpolate.splprep
x_points = np.linspace(0, 2*np.pi, 10)
y_points = np.sin(x_points)
z_points = np.cos(x_points)
path = np.vstack([x_points, y_points, z_points])
tck, u = sc.splprep(path, k=3, s=0)
I wish to get the coefficients of the spline[i]:
For example the latest splins:
sp9 = a9 + b9(x-x4) + c9(x-x4)^2 + d9(x-x4)^3
I know that the tck is (t,c,k) a tuple containing the vector of knots, the B-spline coefficients, and the degree of the spline.
But I don't see how I can get this spline function and plot only it
I tried using this method:
import numpy as np
import scipy.interpolate as sc
x_points = np.linspace(0, 2*np.pi, 10)
y_points = np.sin(x_points)
z_points = np.cos(x_points)
path = np.vstack([x_points, y_points, z_points])
tck, u = sc.splprep(path, k=3, s=0)
p = sc.PPoly.from_spline(tck)
but I'm getting this error on the last line:
p = sc.PPoly.from_spline(tck) File
"C:\Users...\Python38\lib\site-packages\scipy\interpolate\interpolate.py",
line 1314, in from_spline cvals = np.empty((k + 1, len(t)-1),
dtype=c.dtype)
AttributeError: 'list' object has no attribute 'dtype'
The coefficients in the tck tuple are in the b-spline basis. If you want to convert them to the power basis, you can do PPoly.from_spline(tck) .
An obligatory note however: converting between bases incurs numerical errors.
EDIT. First, as it's splprep, you'll need to convert the list-of-arrays c into a proper numpy array and transpose (it's a known wart of splPrep). Then, as it turns out, PPoly.from_spline does not handle multidimensional c (this might be a nice pull request to the scipy repository), so you'll need to e.g. loop over the dimensions. Something along the lines of (continuing from your OP)
t, c, k = tck
cc = np.asarray(c) # cc.shape is (3, 10) now
spl0 = sc.PPoly.from_spline((t, cc.T[0], k))
print(spl0.c) # here are your coefficients for the component 0
My aim is to rotate a 1d numpy array by left.For example desired output for numpy array [1,2,3,4] should be [2,3,4,1].
Here is my approach:
import numpy as np
x = np.array([1,2,3,4])
x1 = x[1:]
x2 = x[:1]
print(np.concatenate(x1,x2))
I am facing an error while concatenating. Why is that?
I tried this approach also:
lst = x[1:] + x[:1]
print(np.array(lst))
Although I am getting the desired output I am also getting a error
DeprecationWarning: elementwise comparison failed; this will raise an error in the future.
I already saw this kind of problem to build AES in python. But I didn't used numpy, only list: the step is called ShiftRows
Anyway, you can use: np.roll(your_array, int_shift)
For example:
>>> x = np.array([1,2,3,4])
>>> shift_x = np.roll(x, 3)
>>> shift_x
>>> array([2, 3, 4, 1])
Copy and pasting this code into the python3 REPL works, but when I run it as a script, I get a type error.
"""Softmax."""
scores = [3.0, 1.0, 0.2]
import numpy as np
from math import e
def softmax(x):
"""Compute softmax values for each sets of scores in x."""
results = []
x = np.transpose(x)
for j in range(len(x)):
exps = [np.exp(s) for s in x[j]]
_sum = np.sum(np.exp(x[j]))
softmax = [i / _sum for i in exps]
results.append(softmax)
final = np.vstack(results)
return np.transpose(final)
# pass # TODO: Compute and return softmax(x)
print(softmax(scores))
# Plot softmax curves
import matplotlib.pyplot as plt
x = np.arange(-2.0, 6.0, 0.1)
scores = np.vstack([x, np.ones_like(x), 0.2 * np.ones_like(x)])
plt.plot(x, softmax(scores).T, linewidth=2)
plt.show()
The error I get running the script via CLI is the following:
bash$ python3 softmax.py
Traceback (most recent call last):
File "softmax.py", line 22, in <module>
print(softmax(scores))
File "softmax.py", line 13, in softmax
exps = [np.exp(s) for s in x[j]]
TypeError: 'numpy.float64' object is not iterable
This kind of crap makes me so nervous about running interpreted code in production with libraries like these, seriously unreliable and undefined behaviour is totally unacceptable IMO.
At the top of your script, you define
scores = [3.0, 1.0, 0.2]
This is the argument in your first call of softmax(scores). When converted to a numpy array, scores is 1-d array with shape (3,).
You pass scores into the function, and then it is converted to a numpy array by the call
x = np.transpose(x)
However, it is still 1-d, with shape (3,). The transpose function swaps dimensions, but it does not add a dimension to a 1-d array. In effect, transpose is a "no-op" when applied to a 1-d array.
