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How do I create a recursive python function for 5n? [closed]

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I only got this far with the answer. I don't understand how recursive functions work. I would really appreciate any help. And also, if someone could explain what base and recursive calls are, that'd be great.
def multi(n):
if n==1:
return 4

Please try this code :
def multi(n):
return (5 + multi(n-1)) if n>0 else 0;
this function is a 5n, here's the explanation:
when n = 1 the call is like this :
output : 5 + multi(0) --> multi(0) = 0, so 5 + 0 = 5
It's something like this :
multi(3)
|
-- 5 + multi(2)
|
--- 5 + (5 + multi(1))
|
---- 5 + (5 + (5 + multi(0)))
|
----- 5 + 5 + 5 + 0
5n --> if n = 3 --> 5(3) = 15
output multi(3) = 15, as same as 5(3)=15.
5(3) = 5+5+5 = 15
5(2) = 5+5 = 10
in recursion, you'll do it like this:
5(0) = 0
5(1) = 5 + 0 = 5
5(2) = 5 + 5 + 0 = 10
...
5(n) = 5 + f(n-1), where f(0) = 0, so n must greater than 0. To prevent infinite loop.
Case for 5(3) :
1. 5(3) = 5 + f(2)
2. f(2) = 5 + f(1)
3. f(1) = 5 + f(0)
4. f(0) = 0
So, you will get 5(3) = 5 + (5 + (5 + 0)) = 15
Hope this help you to understand it :D

Recursive function is a function that makes a call to it self.
for example:
def multi(n):
if n>0:
return n * multi(n-1)
else:
return 0
To prevent infinit recursion you have to define a case where the function does not make a call to it self.(in our example it's in else statement when n==1).

is there possible to resolve this star pyramid [closed]

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Today, i have trying to resolve a small star pyramid :
Input:
5 1
Output:
*
**
***
****
Code:
x = 1
y = 0
m, e = map(int, raw_input().split())
while x < m:
print "\n" * y, "*" * e
m -= 1
e += 1
I did that but there is a better solution?? Thanks =)

I think this can be solved more easily:
stop, first = map(int, raw_input().split())
for i in range(stop - 1):
print '*' * (i + first)

just for fun >:)
class c:
def __init__(s,m,e):
s.e , s.m = sorted([e, m])
s.r = 42
def __iter__(s):
return s
def next(s):
if s.m < s.e:
t = "".join(chr(s.r) for _ in range(s.m))
s.m += 1
return t
else:
raise StopIteration
print "\n".join(c(*map(int,raw_input().split())))

n = int(raw_input())
for i in range(n): print "*"*i
This appears to do what your program intends to do, however I can't quite tell because of the issues I raised in my comment above.

How to make a binomial expander in python [closed]

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For example
(x + y)^4 = x^4 +(4x^3)y + (6x^2)y^2 + + 4xy^3 + y^4
I'm using python 3.3.2 and I don't know where to start, I need a little guidance. I'm not asking for the answer to it, just the general steps to make this program work. I've done a few other programs before, and this is probably pushing the limit on what I can do.

SymPy already do that for you:
>>> import sympy
>>> x, y = sympy.symbols("x y")
>>> formula = (x + y) ** 4
>>> formula
(x + y)**4
>>> formula.expand()
x**4 + 4*x**3*y + 6*x**2*y**2 + 4*x*y**3 + y**4
If you need a visualization as a string with "^", you can do:
>>> str(formula.expand()).replace("**", "^")
'x^4 + 4*x^3*y + 6*x^2*y^2 + 4*x*y^3 + y^4'

The code below can be found on this site.
Usage
python binomialexpansion.py 3
Output
x^3 + 3x^2y + 3xy^2 + y^3
Code
import sys
power = int(eval(sys.argv[1]))
strpower = str(power)
coeffs = []
if power == 0:
print 1
exit()
if (power+1) % 2 == 0:
turningp = (power+1)/2
counter = 1
else:
turningp = (power+2)/2
counter = 2
for i in range(1, power+2):
if i == 1:
sys.stdout.write("x^"+strpower+" + ")
coeffs.append(1)
continue
if i == power+1:
print "y^"+strpower,
coeffs.append(1)
break
if i > turningp:
co = coeffs[turningp-counter]
counter = counter+1
else:
co = ((power-(i-2))*coeffs[i-2])/(i-1)
coeffs.append(co)
sys.stdout.write(str(co))
if power-(i-1) == 1:
sys.stdout.write("x")
else:
sys.stdout.write("x^"+str(power-(i-1)))
if i-1 == 1:
sys.stdout.write("y ")
else:
sys.stdout.write("y^"+str(i-1)+" ")
sys.stdout.write("+ ")

Python float operations [closed]

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How can I print the full float result of a simple operation in python? My code does:
if( int(array_Y[counter2]) == int(round(float(elem[0])))):
if(int(round(float(elem[0]))) == 0):
negatiu_verdader += 1
if(int(round(float(elem[0]))) == 1):
positiu_verdader += 1
counter = counter + 1
counter2 = counter2 + 1
error = float(1.0000- (1.0000 * counter / counter2))
print " ERROR!!!!!!!!!!!!!!!!!!!!!!!! :" + ("{0:.15f}".format(round(error,2)))
But the error is always: 0.420000000000000 or 0.230000000000000 but I would like the error to be: 0.43233213213232.

You are rounding your error down to two decimal places by calling round(error, 2):
>>> round(0.43233213213232, 2)
0.43
Don't do that if you want to show more precision:
>>> format(round(0.43233213213232, 2), '.15f')
'0.430000000000000'
>>> format(0.43233213213232, '.15f')
'0.432332132132320'
You do a lot of redundant work in your code, simplify it down a little:
elem_rounded = int(round(float(elem[0])))
if int(array_Y[counter2]) == elem_rounded:
if not elem_rounded:
negatiu_verdader += 1
elif elem_rounded == 1:
positiu_verdader += 1
counter += 1
counter2 += 1
error = 1.0 - (1.0 * counter / counter2)
print " ERROR!!!!!!!!!!!!!!!!!!!!!!!! :{0:.15f}".format(error)

Should I do anything to make my code more pythonic? [closed]

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This is a very simple piece of code that I wrote but if there is a way to make it more pythonic then I would love to know. Thanks!
def money():
current_salary = float(input("What is your current salary? "))
years = int(input("How many years would you like to look ahead? ")) + 1
amount_of_raise = float(input("What is the average percentage raise you think you will get? "))
amount_of_raise = amount_of_raise * 0.01
while years > 1:
years = years - 1
new_salary = current_salary + (current_salary * amount_of_raise)
current_salary = new_salary
print('Looks like you will be making', new_salary,' in ', years,'years.')
money()

Extended assignment operators
amount_of_raise = amount_of_raise * 0.01
years = years - 1
x = x * y can be shortened to x *= y. Same thing for -.
amount_of_raise *= 0.01
years -= 1
Iteration and counting
while years > 1:
years = years - 1
Counting down causes your printouts to display backwards. I would count up. The Pythonic way to count uses range:
for year in range(1, years + 1):
print('Looks like you will be making', new_salary,' in ', years,'years.')
Computing new salary
new_salary = current_salary + (current_salary * amount_of_raise)
current_salary = new_salary
I'd probably just simplify that to:
current_salary += current_salary * amount_of_raise
Or even better is to give a 5% raise by multiplying by 1.05. In code that is:
current_salary *= 1 + amount_of_raise