I want to write a recursive function in Python for Fibonacci.
x will be the starting point, y will be the subsequent of x and l is the length.
I don't understand what is my error in thinking:
def fib(x, y, l, fibList=None):
fibList = []
z = x + y
x = y
fibList.append(z)
y = z
if len(fibList) <= l:
return fib(x, y, l, fibList)
else:
return(fibList)
The result is:
RecursionError: maximum recursion depth exceeded while calling a Python object
I can solve it with a for loop but not with recursive function.
There are a few problems here. Once you fix the infinite recursion, you still have an issue.
As #Raqha points out, you need to not initialize your list every time the fib function is called, but rather only the first time, when the fibList parameter isn't supplied and so defaults to None.
Your code is not able to generate the first two numbers in the sequence, 0 and 1. You can fix this by simply initializing your list to include these terms, and adjust the logic to only provide N-2 more terms.
The signature of your function could be improved to make it much easier for the caller to use it. The caller only cares about the number of terms he/she wants. The user shouldn't have to know what to enter for the initial x and y values.
Here's a version of your code with the fix for the infinite recursion, the fix for the missing terms, and also with the signature rearranged so that the user can call the function simply and obviously:
def fib(l, x=0, y=1, fibList=None):
if fibList is None:
fibList = [0, 1]
z = x + y
x = y
fibList.append(z)
y = z
if len(fibList) < l-1:
return fib(l, x, y, fibList)
else:
return(fibList)
print(fib(10))
Result:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
On the second line you set fibList = []. This means that every time you call the function recursively it resets the list to be empty so len(fibList) will always equal 1.
Remove that line so the fibList variable is not reset, then it should hit your exit condition correctly. How it is written now it runs forever until it breaks.
At the start of every fib function call you clear your fibList with fibList = []. So the list's length will always be <= 1, thus you run into an infinite recursion loop which isn't good. You need to add something like if fibList is None:
When you first call the function "fib" without providing any 4th statement in the argument list, the value of fibList will initially set to "None". But later when you recursively call the function "fib" again, you provide a fibList in the argument list. So the value isnt "None" anymore. So when adding an if-statement as mentioned above, the function knows when you call the function from the outside (when you call it in your code as "fib(1,2,10)"), or when the function recursively calls itself. And so you dont reset the fibList everytime you call the function, but set it only 1 time at the start.
def fib(x, y, l, fibList=None):
if fibList is None:
fibList = []
z = x + y
...
Related
I intend to make a while loop inside a defined function. In addition, I want to return a value on every iteration. Yet it doesn't allow me to iterate over the loop.
Here is the plan:
def func(x):
n=3
while(n>0):
x = x+1
return x
print(func(6))
I know the reason to such issue-return function breaks the loop.
Yet, I insist to use a defined function. Therefore, is there a way to somehow iterate over returning a value, given that such script is inside a defined function?
When you want to return a value and continue the function in the next call at the point where you returned, use yield instead of return.
Technically this produces a so called generator, which gives you the return values value by value. With next() you can iterate over the values. You can also convert it into a list or some other data structure.
Your original function would like this:
def foo(n):
for i in range(n):
yield i
And to use it:
gen = foo(100)
print(next(gen))
or
gen = foo(100)
l = list(gen)
print(l)
Keep in mind that the generator calculates the results 'on demand', so it does not allocate too much memory to store results. When converting this into a list, all results are caclculated and stored in the memory, which causes problems for large n.
Depending on your use case, you may simply use print(x) inside the loop and then return the final value.
If you actually need to return intermediate values to a caller function, you can use yield.
You can create a generator for that, so you could yield values from your generator.
Example:
def func(x):
n=3
while(n>0):
x = x+1
yield x
func_call = func(6) # create generator
print(next(func_call)) # 7
print(next(func_call)) # 8
How can I define a function in python in such a way that it takes the previous value of my iteration where I define the initial value.
My function is defined as following:
def Deulab(c, yh1, a, b):
Deulab = c- (EULab(c, yh1, a, b)-1)*0.3
return (Deulab,yh1, a,b)
Output is
Deulab(1.01, 1, 4, 2)
0.9964391705626454
Now I want to iterate keeping yh1, a ,b fixed and start with c0=1 and iterate recursively for c.