Then, in the loop that follows, x[j] is a scalar of type numpy.float64, so it does not make sense to write [np.exp(s) for s in x[j]]. x[j] is a scalar, not a sequence, so you can't iterate over it.
In the bottom part of your script, you redefine scores as
x = np.arange(-2.0, 6.0, 0.1)
scores = np.vstack([x, np.ones_like(x), 0.2 * np.ones_like(x)])
Now scores is 2-d array (scores.shape is (3, 80)), so you don't get an error when you call softmax(scores).
Using the following function i am trying to generate index from the data:
Function:
import numpy as np
from sklearn.decomposition import PCA
def pca_index(data,components=1,indx=1):
corrs = np.asarray(data.cov())
pca = PCA(n_components = components).fit(corrs)
trns = pca.transform(data)
index=np.dot(trns[0:indx],pca.explained_variance_ratio_[0:indx])
return index
Index: generation from principal components
index = pca_index(data=mydata,components=3,indx=2)
Following error is being generated when i am calling the function:
Traceback (most recent call last):
File "<ipython-input-411-35115ef28e61>", line 1, in <module>
index = pca_index(data=mydata,components=3,indx=2)
File "<ipython-input-410-49c0174a047a>", line 15, in pca_index
index=np.dot(trns[0:indx],pca.explained_variance_ratio_[0:indx])
ValueError: shapes (2,3) and (2,) not aligned: 3 (dim 1) != 2 (dim 0)
Can anyone help with the error.
According to my understanding there is some error at the following point when i am passing the subscript indices as variable (indx):
trns[0:indx],pca.explained_variance_ratio_[0:**indx**]
In np.dot you are trying to multiply a matrix having dimensions (2,3) with a matrix having dimensions (2,), i.e. a vector.
However, you can only multiply NxM to MxP, e.g. (3,2) to (2,1) or (2,3) to (3,1).
In your example the second matrix have dimensions of (2,) which, in numpy terms, is similar but not the same as (2,1). You can reshape a vector into a matrix with vector.reshape([2,1])
You might also transpose you first matrix, thus converting its dimensions from (2,3) to (3,2).
However, make sure that you multiply appropriate matrices as the result will differ from you might expect.
I am new to theano. I am trying to implement simple linear regression but my program throws following error:
TypeError: ('Bad input argument to theano function with name "/home/akhan/Theano-Project/uog/theano_application/linear_regression.py:36" at index 0(0-based)', 'Expected an array-like object, but found a Variable: maybe you are trying to call a function on a (possibly shared) variable instead of a numeric array?')
Here is my code:
import theano
from theano import tensor as T
import numpy as np
import matplotlib.pyplot as plt
x_points=np.zeros((9,3),float)
x_points[:,0] = 1
x_points[:,1] = np.arange(1,10,1)
x_points[:,2] = np.arange(1,10,1)
y_points = np.arange(3,30,3) + 1
X = T.vector('X')
Y = T.scalar('Y')
W = theano.shared(
value=np.zeros(
(3,1),
dtype=theano.config.floatX
),
name='W',
borrow=True
)
out = T.dot(X, W)
predict = theano.function(inputs=[X], outputs=out)
y = predict(X) # y = T.dot(X, W) work fine
cost = T.mean(T.sqr(y-Y))
gradient=T.grad(cost=cost,wrt=W)
updates = [[W,W-gradient*0.01]]
train = theano.function(inputs=[X,Y], outputs=cost, updates=updates, allow_input_downcast=True)
for i in np.arange(x_points.shape[0]):
print "iteration" + str(i)
train(x_points[i,:],y_points[i])
sample = np.arange(x_points.shape[0])+1
y_p = np.dot(x_points,W.get_value())
plt.plot(sample,y_p,'r-',sample,y_points,'ro')
plt.show()
What is the explanation behind this error? (didn't got from the error message). Thanks in Advance.
There's an important distinction in Theano between defining a computation graph and a function which uses such a graph to compute a result.
When you define
out = T.dot(X, W)
predict = theano.function(inputs=[X], outputs=out)
you first set up a computation graph for out in terms of X and W. Note that X is a purely symbolic variable, it doesn't have any value, but the definition for out tells Theano, "given a value for X, this is how to compute out".
On the other hand, predict is a theano.function which takes the computation graph for out and actual numeric values for X to produce a numeric output. What you pass into a theano.function when you call it always has to have an actual numeric value. So it simply makes no sense to do
y = predict(X)
because X is a symbolic variable and doesn't have an actual value.
The reason you want to do this is so that you can use y to further build your computation graph. But there is no need to use predict for this: the computation graph for predict is already available in the variable out defined earlier. So you can simply remove the line defining y altogether and then define your cost as
cost = T.mean(T.sqr(out - Y))
The rest of the code will then work unmodified.