I have a non linear function with two variables, I want to solve the equation . But the solution itself is an equation. How do i evaluate the function at a certain point
CODE:
import sympy as sp
sp.init_printing()
x1,x2,y1,y2 = sp.symbols('x1,x2,y1,y2')
x1,y1=-2,3
f = sp.Eq((x1-x2)**2 + (y1-y2)**2,1)
a = sp.solve([f],(x2,y2))
now i want a a few solution pairs of the function 'a' .
Thanks in advance
:)
You have a single equation in two unknowns: pick a value for one and solve for the other. Here, we pick values for y2 and solve for x2 and pair each solution with the value of i. There is one solution for y2 = 2 and 4 and 2 solutions when it is 3
>>> [(j,i) for i in range(2,5) for j in sp.solve(f.subs(y2,i),x2)]
[(-2, 2), (-3, 3), (-1, 3), (-2, 4)]
Realizing that the equation represents a circle centered at (-2, 3) also permits one to use SymPy's Circle to give you an arbitrary Point in terms of a parameter:
>>> from sympy import Circle
>>> from sympy.abc import t
>>> Circle((-2,3),1).arbitrary_point(t)
Point2D(cos(t) - 2, sin(t) + 3)
Substitute a value for t to get the corresponding point
>>> _.subs(t,pi)
Point2D(-3, 3)
I want to solve the following equation for x with SymPy:
(Note that the equation can be simplified as mentioned in the comments, I copied it verbatim from an example in a legal document.)
According to my understanding, this translates to the following SymPy expression:
from sympy import Sum, solve
from sympy.abc import k, x
solve(350 - 18500 + Sum(182.94 * (1/(1+x)**(k/12)), (k, 1, 120)), x)
However, when I run this, the result is empty:
[]
What am I doing wrong?
solve probably shouldn't give [] but you will get better results from nsolve for this expression using a guess for x near 0:
>>> from sympy.abc import k, x
>>> from sympy import nsolve
eq = 350 - 18500 + Sum(182.94 * (1/(1+x)**(k/12)), (k, 1, 120))
>>> nsolve(eq, 0)
0.0397546543274819
>>> eq.subs(x,_).round(2)
0
I have an image of land cover and I segmented it using K-means clustering. Now I want to calculate the accuracy of my segmentation algorithm. I read somewhere that dice co-efficient is the substantive evaluation measure. But I am not sure how to calculate it.
I use Python 2.7
Are there any other effective evaluation methods? Please give a summary or a link to a source. Thank You!
Edits:
I used the following code for measuring the dice similarity for my original and the segmented image but it seems to take hours to calculate:
for i in xrange(0,7672320):
for j in xrange(0,3):
dice = np.sum([seg==gt])*2.0/(np.sum(seg)+np.sum(gt)) #seg is the segmented image and gt is the original image. Both are of same size
Please refer to Dice similarity coefficient at wiki
A sample code segment here for your reference. Please note that you need to replace k with your desired cluster since you are using k-means.
import numpy as np
k=1
# segmentation
seg = np.zeros((100,100), dtype='int')
seg[30:70, 30:70] = k
# ground truth
gt = np.zeros((100,100), dtype='int')
gt[30:70, 40:80] = k
dice = np.sum(seg[gt==k])*2.0 / (np.sum(seg) + np.sum(gt))
print 'Dice similarity score is {}'.format(dice)
If you are working with opencv you could use the following function:
import cv2
import numpy as np
#load images
y_pred = cv2.imread('predictions/image_001.png')
y_true = cv2.imread('ground_truth/image_001.png')
# Dice similarity function
def dice(pred, true, k = 1):
intersection = np.sum(pred[true==k]) * 2.0
dice = intersection / (np.sum(pred) + np.sum(true))
return dice
dice_score = dice(y_pred, y_true, k = 255) #255 in my case, can be 1
print ("Dice Similarity: {}".format(dice_score))
In case you want to evaluate with this metric within a deep learning model using tensorflow you can use the following:
def dice_coef(y_true, y_pred):
y_true_f = tf.reshape(tf.dtypes.cast(y_true, tf.float32), [-1])
y_pred_f = tf.reshape(tf.dtypes.cast(y_pred, tf.float32), [-1])
intersection = tf.reduce_sum(y_true_f * y_pred_f)
return (2. * intersection + 1.) / (tf.reduce_sum(y_true_f) + tf.reduce_sum(y_pred_f) + 1.)
This is an important clarification if what you're using has more than 2 classes (aka, a mask with 1 and 0).
If you are using multiple classes, make sure to specify that the prediction and ground truth also equal the value which you want. Otherwise you can end up getting DSC values greater than 1.