The most pythonic way of doing this is to define an interating generator:
def iterates(f,x):
while True:
yield x
x = f(x)
#test:
def f(x):
return 3.2*x*(1-x)
orbit = iterates(f,0.1)
for _ in range(10):
print(next(orbit))
Output:
0.1
0.2880000000000001
0.6561792000000002
0.7219457839595519
0.6423682207442558
0.7351401271107676
0.6230691859914625
0.7515327214700762
0.5975401280955426
0.7695549549155365
You can use the generator until some stop criterion is met. For example, in fixed-point iteration you might iterate until two successive iterates are within some tolerance of each other. The generator itself will go on forever, so when you use it you need to make sure that your code doesn't go into an infinite loop (e.g. don't simply assume convergence).
It sound like you are after recursion.
Here is a basic example
def f(x):
x += 1
if x < 10:
x = f(x)
return x
print (f(4))
In this example a function calls itself until a criteria is met.
CodeCupboard has supplied an example which should fit your needs.
This is a bit of a more persistent version of that, which would allow you to go back to where you were with multiple separate function calls
class classA:
#Declare initial values for class variables here
fooResult = 0 #Say, taking 0 as an initial value, not unreasonable!
def myFoo1(x):
y = 2*x + fooResult #A simple example function
classA.fooResult = y #This line is updating that class variable, so next time you come in, you'll be using it as part of calc'ing y
return y #and this will return the calculation back up to wherever you called it from
#Example call
rtn = classA.myFoo1(5)
#rtn1 will be 10, as this is the first call to the function, so the class variable had initial state of 0
#Example call2
rtn2 = classA.myFoo1(3)
#rtn2 will be 16, as the class variable had a state of 10 when you called classA.myFoo1()
So if you were working with a dataset where you didn't know what the second call would be (i.e. the 3 in call2 above was unknown), then you can revisit the function without having to worry about handling the data retention in your top level code. Useful for a niche case.
Of course, you could use it as per:
list1 = [1,2,3,4,5]
for i in list1:
rtn = classA.myFoo1(i)
Which would give you a final rtn value of 30 when you exit the for loop.
Say I have these two functions:
def s(x,y,z):
if x <= 0:
return y
return z
def f(a,b):
return s(b, a+1, f(a,b-1)+1)
If I were to try and find f(5,2) in my head, it would go like this:
f(5,2) = s(2,6,f(5,1)+1)
f(5,1) = s(1,6,f(5,0)+1)
f(5,0) = s(0,6,f(5,-1)+1) = 6
f(5,1) = 7
f(5,2) = 8
I never evaluate f(5,-1) because it is not needed. The s function is going to return 6, since argument x is zero, thus evaluation of argument z is unnecessary.
If I were however to try and run this in python, it would keep recursing forever or or until I get a maximum recursion depth error, presumably because python wants to evaluate all the arguments before executing the s function.
My question is, how would I go about implementing these functions, or any similar scenario, in such a way that the recursion stops when it is no longer needed? Would it be possible to delay the evaluation of each argument until it is used in the function?
Your mind is working with 'insider knowledge' of how s() works. Python can't, so it can only follow the strict rules that all argument expressions to a call will have to be evaluated before the call can be made.
Python is a highly dynamic language, and at every step of execution, both s and f can be rebound to point to a different object. This means Python can't optimise recursion or inline function logic. It can't hoist the if x <= 0 test out of s() to avoid evaluating the value for z first.
If you as a programmer know the third expression needs to be avoided in certain circumstances, you need to make this optimisation yourself. Either merge the logic in s into f manually:
def f(a, b):
if b <= 0:
return a + 1
return f(a, b - 1) + 1
or postpone evaluating of the third expression until s() has determined if it needs to be calculated at all, by passing in a callable and make s responsible for evaluating it:
def s(x, y, z):
if x <= 0:
return y
return z() # evaluate the value for z late
def f(a, b):
# make the third argument a function so it is not evaluated until called
return s(b, a+1, lambda: f(a, b - 1) + 1)
When a function is called, all arguments are fully evaluated before they're passed to the function. In other words, f(5,-1) is being executed before s is even started.
Fortunately there's an easy way to evaluate expressions on demand: functions. Instead of passing the result of f(a,b-1) to z, pass it a function that computes that result:
def s(x,y,z):
if x <= 0:
return y
return z() # z is a function now
def f(a,b):
return s(b, a+1, lambda:f(a,b-1)+1)
print(f(5,2)) # output: 8
I am using python so can someone tell me how to call same function twice in python but when you call it the second time it should be changed an already have value stored in it when you called the function the first time, so I basically mean you are calling the function first time and right after that you are calling it again but with the return value from the first time you called that function.
