The sigmoid function is defined as sigmoid(x) = 1/(1+e-x). If the score is defined by 4x1 + 5x2 - 9 = score, then which of the following points has exactly a 50% probability of being blue or red? (Choose all that are correct.)
answers:
(1,1) - (2,4) - (5,-5), (-4,5)
can someone explains how to solve this question?
from math import e
#Equation: 4x1+5x2-9=score
#weights are 4,5 and bias is -9, find the sigmoid(x) = 1/(1+e-x).
features=[(1,1),(2,4),(5,-5),(-4,5)]
def linear_func(features):
for x in features:
score= 4*x[0]+ 5*x[1]-9
y = 1 / (1 + e - score)
print("X",x,score,y)
#call the function
linear_func(features)
Im taking the same udacity course. You asked the wrong question, got the wrong answer.
its 1/1(1+e^-x)
the correct code is
from math import e,pow
#Equation: 4x1+5x2-9=score
#weights are 4,5 and bias is -9, find the sigmoid(x) = 1/(1+e-x).
features=[(1,1),(2,4),(5,-5),(-4,5)]
def linear_func(features):
for x in features:
score= 4*x[0]+ 5*x[1]-9
y = 1 / (1 + pow(e,- score))
print("X",x,score,y)
#call the function
linear_func(features)
answer is 1,1 and -4,5
answer on udacity:
That's correct. Intuitively, if you look at the plot for a sigmoid function, the probability would always be 50% if the score evaluates to zero.
Related
I have two points on a map -
val point1 : LatLng(13.3016139,77.4219107)
val point2 : LatLng(14.1788932,77.7613413)
I want to calculate and find 100 equidistant points along a straight line between these two coordinates. How do I do that?
ps. I'm sure this has been asked before, I just can't find it.
Equidistant, and more importantly, straight by which projection?
usually, to find a distance in cartesian space one would use something like the
Haversine formula to find a value, as previously answered in stack answer: How to convert latitude or longitude to meters?
As for the equidistant part, once you have the distance decided as per your taste of the shape and radius of Earth at given points, a simple division will do. .
python 3.7
>>> dist = 5427 #just some number
>>> nbr_o_points = 101
>>> points = [(dist/nbr_o_points)*(i+1) for i in range(nbr_o_points)]
>>> [f'{p:.2f}' for p in points]
['53.73', '107.47', '161.20',..., '5319.53', '5373.27', '5427.00']
Now to transfer these distances from point a to b back onto the desired projection... This is not part of your question... Stack - how-to-determine-vector-between-two-lat-lon-points might help.
take the vector and multiply by the dists in points in order to get your coordinates.
This is how I solved it -
fun findEquidistantPoints(latLng1: LatLng, latLng2: LatLng, pointCount: Int): ArrayList<LatLng> {
if (pointCount < 0)
throw IllegalArgumentException("PointCount cannot be less than 0")
val points = ArrayList<LatLng>()
val displacement = latLng1.displacementFromInMeters(latLng2)
val distanceBetweenPoints = displacement / (pointCount + 1)
for (i in 1..pointCount) {
val t = (distanceBetweenPoints * i) / displacement
points.add(LatLng(
(1 - t) * latLng1.latitude + t * latLng2.latitude,
(1 - t) * latLng1.longitude + t * latLng2.longitude
))
}
return points
}
If you are given:
a probability distribution that a robot sensor detects an object , given that it is in a location p(z|x).
prior probabilities that the robot is in any location
An actual observation made by the robots sensor
and asked to update the probability distribution given this observation, what method would one use?
I am not sure if i should be using a bayes filter, kalman filter, or if i am over thinking this problem.
as an example:
if a robot can move along a the number line from 1-7, with a standing poll at x=4. The robot can tell if the poll is to its left, right or in front of it (z=-1,1,0 respectfully).
p(z|x) x=1 x=2 x=3 x=4 x=5 x=6 x=7
z= -1 0 0 0 .25 .5 .5 .5
z = 0 ... (its an example so im leaving this off)
z = 1 ...
𝑝(𝑥 = 1) = 0.1; 𝑝(𝑥 = 2) = 0.2; 𝑝(𝑥 = 3) = 0.2; 𝑝(𝑥 = 4) = 0.2;
𝑝(𝑥 = 5) = 0.2; 𝑝(𝑥 = 6) = 0.1; 𝑝(𝑥 = 7) = 0.0
than the robot sensor outputs z=-1 . What method would i use to update the above table
i am not an expert in this but i have seen a similar kind of topic discussed in this link. i hope this mighthelp you. good luck.
http://bilgin.esme.org/BitsAndBytes/KalmanFilterforDummies
This question is related to matlab: find the index of common values at the same entry from two arrays.
