How do I handle mixed input to a Python function, ie: 5x?

[TMS]

I'm working through the bisection method to find the root of a function. I'm asking patience (yet again) because I haven't used this method for a while. I found the formula on wikipedia.

Pseudocode is as follows:

start loop
Do While (xR - xL) > epsilon

'calculate midpoint of domain
xM = (xR + xL) / 2

'Find f(xM)
If ((f(xL) * f(xM)) > 0 Then
'throw away left half
xL = xM
else
'throw away right half
xR = xM
End If
Loop

I'm trying to make my way through this and I have some questions about how Python handles an equation like 5x^8 - 3x^4 + x - 2? If I try it from the shell I get an error message from the x since its with the 5, so I assume it confused about what I'm asking.

A friend told me to use eval(function) but I'm not sure how to use it in my code, or if it is even the best way to go.

Mind you, this is a starting point. There will be MANY more questions before I'm done with this assignment.

The function is to look like this:

def root_find(f, xLo, xHi, eps)

So, if I test the above so that f = 5x**8 - 3x**4 + x - 2. by saying y0 = f(xLo) I will get an error, right? How do I format the input so the function will use it? Am I even asking the right question? (I'm so lost!)


thank you for helping

[bartonc]

I'm working through the bisection method to find the root of a function. I'm asking patience (yet again) because I haven't used this method for a while. I found the formula on wikipedia.

Pseudocode is as follows:

start loop
Do While (xR - xL) > epsilon

'calculate midpoint of domain
xM = (xR + xL) / 2

'Find f(xM)
If ((f(xL) * f(xM)) > 0 Then
'throw away left half
xL = xM
else
'throw away right half
xR = xM
End If
Loop

I'm trying to make my way through this and I have some questions about how Python handles an equation like 5x^8 - 3x^4 + x - 2? If I try it from the shell I get an error message from the x since its with the 5, so I assume it confused about what I'm asking.

A friend told me to use eval(function) but I'm not sure how to use it in my code, or if it is even the best way to go.

Mind you, this is a starting point. There will be MANY more questions before I'm done with this assignment.

The function is to look like this:

def root_find(f, xLo, xHi, eps)

So, if I test the above so that f = 5x**8 - 3x**4 + x - 2. by saying y0 = f(xLo) I will get an error, right? How do I format the input so the function will use it? Am I even asking the right question? (I'm so lost!)


thank you for helping

I always start off by assuming the x == 1 in order to simplify things at the outset. This makes 5x == 5, etc. Then, after the function is working, you can add the x to the parameter list and do things like (5 * x)**8. Of course, if you are trying to solve for x then things are a lot more complicated, but if you can work it out on paper you should be able to forulate the appropriate algorithm.
Whether you use eval() or not (eval() is just another way of giving the interpeter text to work on) you basically have two things that go into the formula: variables which have names and values (ie a = 1) and literals (ie "hello") all of which must be assigned before they can be evaluated.

[TMS]

thank you... as always you are a life saver!!! And you make it look so easy (never mind that it is).

TMS

[bartonc]

thank you... as always you are a life saver!!! And you make it look so easy (never mind that it is).

TMS

Any time... really!

[TMS]

ok, wait.

If my function is (and I'm only trying to see that 5*x is handled properly)

def root_find(f, xLo, xHi, eps):
    y1 = f(xLo)
print y1
 
 

and I test it with:

s = root_find(5*x, -1, 1, .0001)
print s
Nothing happens. It doesn't print, it doesn't do anything. Is there something else I must do to get it to process the function that is entered for f? Again, thank you for your patience with me.

TMS

[bartonc]

ok, wait.

If my function is (and I'm only trying to see that 5*x is handled properly)

def root_find(f, xLo, xHi, eps):
    y1 = f(xLo)
print y1
 
 

and I test it with:

s = root_find(5*x, -1, 1, .0001)
print s
Nothing happens. It doesn't print, it doesn't do anything. Is there something else I must do to get it to process the function that is entered for f? Again, thank you for your patience with me.

TMS

That's funny. You should have gotten two errors. First:

>>> def root_find(f, xLo, xHi, eps):
... y1 = f(xLo)
... print y1
...
>>> s = root_find(5*x, -1, 1, .0001)
>>> s = root_find(5*x, -1, 1, .0001)
File "<console>", line 1, in ?
''' exceptions.NameError : name 'x' is not defined '''

because x has not yet been assigned a value.
>>> x = 1
Fixed error #1. Then:

>>> s = root_find(5*x, -1, 1, .0001)
File "<console>", line 1, in ?
File "<console>", line 3, in root_find
''' exceptions.TypeError : 'int' object is not callable '''
>>>

Because
... y1 = f(xLo)
syntax says "call f with xLo as the argument".
you want f**xLo, right?

also you missed one indent and the return:

def root_find(f, xLo, xHi, eps):
    y1 = f**xLo
    return y1
 

If you don't put the return in there, the function will return None.

