Coding problems

[ritesh247]

I got a list of numbers lets say A = [0.456, 0.4344, 1, -1, 4, 5, 6,8]

how can the function mentioned below be fixed in python using the list of numbers mentioned above and get the six different solutions of 'xf' for lets say the code mentioned below:

c=2
xfmin = 0.432*c
xfmax = 0.528*c
def xf(a, b):                                           
    return a+(b-a)*A
xfrn= xf(xfmin, xfmax)
print "xf=", xfrn

[bvdet]

Please use code tags when posting code. See posting guidelines here.

Add an argument to the function and iterate on the list in a for loop:

A = [0.456, 0.4344, 1, -1, 4, 5, 6,8]
 
def xf(a, b, item):
    return a+(b-a)*item
 
c=2
xfmin = 0.432*c
xfmax = 0.528*c
 
for item in A:
    xfrn= xf(xfmin, xfmax, item)
    print "xf =", xfrn

BV - Moderator

[ritesh247]

thank you very much for the reply,

since am a beginner in python I also wanted ask you another question:

let say if you got if you got a sequence of numbers lets say

A = [3.4, 5.6, 7.7, 9.7]
B = [3.5, 4.7.-3.3, 8.7]
C = [8.5, 6.7.-8.3, 4.7]
 
Z = A*B*C

the value of Z should have four answers


I tried using for loops, lets say for

A = [3.4, 5.6, 7.7, 9.7]
B = [3.5, 4.7.-3.3, 8.7]
C = [8.5, 6.7.-8.3, 4.7]
for i in A:
for j in B:
for k in C:
Z = i*j*k

print z # four possible answers

but it didn't do what I wanted it to do.
Could you please help me out solve this problem.

[bvdet]

Since the lists are the same length (4) and you want four answers, one could assume you want to multiply the number at index i of list A by the corresponding index of the other two lists. Iterate on A using built-in function enumerate().

for i, item in enumerate(A):
    print item*B[i]*C[i]

[ritesh247]

Is there any other way of doing this other than using enumerate

lets say if the scenario was such that

 A = [3.4, 5.6, 7.7, 9.7]
 B = [3.5, 4.7.-3.3, 8.7]
 C = [8.5, 6.7.-8.3, 4.7]
 
f=2
g=4
 
z=A*B
y=A*C*f
u=C*g
 
ww=z*y*u

is there any possibility of obtaining four solutions for ww??

and also lets say if you had to use each one of the column of A B and C at each time, is it possible, cause I have tried different methods but am failing to get what am trying to achieve

[Glenton]

Even better is to use numpy arrays. Import numpy, convert a,b,c to numpy arrays, and then z=a*b*c does what you want it to do.

[ritesh247]

I tried using numpy arrays but i was getting this error

Traceback (most recent call last):

File "C:\Python25\GEM SA.py", line 74, in <module>
A = matrix([[a11,a12],[a21,a22]])
File "C:\Python25\Lib\site-packages\numpy\core\defmatrix.py", line 217, in __new__
arr = N.array(data, dtype=dtype, copy=copy)
ValueError: setting an array element with a sequence.

I would be really thankful if someone can help me out as am struggling to get the end result. I am trying to figure out how I can use each value from the each array to get an output result, in total i will get three different results, is there any method available to solve such a scenario

xfrn=array([0.937143,0.885173,1.0329])
 
 
theta_freqrn= array([10.1579,9.99248,10.9348])
 
kappa_freqrn= array([4.95614, 4, 2])
 
arn = array([6.32918, 5, 2])
 
Mthetadotrn= array([-1.15940, 1, 4])
 