This is the extra ==k at the end of each [] statement:
import numpy as np
k=1
# segmentation
seg = np.zeros((100,100), dtype='int')
seg[30:70, 30:70] = k
# ground truth
gt = np.zeros((100,100), dtype='int')
gt[30:70, 40:80] = k
dice = np.sum(seg[gt==k]==k)*2.0 / (np.sum(seg[seg==k]==k) + np.sum(gt[gt==k]==k))
print 'Dice similarity score is {}'.format(dice)
I would like to convert a NumPy array to a unit vector. More specifically, I am looking for an equivalent version of this normalisation function:
def normalize(v):
norm = np.linalg.norm(v)
if norm == 0:
return v
return v / norm
This function handles the situation where vector v has the norm value of 0.
Is there any similar functions provided in sklearn or numpy?
If you're using scikit-learn you can use sklearn.preprocessing.normalize:
import numpy as np
from sklearn.preprocessing import normalize
x = np.random.rand(1000)*10
norm1 = x / np.linalg.norm(x)
norm2 = normalize(x[:,np.newaxis], axis=0).ravel()
print np.all(norm1 == norm2)
# True
I agree that it would be nice if such a function were part of the included libraries. But it isn't, as far as I know. So here is a version for arbitrary axes that gives optimal performance.
import numpy as np
def normalized(a, axis=-1, order=2):
l2 = np.atleast_1d(np.linalg.norm(a, order, axis))
l2[l2==0] = 1
return a / np.expand_dims(l2, axis)
A = np.random.randn(3,3,3)
print(normalized(A,0))
print(normalized(A,1))
print(normalized(A,2))
print(normalized(np.arange(3)[:,None]))
print(normalized(np.arange(3)))
This might also work for you
import numpy as np
normalized_v = v / np.sqrt(np.sum(v**2))
but fails when v has length 0.
In that case, introducing a small constant to prevent the zero division solves this.
As proposed in the comments one could also use
v/np.linalg.norm(v)
To avoid zero division I use eps, but that's maybe not great.
def normalize(v):
norm=np.linalg.norm(v)
if norm==0:
norm=np.finfo(v.dtype).eps
return v/norm
If you have multidimensional data and want each axis normalized to its max or its sum:
def normalize(_d, to_sum=True, copy=True):
# d is a (n x dimension) np array
d = _d if not copy else np.copy(_d)
d -= np.min(d, axis=0)
d /= (np.sum(d, axis=0) if to_sum else np.ptp(d, axis=0))
return d
Uses numpys peak to peak function.
a = np.random.random((5, 3))
b = normalize(a, copy=False)
b.sum(axis=0) # array([1., 1., 1.]), the rows sum to 1
c = normalize(a, to_sum=False, copy=False)
c.max(axis=0) # array([1., 1., 1.]), the max of each row is 1
If you don't need utmost precision, your function can be reduced to:
v_norm = v / (np.linalg.norm(v) + 1e-16)
You mentioned sci-kit learn, so I want to share another solution.
sci-kit learn MinMaxScaler
In sci-kit learn, there is a API called MinMaxScaler which can customize the the value range as you like.
It also deal with NaN issues for us.
NaNs are treated as missing values: disregarded in fit, and maintained
in transform. ... see reference [1]
Code sample
The code is simple, just type
# Let's say X_train is your input dataframe
from sklearn.preprocessing import MinMaxScaler
# call MinMaxScaler object
min_max_scaler = MinMaxScaler()
# feed in a numpy array
X_train_norm = min_max_scaler.fit_transform(X_train.values)
# wrap it up if you need a dataframe
df = pd.DataFrame(X_train_norm)
Reference
[1] sklearn.preprocessing.MinMaxScaler
There is also the function unit_vector() to normalize vectors in the popular transformations module by Christoph Gohlke:
import transformations as trafo
import numpy as np
data = np.array([[1.0, 1.0, 0.0],
[1.0, 1.0, 1.0],
[1.0, 2.0, 3.0]])
print(trafo.unit_vector(data, axis=1))
If you work with multidimensional array following fast solution is possible.