>>> s = root_find(5*x, -1, 1, .0001)
>>> print s
0.2
>>>

By the way, are using IDLE to edit and run this?

[TMS]

yes. I am just running the module.

What I'm trying to do is start with 5*x and then go to 5*x**3. So, xLo should be x for one iteration, then when that works, xHi would be next. In otherwords, I'm starting really small to get the find_root algorithm going.

When I say f(xLo) I am wanting it to map xLo into x, so if f = 5x and xLo = -1, I should get a -5 back. It is my understanding of the bisection method that you solve using xLo in the equation, then try it using xHi. Then you divide the two answers by 2 to get the mid. That allows you to get closer to epsilon. Am I way off base?

tms

[bartonc]

yes. I am just running the module.

What I'm trying to do is start with 5*x and then go to 5*x**3. So, xLo should be x for one iteration, then when that works, xHi would be next. In otherwords, I'm starting really small to get the find_root algorithm going.

When I say f(xLo) I am wanting it to map xLo into x, so if f = 5x and xLo = -1, I should get a -5 back. It is my understanding of the bisection method that you solve using xLo in the equation, then try it using xHi. Then you divide the two answers by 2 to get the mid. That allows you to get closer to epsilon. Am I way off base?

tms

No you're not off base. I had read "5x^8 - 3x^4 + x - 2" not the algorithm when I wrote that. Psuedocode translates very nicely into python. In fact, I write in python now where I used to use psuedocode. Although the parentheses are a bit confusing, in the example,

start loop
Do While (xR - xL) > epsilon

'calculate midpoint of domain
xM = (xR + xL) / 2

'Find f(xM)
If ((f(xL) * f(xM)) > 0 Then
'throw away left half
xL = xM
else
'throw away right half
xR = xM
End If
Loop

the python would look something like this:

def f(arg):
    arg = arg * 1   # re-assignment is perfectly fine.
    return arg   # do any math you want to here
 
def FindRoot(f, x, L, R, e):
    xL = x * L
    xR = x * R
    while (xR  - xL) > e:
        xM = (xR  + xL) / 2.0   # python is a bit funny aboud dividing ints vs floats
        if (f(xL) * f(xM) > 0:
            xL = xM
        else:
            xR = xM
 
print FindRoot(f, 5, -1, 1, .0001)

[bvdet]

A recent thread http://www.thescripts.com/forum/thread583573.html may provide a partial solution to you. I wrote a simple script that incorporates some of the functions and classes from it:

# 5x^3-3x^2+x-2
d = Doublet(Monome(5,3,0),Doublet(Monome(-3,2,0),Doublet(Monome(1,1,0),Doublet(Monome(-2,0,0), None))))
print d
 
def solve(d, vx, vy = 1.0):
    r = 0.0
    for m in iterDoub(d):
        r += m.coeff*float(vx)**m.expoX*float(vy)**m.expoY
    return r        
 
def bisection(d, U, L, e):
    if solve(d, U) > 0 and solve(d, L) < 0:
        while abs(U - L) > e:
            M = (U+L)/2
            if solve(d, M)*solve(d,U) > 0:
                U = M
            else:
                L = M
            print U, L
        return U,L
    else:
        return "Invalid arguments. f(U) must be > 0 and f(L) must be < 0."
 
print bisection(d, 1.0, -1.0, 0.00001)

Output:

>>> 5(x^3)-3(x^2)+x-2
1.0 0.0
1.0 0.5
1.0 0.75
1.0 0.875
0.9375 0.875
0.90625 0.875
0.890625 0.875
0.890625 0.8828125
0.88671875 0.8828125
0.88671875 0.884765625
0.8857421875 0.884765625
0.88525390625 0.884765625
0.885009765625 0.884765625
0.884887695313 0.884765625
0.884887695313 0.884826660156
0.884857177734 0.884826660156
0.884857177734 0.884841918945
0.884857177734 0.88484954834
(0.884857177734375, 0.88484954833984375)

[TMS]

Ok, but I'm still confused about how Python handles the 5x**8. The x is not defined, so there is an error. You defined f with a function, but later you added a variable to FindRoot(f, x, L, R, e). My function is assigned by the teacher and is as follows:
find_root(f, xLo, xHi, eps). I can only have 4 variables. Does your def f(arg) attempt to solve that problem?

def f(arg):
    arg = arg * 1   # re-assignment is perfectly fine.
    return arg   # do any math you want to here
 
def FindRoot(f, x, L, R, e):
    xL = x * L
    xR = x * R
    while (xR  - xL) > e:
        xM = (xR  + xL) / 2.0   # python is a bit funny aboud dividing ints vs floats
        if (f(xL) * f(xM) > 0:
            xL = xM
        else:
            xR = xM
 
print FindRoot(f, 5, -1, 1, .0001)

[TMS]

A recent thread http://www.thescripts.com/forum/thread583573.html may provide a partial solution to you.