Suppose that I have an 1000 by 10000 matrix that contains value 0,1,and 2. Each row are treated as a sample. I want to calculate the pairwise distance between those samples according to the formula d = 1-1/(2p)sum(a/c+b/d) where a,b,c,d can treated as as the row vector of length 10000 according to some definition and p=10000. c and d are probabilities such that c+d=1.
An example of how to find the values of a,b,c,d: suppose we want to find d between sample i and bj, then I look at row i and j.
If kth entry of row i and j has value 2 and 2, then a=2,b=0,c=1,d=0 (I guess I will assign 0/0=0 in this case).
If kth entry of row i and j has value 2 and 1 or vice versa, then a=1,b=0,c=3/4,d=1/4.
The similar assignment will give to the case for 2,0(a=0,b=0,c=1/2,d=1/2),1,1(a=1,b=1,c=1/2,d=1/2),1,0(a=0,b=1,c=1/4,d=3/4),0,0(a=0,b=2,c=0,d=1).
The matlab code I have so far is using for loops for i and j, then find the cases above by using find, then create two arrays for a/c and b/d. This is extremely slow, is there a way that I can improve the efficiency?
Edit: the distance d is the formula given in this paper on page 13.
Provided those coefficients are fixed, then I think I've successfully vectorised the distance function. Figuring out the formulae was fun. I flipped things around a bit to minimise division, and since I wasn't aware of pdist until #horchler's comment, you get it wrapped in loops with the constants factored out:
% m is the data
[n p] = size(m, 1);
distance = zeros(n);
for ii=1:n
for jj=ii+1:n
a = min(m(ii,:), m(jj,:));
b = 2 - max(m(ii,:), m(jj,:));
c = 4 ./ (m(ii,:) + m(jj,:));
c(c == Inf) = 0;
d = 1 - c;
distance(ii,jj) = sum(a.*c + b.*d);
% distance(jj,ii) = distance(ii,jj); % optional for the full matrix
end
end
distance = 1 - (1 / (2 * p)) * distance;
I need help plotting a moving average on top of the data I am already able to plot (see below)
I am trying to make m (my moving average) equal to the length of y (my data) and then within my 'for' loop, I seem to have the right math for my moving average.
What works: plotting x and y
What doesn't work: plotting m on top of x & y and gives me this error
RuntimeWarning: invalid value encountered in double_scalars
My theory: I am setting m to np.arrays = y.shape and then creating my for loop to make m equal to the math set within the loop thus replacing all the 0's to the moving average
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as mdates
import csv
import math
def graph():
date, value = np.loadtxt("CL1.csv", delimiter=',', unpack=True,
converters = {0: mdates.strpdate2num('%d/%m/%Y')})
fig = plt.figure()
ax1 = fig.add_subplot(1,1,1, axisbg = 'white')
plt.plot_date(x=date, y=value, fmt = '-')
y = value
m = np.zeros(y.shape)
for i in range(10, y.shape[0]):
m[i-10] = y[i-10:1].mean()
plt.plot_date(x=date, y=value, fmt = '-', color='g')
plt.plot_date(x=date, y=m, fmt = '-', color='b')
plt.title('NG1 Chart')
plt.xlabel('Date')
plt.ylabel('Price')
plt.show()
graph ()
I think that lmjohns3 answer is correct, but you have a couple of problems with your moving average function. First of all, there is the indexing problem the lmjohns3 pointed out. Take the following data for example:
In [1]: import numpy as np
In [2]: a = np.arange(10)
In [3]: a
Out[3]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
Your function gives the following moving average values:
In [4]: for i in range(3, a.shape[0]):
...: print a[i-3:i].mean(),
1.0 2.0 3.0 4.0 5.0 6.0 7.0
The size of this array (7) is too small by one number. The last value in the moving average should be (7+8+9)/3=8. To fix that you could change your function as follows:
In [5]: for i in range(3, a.shape[0] + 1):
...: print a[i-3:i].sum()/3,
1 2 3 4 5 6 7 8
The second problem is that in order to plot two sets of data, the total number of data points needs to be the same. Your function returns a new set of data that is smaller than the original data set. (You maybe didn't notice because you preassigned a zeros array of the same size. Your for loop will always produce an array with a bunch of zeros at the end.)
The convolution function gives you the correct data, but it has two extra values (some at each end) because of the same argument, which ensures that the new data array has the same size as the original.
In [6]: np.convolve(a, [1./3]*3, 'same')
Out[6]:
array([ 0.33333333, 1. , 2. , 3. , 4. ,
5. , 6. , 7. , 8. , 5.66666667])
As an alternate method, you could vectorize your code by using Numpy's cumsum function.