Assuming you have a function that has both a parameter and a return value:
def myFunction(input):
# do something with input
return input
To have a second instance of the function use the result of the first instance, you can simply nest the functions:
result = myFunction(myFunction(value))
you can create function that will apply n-times
def multf(f, n, x):
if n == 0:
return x
return multf(f, n-1, f(x))
so here we apply lambda sqr 3 times, it becomes f(f(f(x)))
sqr = lambda x: x**2
print(multf(sqr,3,2))
256
i would like to perform a calculation using python, where the current value (i) of the equation is based on the previous value of the equation (i-1), which is really easy to do in a spreadsheet but i would rather learn to code it
i have noticed that there is loads of information on finding the previous value from a list, but i don't have a list i need to create it! my equation is shown below.
h=(2*b)-h[i-1]
can anyone give me tell me a method to do this ?
i tried this sort of thing, but that will not work as when i try to do the equation i'm calling a value i haven't created yet, if i set h=0 then i get an error that i am out of index range
i = 1
for i in range(1, len(b)):
h=[]
h=(2*b)-h[i-1]
x+=1
h = [b[0]]
for val in b[1:]:
h.append(2 * val - h[-1]) # As you add to h, you keep up with its tail
for large b list (brr, one-letter identifier), to avoid creating large slice
from itertools import islice # For big list it will keep code less wasteful
for val in islice(b, 1, None):
....
As pointed out by #pad, you simply need to handle the base case of receiving the first sample.
However, your equation makes no use of i other than to retrieve the previous result. It's looking more like a running filter than something which needs to maintain a list of past values (with an array which might never stop growing).
If that is the case, and you only ever want the most recent value,then you might want to go with a generator instead.
def gen():
def eqn(b):
eqn.h = 2*b - eqn.h
return eqn.h
eqn.h = 0
return eqn
And then use thus
>>> f = gen()
>>> f(2)
4
>>> f(3)
2
>>> f(2)
0
>>>
The same effect could be acheived with a true generator using yield and send.
First of, do you need all the intermediate values? That is, do you want a list h from 0 to i? Or do you just want h[i]?
If you just need the i-th value you could us recursion:
def get_h(i):
if i>0:
return (2*b) - get_h(i-1)
else:
return h_0
But be aware that this will not work for large i, as it will exceed the maximum recursion depth. (Thanks for pointing this out kdopen) In that case a simple for-loop or a generator is better.
Even better is to use a (mathematically) closed form of the equation (for your example that is possible, it might not be in other cases):
def get_h(i):
if i%2 == 0:
return h_0
else:
return (2*b)-h_0
In both cases h_0 is the initial value that you start out with.
h = []
for i in range(len(b)):
if i>0:
h.append(2*b - h[i-1])
else:
# handle i=0 case here
You are successively applying a function (equation) to the result of a previous application of that function - the process needs a seed to start it. Your result looks like this [seed, f(seed), f(f(seed)), f(f(f(seed)), ...]. This concept is function composition. You can create a generalized function that will do this for any sequence of functions, in Python functions are first class objects and can be passed around just like any other object. If you need to preserve the intermediate results use a generator.
def composition(functions, x):
""" yields f(x), f(f(x)), f(f(f(x)) ....
for each f in functions
functions is an iterable of callables taking one argument
"""
for f in functions:
x = f(x)
yield x
Your specs require a seed and a constant,
seed = 0
b = 10
The equation/function,
def f(x, b = b):
return 2*b - x
f is applied b times.
functions = [f]*b
Usage
print list(composition(functions, seed))
If the intermediate results are not needed composition can be redefined as
def composition(functions, x):
""" Returns f(x), g(f(x)), h(g(f(x)) ....
for each function in functions
functions is an iterable of callables taking one argument
"""
for f in functions:
x = f(x)
return x
print composition(functions, seed)
Or more generally, with no limitations on call signature:
def compose(funcs):
'''Return a callable composed of successive application of functions
funcs is an iterable producing callables
for [f, g, h] returns f(g(h(*args, **kwargs)))
'''
def outer(f, g):
def inner(*args, **kwargs):
return f(g(*args, **kwargs))
return inner
return reduce(outer, funcs)
def plus2(x):
return x + 2
def times2(x):
return x * 2
def mod16(x):
return x % 16
funcs = (mod16, plus2, times2)
eq = compose(funcs) # mod16(plus2(times2(x)))
print eq(15)
While the process definition appears to be recursive, I resisted the temptation so I could stay out of maximum recursion depth hades.
I got curious, searched SO for function composition and, of course, there are numerous relavent Q&A's.