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I want to estimate the point of fracture (x_F) (red circle) via ternary operator to restrict the range of my plot to it.
Example plot
To achieve a restriction to the X(Y_max)-value the stats command in combination with ternary-operator seems to be sufficient:
stats 'foo.csv' u 5 nooutput name 'Y_'
stats 'foo.csv' u 4 every ::Y_index_max::Y_index_max nooutput
X_max = STATS_max
plot 'foo.csv' u 4:(($4 <= X_max) ? $5 : 1/0) w l notitle
I cannot use the X_max-variable, because there a several points beyond the point of fracture (x_n > x_F) due to measurement errors. My idea was to compare the x-entries $4 to one another and to save the first point which satisfies $4_prev > $4_curr and to save it as x_F=$4_prev.
A simple delta-function seems to do the trick: delta(x)=(D=x-old_D,old_D=x,D) and old_D=NaN in combination with the ternary operator (delta($4)>0 ? $5 : 1/0) whereas $5 is the y-value, which will be plotted as long as the difference of two sequent x-values is positive.
You want to discard any data point after dx has become negative for the first time, right? You'll need a flag variable, i called it broken, which is set after the first occurrence of dx < 0:
broken = 0
plot dataf us (lastx=thisx, thisx=$4): \
(broken ? 1/0 :($4 < lastx) ? $5 : (broken=1, 1/0))
This uses the comma as "serial evaluator", same as in C, etc.
(Not tested now, as i don't have a suitable data set at hand and was too lazy to create one.)
Update: You can put the assignment broken=0 into the plot
plot broken=0, dataf us ....
, to be able to replot, zoom that plot etc.
I have a symbolic matrix that I want to differentiate. I have to substitute numeric values to some of the vars and then to solve with respect to 6 unknowns. My problem is that defining the element of matrix A by lambda and subistituting with subs doesn't change any value in the matrix. When I want retrieve the type of matrix in fact it's shown that it's immutable, which seems quite odd. Here's the code:
def optimalF1():
x,y,z=symbols('x y z', Real=True)
phi,theta,psi=symbols('phi theta psi')
b1x,b1y=symbols('b1x b1y')
b2x,b2y=symbols('b2x b2y')
b3x,b3y=symbols('b3x b3y')
b4x,b4y=symbols('b4x b4y')
b5x,b5y=symbols('b5x b5y')
b6x,b6y=symbols('b6x b6y')
bMat=sym.Matrix(([b1x,b2x,b3x,b4x,b5x,b6x],
[b1y,b2y,b3y,b4y,b5y,b6y],[0,0,0,0,0,0]))
mov=np.array([[x],[y],[z]])
Pi=np.repeat(mov,6,axis=1)
sym.pprint(Pi)
print 'shape of thing Pi', np.shape(Pi)
p1x,p1y,p1z=symbols('p1x,p1y,p1z')
p2x,p2y,p2z=symbols('p2x,p2y,p2z')
p3x,p3y,p3z=symbols('p3x,p3y,p3z')
p4x,p4y,p4z=symbols('p4x,p4y,p4z')
p5x,p5y,p5z=symbols('p5x,p5y,p5z')
p6x,p6y,p6z=symbols('p6x,p6y,p6z')
#legs symbolic array
l1,l2,l3,l4,l5,l6=symbols('l1,l2,l3,l4,l5,l6')
piMat=Matrix(([p1x,p2x,p3x,p4x,p5x,p6x],[p1y,p2y,p3y,\
p4y,p5y,p6y],[p1z,p2z,p3z,p4z,p5z,p6z]))
piMat=piMat.subs('p1z',0)
piMat=piMat.subs('p2z',0)
piMat=piMat.subs('p3z',0)
piMat=piMat.subs('p4z',0)
piMat=piMat.subs('p5z',0)
piMat=piMat.subs('p6z',0)
sym.pprint(piMat)
legStroke=np.array([[l1],[l2],[l3],[l4],[l5],[l6]])
'''redefine the Eul matrix
copy values of Pi 6 times by using np.repeat
'''
r1=[cos(phi)*cos(theta)*cos(psi)-sin(phi)*sin(psi),\
-cos(phi)*cos(theta)*sin(psi)-sin(phi)*cos(psi),\
cos(phi)*sin(theta)]
r2=[sin(phi)*cos(theta)*cos(psi)+cos(phi)*sin(psi),\
-sin(phi)*cos(theta)*sin(psi)+cos(phi)*cos(psi),\
sin(phi)*sin(theta)]
r3= [-sin(theta)*cos(psi),sin(theta)*sin(psi),cos(theta)]
EulMat=Matrix((r1,r2,r3))
print(EulMat)
uvw=Pi+EulMat*piMat
print 'uvw matrix is:\n', uvw, np.shape(uvw)
# check thisout -more elegant and compact form
A=Matrix(6,1,lambda j,i:((uvw[0,j]- \
bMat[0,j])**2+(uvw[1,j]-bMat[1,j])**2+\
(uvw[2,j]-bMat[2,j])**2)-legStroke[j]**2)
print'A matrix before simplification:\n ', A
B=simplify(A)
B=B.subs({'x':1.37,'y':0,'z':0,theta:-1.37,phi:0})
print'A matrix form after substituting:\n',B
So comparing A and B leads to the same output. I don't understand why!