I looked over that link. There is a lot of information, but I understand very little of it. For example, the Doublet(Monome...) part. Doublet is part of what module? It isn't defined so I get an error message.

Again, I'm starting at the very beginning of this assignment, and not necessarily trying to write the functions complete yet (its due on Wednesday!!!). I want to understand this:
If my teacher runs my module and find_root is defined like this:

find_root(f, xLo, xHi, eps):

and the teacher enters this:

find_root(5x**3 + 2x**2-x+1, -3, 4, .0001)

how do I help Python understand that x is defined by xLo and then xHi. In order to solve for the root, if I understand bisection correctly, I use xLo to find y0, and xHi to find y1 by using it like this:

y0 = 5(xLo)**3 + 2(xLo)**2 - (xLo) + 1
y1 = 5(xHi)**3 + 2(xHi)**2 - (xHi) + 1

If it is still higher than eps, then I divide by 2... etc. But in order to begin I have to map xLo into x. And in order to do that, I have to get Python to know what to do with an x that is undefined.

Perhaps I am not reducing my question far enough? I do appreciate all the help, and I definitely appreciate the help on the functions. I just can't get there until I understand how to communicate with Python what to do with a variable that will be defined through the process.

Thank you, thank you... for your patience.

[bartonc]

Ok, but I'm still confused about how Python handles the 5x**8. The x is not defined, so there is an error. You defined f with a function, but later you added a variable to FindRoot(f, x, L, R, e). My function is assigned by the teacher and is as follows:
find_root(f, xLo, xHi, eps). I can only have 4 variables. Does your def f(arg) attempt to solve that problem?

def f(arg):
    arg = arg * 1   # re-assignment is perfectly fine.
    return arg   # do any math you want to here
 
def FindRoot(f, x, L, R, e):
    xL = x * L
    xR = x * R
    while (xR  - xL) > e:
        xM = (xR  + xL) / 2.0   # python is a bit funny aboud dividing ints vs floats
        if (f(xL) * f(xM) > 0:
            xL = xM
        else:
            xR = xM
 
print FindRoot(f, 5, -1, 1, .0001)

What I've done here is give an example of how this MIGHT be done. Hence "someting like this". Please make an attempt at writing the function according to your assignment and I'll help you debug it. I believe that you will learn much more this way rather than just being given the answer.

[TMS]

What I've done here is give an example of how this MIGHT be done. Hence "someting like this". Please make an attempt at writing the function according to your assignment and I'll help you debug it. I believe that you will learn much more this way rather than just being given the answer.

I don't want you to write the function (arggggg) I've been working on this since Wednesday night. I can't find a way for python to interpret the x when I put 5x as the value of f. That is all I'm looking for. I can get through the pseudocode... its this one thing I'm hung up on. I've been reading, searching... I don't understand how to tell Python that I will eventually define x when it is in the format of 5x**8. Thats all I've been asking all along. If I sound exasperated its because I am. Believe me, I don't go to the boards unless I'm really stuck. I even emailed my teacher (before I tried the board) and he hasn't responded.

I must be asking the question wrong.

[bartonc]

I don't want you to write the function (arggggg) I've been working on this since Wednesday night. I can't find a way for python to interpret the x when I put 5x as the value of f. That is all I'm looking for. I can get through the pseudocode... its this one thing I'm hung up on. I've been reading, searching... I don't understand how to tell Python that I will eventually define x when it is in the format of 5x**8. Thats all I've been asking all along. If I sound exasperated its because I am. Believe me, I don't go to the boards unless I'm really stuck. I even emailed my teacher (before I tried the board) and he hasn't responded.

I must be asking the question wrong.

how about

fiveX = 5 * x

?

[TMS]

how about

fiveX = 5 * x

?

no, because x is undefined.