In [7]: (cs[3-1:] - np.append(0,cs[:-3]))/3.
Out[7]: array([ 1., 2., 3., 4., 5., 6., 7., 8.])
(This last one is a modification of the answer in a previous post.)
The trick might be that you should drop the first values of your date array. For example use the following plotting call, where n is the number of points in your average:
plt.plot_date(x=date[n-1:], y=m, fmt = '-', color='b')
The problem here lives in your computation of the moving average -- you just have a couple of off-by-one problems in the indexing !
y = value
m = np.zeros(y.shape)
for i in range(10, y.shape[0]):
m[i-10] = y[i-10:1].mean()
Here you've got everything right except for the :1]. This tells the interpreter to take a slice starting at whatever i-10 happens to be, and ending just before 1. But if i-10 is larger than 1, this results in the empty list ! To fix it, just replace 1 with i.
Additionally, your range needs to be extended by one at the end. Replace y.shape[0] with y.shape[0]+1.
Alternative
I just thought I'd mention that you can compute the moving average more automatically by using np.convolve (docs) :
m = np.convolve(y, [1. / 10] * 10, 'same')
In this case, m will have the same length as y, but the moving average values might look strange at the beginning and end. This is because 'same' effectively causes y to be padded with zeros at both ends so that there are enough y values to use when computing the convolution.
If you'd prefer to get only moving average values that are computed using values from y (and not from additional zero-padding), you can replace 'same' with 'valid'. In this case, as Ryan points out, m will be shorter than y (more precisely, len(m) == len(y) - len(filter) + 1), which you can address in your plot by removing the first or last elements of your date array.
Okay, either I'm going nuts or it actually worked - I compared my chart vs. another chart and it seemed to have worked.
Does this make sense?
m = np.zeros(y.shape)
for i in range(10, y.shape[0]):
m[i-10] = y[i-10:i].mean()
plt.plot_date(x=date, y=m, fmt = '-', color='r')
Here is the graph I currently have
:
The Dotted Blue line represented the y value that corresponds to the x value I am looking for. I am trying to find the x values of the line's intersections with the blue curve(Upper).Since the interesections do not fall on a point that has already been defined, we need to interpolate a point that falls onto the Upper plot.
Here is the information I have:
LineValue - The y value of the intersection and the value of the dotted line( y = LineValue)
Frequency - an array containing the x value coordinates seen on this plot. The interpolated values of Frequency that corresponds to LineValue are what we are looking for
Upper/Lower - arrays containing the y value info for this graph
This solution is an improvement on Amro's answer. Instead of using fzero you can simply calculate the intersection of the line by looking for transition in the first-difference of the series created by a logical comparison to LineValue. So, using Amro's sample data:
>> x = linspace(-100,100,100);
>> y = 1-2.*exp(-0.5*x.^2./20)./(2*pi) + randn(size(x))*0.002;
>> LineValue = 0.8;
Find the starting indices of those segments of consecutive points that exceed LineValue:
>> idx = find(diff(y >= LineValue))
idx =
48 52
You can then calculate the x positions of the intersection points using weighted averages (i.e. linear interpolation):
>> x2 = x(idx) + (LineValue - y(idx)) .* (x(idx+1) - x(idx)) ./ (y(idx+1) - y(idx))
x2 =
-4.24568579887939 4.28720287203057
Plot these up to verify the results:
>> figure;
>> plot(x, y, 'b.-', x2, LineValue, 'go', [x(1) x(end)], LineValue*[1 1], 'k:');
The advantages of this approach are:
The determination of the intersection points is vectorized so will work regardless of the number of intersection points.
Determining the intersection points arithmetically is presumably faster than using fzero.
Example solution using FZERO:
%# data resembling your curve
x = linspace(-100,100,100);
f = #(x) 1-2.*exp(-0.5*x.^2./20)./(2*pi) + randn(size(x))*0.002;
VALUE = 0.8;
%# solve f(x)=VALUE
z1 = fzero(#(x)f(x)-VALUE, -10); %# find solution near x=-10
z2 = fzero(#(x)f(x)-VALUE, 10); %# find solution near x=+10
%# plot
plot(x,f(x),'b.-'), hold on
plot(z1, VALUE, 'go', z2, VALUE, 'go')
line(xlim(), [VALUE VALUE], 'Color',[0.4 0.4 0.4], 'LineStyle',':')
hold off
Are the step sizes in your data series the same?
Is the governing equation assumed to be cubic, sinuisoidal, etc..?
doc interpl
Find the zero crossings