When you use subs with variables that have assumptions, you have to use the symbols not strings. Using strings causes a new generic symbol to be created which does not match the symbol having assumptions so the subs fails.
>>> var('x')
x
>>> var('y',real=True)
y
>>> (x+y).subs('x',1).subs('y',2)
y + 1
Note, too, that to make real symbols you should use real=True not Real=True (lower case r).
I need to find ranges in order to create a Uniform histogram
i.e: ages
to 4 ranges
data_set = [18,21,22,24,27,27,28,29,30,32,33,33,42,42,45,46]
is there a function that gives me the ranges so the histogram is uniform?
in this case
ranges = [(18,24), (27,29), (30,33), (42,46)]
This example is easy, I'd like to know if there is an algorithm that deals with complex data sets as well
thanks
You are looking for the quantiles that split up your data equally. This combined with cutshould work. So, suppose you want n groups.
set.seed(1)
x <- rnorm(1000) # Generate some toy data
n <- 10
uniform <- cut(x, c(-Inf, quantile(x, prob = (1:(n-1))/n), Inf)) # Determine the groups
plot(uniform)
Edit: now corrected to yield the correct cuts in the ends.
Edit2: I don't quite understand the downvote. But this also works in your example:
data_set = c(18,21,22,24,27,27,28,29,30,32,33,33,42,42,45,46)
n <- 4
groups <- cut(data_set, breaks = c(-Inf, quantile(data_set, prob = 1:(n-1)/n), Inf))
levels(groups)
With some minor renaming nessesary. For slightly better level names, you could also put in min(x) and max(x) instead of -Inf and Inf.
I am new to programming and I am trying to create a program that will take 3 random numbers X Y and Z and will sort them into ascending order X being the lowest and Z the highest using Min, Max functions and a Variable (tmp)
I know that there is a particular strategy that I need to use that effects the (X,Y) pair first then (Y,Z) then (X,Y) again but I can't grasp the logic.
The closest I have got so far is...
y=min(y,z)
x=min(x,y)
tmp=max(y,z)
z=tmp
tmp=max(x,y)
y=tmp
x=min(x,y)
tmp=max(x,y)
y=tmp
I've tried so many different combinations but it seems that the problem is UNSOLVABLE can anybody else help?
You need to get sort the X,Y Pair first
tmp=min(x,y)
y=max(x,y)
x=tmp
Then sort the Y,Z pair
tmp = min(y,z)
z=max(y,z)
y=tmp
Then, resort the X,Y pair (in case the original Z was the lowest value...
tmp=min(x,y)
y=max(x,y)
x=tmp
If the commands you have mentioned are the only ones available on the website, and you can only use each one once try:
# Sort X,Y pair
tmp=max(x,y)
x=min(x,y)
y=tmp
# Sort Y,Z pair
tmp=max(y,z)
y=min(y,z)
z=tmp
# Sort X,Y pair again.
tmp=max(x,y)
x=min(x,y)
y=tmp
Hope that helps.
I'm not sure if I understood your question correctly, but you are over righting your variables. Or are you trying to solve some homework with the restriction to only use min() and max() functions?
What about using a list?
tmp = [x, y, z]
tmp.sort()
x, y, z = tmp
Is there a way to convert Math Equations to a Text?
ex.
3π * 3 = x
out put will be
3(3.141592) multiplied by 3 equals to x
By default Its not possible, You can write a code in such way that it will get the correct string(word) for different symbols and numbers. You can use the database for storing the words or you can do it with your functions.