My teacher will be testing my assignment. He will use something like this:

find_root(5x**3+2x**2+x+1, -3, 4, .0006)

I have to find a way to process that statement, where f is defined as 5x**3+2x**2+x+1, but will (through the function) define x with xLo and then later with xHi. I get an error saying that x is not defined. It won't let me pass this right now because I'm missing something, obviously :(

It is probably so obvious to python programmers who have been programming in python for a while, but I'm not getting it, which is why I'm being so persistent. I really want to understand what I'm missing.

[TMS]

ok, I'm really not trying to be rude. I am hoping you understand that I can't change the way the function find_root(f, xLo, xHi,eps) is because its the assignment. It the way the teacher told us that we have to do it. Then, he said, try it with... 5x**8 + 3x**4 -2 or some such equation. The requirements specify that I have to define f with the above 'type' of equation. I am limited in this respect. All I want help with is how to tell python what I mean when an integer and a variable are entered together.

Is that more clear?

[bartonc]

no, because x is undefined.

My teacher will be testing my assignment. He will use something like this:

find_root(5x**3+2x**2+x+1, -3, 4, .0006)

I have to find a way to process that statement, where f is defined as 5x**3+2x**2+x+1, but will (through the function) define x with xLo and then later with xHi. I get an error saying that x is not defined. It won't let me pass this right now because I'm missing something, obviously :(

It is probably so obvious to python programmers who have been programming in python for a while, but I'm not getting it, which is why I'm being so persistent. I really want to understand what I'm missing.

There are two ways that I see for the call
find_root(5x**3+2x**2+x+1, -3, 4, .0006)
to be made (this will never be interpreted directly by python). The first is that have a long argument list with a variable for each factor, which violates the "4 arguments only" rule. The second would be to use text processing on a literal first argument as in
find_root("5x**3+2x**2+x+1", -3, 4, .0006)
but that seems like a very complex asignment for a beginning coarse.
How 'bout you put the text of the asignment here?

[bvdet]

I hope these snippets will help

>>> s = '5*(x**3)-3*(x**2)+x-2'
>>> x = 4
>>> eval(s)
274

>>> def solve2(s, x, y=1.0):
...     return eval(s)
... 
>>> s = '5*(x**3)-3*(x**2)+x-2'
>>> solve2(s, 4.0)
274.0
>>> 

[bartonc]

I hope these snippets will help

>>> s = '5*(x**3)-3*(x**2)+x-2'
>>> x = 4
>>> eval(s)
274

>>> def solve2(s, x, y=1.0):
...     return eval(s)
... 
>>> s = '5*(x**3)-3*(x**2)+x-2'
>>> solve2(s, 4.0)
274.0
>>> 

Nice one BV! Although I'm not sure the paren. grouping is right for the problem.

[TMS]

I hope these snippets will help

>>> s = '5*(x**3)-3*(x**2)+x-2'
>>> x = 4
>>> eval(s)
274

>>> def solve2(s, x, y=1.0):
...     return eval(s)
... 
>>> s = '5*(x**3)-3*(x**2)+x-2'
>>> solve2(s, 4.0)
274.0
>>> 

awesome. by George (or whomever you are) I think you've got it! So, now if the teacher won't be formatting the input that way, with the quote, I could simply do this:

f = 'f'
 
def find_root(f, xLo, xHi, eps):
    return eval(f)
 
s = find_root('5*(x**3)-3*(x**2)+x-2', -3, 1, .001)
print s
 

which gives me -11 when I run the module. And I don't think thats right. I think it should be -113 because 5*(-3**3) - 3*(-3**2) -3 -2 evaluates to -113. So, is having the 2 variables confusing it? How does eval work?

thank you very much. This is going to help so much.

tms

[TMS]

Ok, nevermind. I think I figured it out.

If I do this:

f = 'f'
 
def find_root(f, xLo, xHi, eps):
    x = xHi
    return eval(f)
 
s = find_root('5*(x**3)', -3, 1, .001)
print s
 
 

I can control which variable is being evaluated. I think I can take it from here. Thank you so much, everyone for your patience.

TMS

[bartonc]

How does eval work?

thank you very much. This is going to help so much.

tms

In section 2.3.4 of the (version 2.4.4) docs it says:
All numeric types (except complex) support the following operations, sorted by ascending priority (operations in the same box have the same priority; all numeric operations have a higher priority than comparison operations):


Operation Result Notes
________________________
x + y sum of x and y
________________________

x - y difference of x and y
________________________

x * y product of x and y
________________________

x / y quotient of x and y (1)
________________________

x // y (floored) quotient of x and y (5)
________________________

x % y remainder of x / y (4)
________________________

-x x negated
________________________

+x x unchanged
________________________

abs(x) absolute value or magnitude of x
________________________

int(x) x converted to integer (2)
________________________

long(x) x converted to long integer (2)
________________________

float(x) x converted to floating point
________________________

complex(re,im) a complex number with real part re, imaginary part im. im defaults to zero.
________________________

c.conjugate() conjugate of the complex number c
________________________

divmod(x, y) the pair (x // y, x % y) (3)(4)
________________________

pow(x, y) x to the power y
________________________

x ** y x to the power y

Notes:

(1)
For (plain or long) integer division, the result is an integer. The result is always rounded towards minus infinity: 1/2 is 0, (-1)/2 is -1, 1/(-2) is -1, and (-1)/(-2) is 0. Note that the result is a long integer if either operand is a long integer, regardless of the numeric value.