Say we have 2D array, which we want to normalize by last axis, while some rows have zero norm.
import numpy as np
arr = np.array([
[1, 2, 3],
[0, 0, 0],
[5, 6, 7]
], dtype=np.float)
lengths = np.linalg.norm(arr, axis=-1)
print(lengths) # [ 3.74165739 0. 10.48808848]
arr[lengths > 0] = arr[lengths > 0] / lengths[lengths > 0][:, np.newaxis]
print(arr)
# [[0.26726124 0.53452248 0.80178373]
# [0. 0. 0. ]
# [0.47673129 0.57207755 0.66742381]]
If you want to normalize n dimensional feature vectors stored in a 3D tensor, you could also use PyTorch:
import numpy as np
from torch import FloatTensor
from torch.nn.functional import normalize
vecs = np.random.rand(3, 16, 16, 16)
norm_vecs = normalize(FloatTensor(vecs), dim=0, eps=1e-16).numpy()
If you're working with 3D vectors, you can do this concisely using the toolbelt vg. It's a light layer on top of numpy and it supports single values and stacked vectors.
import numpy as np
import vg
x = np.random.rand(1000)*10
norm1 = x / np.linalg.norm(x)
norm2 = vg.normalize(x)
print np.all(norm1 == norm2)
# True
I created the library at my last startup, where it was motivated by uses like this: simple ideas which are way too verbose in NumPy.
Without sklearn and using just numpy.
Just define a function:.
Assuming that the rows are the variables and the columns the samples (axis= 1):
import numpy as np
# Example array
X = np.array([[1,2,3],[4,5,6]])
def stdmtx(X):
means = X.mean(axis =1)
stds = X.std(axis= 1, ddof=1)
X= X - means[:, np.newaxis]
X= X / stds[:, np.newaxis]
return np.nan_to_num(X)
output:
X
array([[1, 2, 3],
[4, 5, 6]])
stdmtx(X)
array([[-1., 0., 1.],
[-1., 0., 1.]])
For a 2D array, you can use the following one-liner to normalize across rows. To normalize across columns, simply set axis=0.
a / np.linalg.norm(a, axis=1, keepdims=True)
If you want all values in [0; 1] for 1d-array then just use
(a - a.min(axis=0)) / (a.max(axis=0) - a.min(axis=0))
Where a is your 1d-array.
An example:
>>> a = np.array([0, 1, 2, 4, 5, 2])
>>> (a - a.min(axis=0)) / (a.max(axis=0) - a.min(axis=0))
array([0. , 0.2, 0.4, 0.8, 1. , 0.4])
Note for the method. For saving proportions between values there is a restriction: 1d-array must have at least one 0 and consists of 0 and positive numbers.
A simple dot product would do the job. No need for any extra package.
x = x/np.sqrt(x.dot(x))
By the way, if the norm of x is zero, it is inherently a zero vector, and cannot be converted to a unit vector (which has norm 1). If you want to catch the case of np.array([0,0,...0]), then use
norm = np.sqrt(x.dot(x))
x = x/norm if norm != 0 else x
I have the following system of 3 nonlinear equations that I need to solve:
-xyt + HF = 0
-2xzt + 4yzt - xyt + 4z^2t - M1F = 0
-2xt + 2yt + 4zt - 1 = 0
where x, HF, and M1F are known parameters. Therefore, y,z, and t are the parameters to be calculated.
Attemp to solve the problem:
def equations(p):
y,z,t = p
f1 = -x*y*t + HF
f2 = -2*x*z*t + 4*y*z*t - x*y*t + 4*t*z**2 - M1F
f3 = -2*x*t + 2*y*t + 4*z*t - 1
return (f1,f2,f3)
y,z,t = fsolve(equations)
print equations((y,z,t))
But the thing is that if I want to use scipy.optimize.fsolve then I should input an initial guess. In my case, I do not have any initial conditions.
Is there another way to solve 3 nonlinear equations with 3 unknowns in python?
Edit:
It turned out that I have a condition! The condition is that HF > M1F, HF > 0, and M1F > 0.
#Christian, I don't think the equation system can be linearize easily, unlike the post you suggested.
Powell's Hybrid method (optimize.fsolve()) is quite sensitive to initial conditions, so it is very useful if you can come up with a good initial parameter guess. In the following example, we firstly minimize the sum-of-squares of all three equations using Nelder-Mead method (optimize.fmin(), for small problem like OP, this is probably already enough). The resulting parameter vector is then used as the initial guess for optimize.fsolve() to get the final result.
>>> from numpy import *
>>> from scipy import stats
>>> from scipy import optimize
>>> HF, M1F, x=1000.,900.,10.
>>> def f(p):
return abs(sum(array(equations(p))**2)-0)
>>> optimize.fmin(f, (1.,1.,1.))
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 131
Function evaluations: 239
array([ -8.95023217, 9.45274653, -11.1728963 ])
>>> optimize.fsolve(equations, (-8.95023217, 9.45274653, -11.1728963))
array([ -8.95022376, 9.45273632, -11.17290503])
>>> pr=optimize.fsolve(equations, (-8.95023217, 9.45274653, -11.1728963))
>>> equations(pr)
(-7.9580786405131221e-13, -1.2732925824820995e-10, -5.6843418860808015e-14)
The result is pretty good.