 
 
s = 7.5
c = 2.0
m = 100.0
 
rho = 1.225
xcm = 0.5*c
 
 
e = xfrn/c-0.25    
 
 
vel   = zeros((3500,1),float)
freq1 = zeros((3500,1),float)
freq2 = zeros((3500,1),float)
freq3 = zeros((3500,1),float)
freq4 = zeros((3500,1),float)
damp1 = zeros((3500,1),float)
damp2 = zeros((3500,1),float)
damp3 = zeros((3500,1),float)
damp4 = zeros((3500,1),float)
 
vstart = 1.0
vinc = 0.1
for icount in range(3500):            
    V=vstart + icount*vinc
 
    a11 = (m*s*s*s*c)/3.0
    a12 = m*s*s*(c*c/2.0-c*xfrn)/2.0
    a21 = a12
    a22 = m*s*(c*c*c/3.0-c*c*xfrn+xfrn*xfrn*c)
 
    A = matrix([[a11,a12],[a21,a22]])
    Ainv = linalg.inv(A)
 
    k1 = ((kappa_freqrn*arn)**2)*a11 
    k2 = ((theta_freqrn*arn)**2)*a22  
 
    c11 = (rho*V*c*s*s*s*arn)/6.0
    c12 = 0
    c21 = -(rho*V*c*c*s*s*e*arn)/4.0
    c22 = -(rho*V*c*c*c*s*Mthetadotrn)/8.0
 
    C = matrix([[c11,c12],[c21,c22]])
 
    AC = -Ainv*C 
 
    k11 = k1
    k12 = (rho*V*V*c*s*s*arn)/4.0
    k21 = 0
    k22 = k2 - (rho*V*V*c*c*s*e*arn)/2.0
    K = matrix([[k11,k12],[k21,k22]])
 
    AK = -Ainv*K
 
    matrix_bind_columns = concatenate((AK,AC),axis=1)
    M = concatenate((matrix([[0.0,0.0,1.0,0.0],[0.0,0.0,0.0,1.0]]),matrix_bind_columns),axis=0)
 
    Meig = linalg.eig(M)[0]
 
    vel[icount] = V
    freq1[icount] = abs(Meig[0])
    damp1[icount] = -100.0*real(Meig[0])/freq1[icount]
    freq1[icount] = freq1[icount]/2*math.pi
    freq2[icount] = abs(Meig[1])
    damp2[icount] = -100.0*real(Meig[1])/freq2[icount]
    freq2[icount] = freq2[icount]/2*math.pi
    freq3[icount] = abs(Meig[2])
    damp3[icount] = -100.0*real(Meig[2])/freq3[icount]
    freq3[icount] = freq3[icount]/2*math.pi
    freq4[icount] = abs(Meig[3])
    damp4[icount] = -100.0*real(Meig[3])/freq4[icount]
    freq4[icount] = freq4[icount]/2*math.pi
 
    if damp3[icount] < 0:           goes below zero
        break
 
ab = (damp3[icount-1])
 
 
bc = (damp3[icount])
 
 
Vf_min = V-0.1
 
 
Vf_max = V
 
 
def flutter_speed(ab,bc,Vf_min,Vf_max):                  
    return (Vf_min+(Vf_max - Vf_min)*(-(ab)/(bc - ab)))
Vf = flutter_speed(ab,bc,Vf_min,Vf_max)
print Vf

[ritesh247]

I am wondering if anyone can help me fix the array problem out?? If someone can give me feedback on where I am going wrong??

[Glenton]

Do you mean just for the multiplication?

In [1]: from numpy import *
 
In [2]: A=array([3.4,5.6,7.7,9.7])
 
In [3]: B=array([3.5,4.7,-3.3,8.7])
 
In [4]: C=array([8.5,6.7,-8.3,4.7])
 
In [5]: f=2
 
In [6]: g=4
 
In [7]: z=A*B
 
In [8]: y=A*C*f
 
In [9]: u=C*g
 
In [10]: ww=z*y*u
 
In [11]: z
Out[11]: array([ 11.9 ,  26.32, -25.41,  84.39])
 
In [12]: y
Out[12]: array([  57.8 ,   75.04, -127.82,   91.18])
 
In [13]: u
Out[13]: array([ 34. ,  26.8, -33.2,  18.8])
 
In [14]: ww
Out[14]: array([  23385.88   ,   52931.41504, -107830.48584,  144659.98776])
 
 

If it's something else, maybe start a new post.