(2)
Conversion from floating point to (long or plain) integer may round or truncate as in C; see functions floor() and ceil() in the math module for well-defined conversions.

(3)
See section 2.1, ``Built-in Functions,'' for a full description.

(4)
Complex floor division operator, modulo operator, and divmod().

Deprecated since release 2.3. Instead convert to float using abs() if appropriate.


(5)
Also referred to as integer division. The resultant value is a whole integer, though the result's type is not necessarily int.

You can change the order of evaluation with the parentheses as in:
(5*x)**3 multiplies 5 times x first then raises to the power of 3,
5*(x**3) raises x to the power of 3 then multiplies by 5

[bvdet]

I looked over that link. There is a lot of information, but I understand very little of it. For example, the Doublet(Monome...) part. Doublet is part of what module? It isn't defined so I get an error message.

Again, I'm starting at the very beginning of this assignment, and not necessarily trying to write the functions complete yet (its due on Wednesday!!!). I want to understand this:
If my teacher runs my module and find_root is defined like this:

find_root(f, xLo, xHi, eps):

and the teacher enters this:

find_root(5x**3 + 2x**2-x+1, -3, 4, .0001)

how do I help Python understand that x is defined by xLo and then xHi. In order to solve for the root, if I understand bisection correctly, I use xLo to find y0, and xHi to find y1 by using it like this:

y0 = 5(xLo)**3 + 2(xLo)**2 - (xLo) + 1
y1 = 5(xHi)**3 + 2(xHi)**2 - (xHi) + 1

If it is still higher than eps, then I divide by 2... etc. But in order to begin I have to map xLo into x. And in order to do that, I have to get Python to know what to do with an x that is undefined.

Perhaps I am not reducing my question far enough? I do appreciate all the help, and I definitely appreciate the help on the functions. I just can't get there until I understand how to communicate with Python what to do with a variable that will be defined through the process.

Thank you, thank you... for your patience.

Yeah, there is a lot of stuff in that thread that is beyond the scope of your assignment. I was trying to give you an example of the calculation to determine the root. Monome, Doublet (the names are French!) and iterDoub are defined at various places in the thread. They are used to create an instance that stores all of the various factors and exponents in a multinomial equation and format the information for printing. It's actually kinda neat.

5x**3 + 2x**2-x+1

If the above statement is the required format for the argument, then you will need to modify the string so it can be 'eval'ed properly. I believe the argument must be entered as a string, i.e. enclosed in single or double quotes. Even if 'x' is a defined variable, the above equation cannot be evaluated as it is.

'5*x**3 + 2*x**2-x-1' can be evaluated if 'x' has been defined as in:

x = 2.5

If you can define the input format, this should work:

find_root('5*x**3+2*x**2-x+1', -3, 4, .0001)
 
def find_root(s, xL, xH, e):
    ....do some stuff...
    intermediate_results = eval(s)
    ....do more stuff....

[TMS]

5*x**3 + 2*x**2-x-1' can be evaluated if 'x' has been defined as in:

x = 2.5

If you can define the input format, this should work:

find_root('5*x**3+2*x**2-x+1', -3, 4, .0001)
 
def find_root(s, xL, xH, e):
    ....do some stuff...
    intermediate_results = eval(s)
    ....do more stuff....

It does work. Thank you. By using eval and formatting the input and utilizing the pseudocode that wikipedia provided for the bisection method I was able to finish the first part of the homework assignment. It works quite well. I will check with my teacher to make sure its what he was looking for.
The second part requires me to "use the rootfinder module to implement a script findroot that concatenates all of its command line arguments (except, of course the program name) into an expression for f() so that you have a rootsolver that works fromt the command line."

????

Well, I'm calling it a day for the first part. I have some AI homework to work on.

thank you again for your help, and everone else too! Especially for being patient with me while I figure out how to ask the question, lol.

TMS

[bvdet]

You are welcome. I am glad to help, and I'm sure Barton is too. Keep with it!