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I am wondering when use the recursion and iterative method please tell me which approch best and why ?
I don't know about the Data structure and Algorithm.
This is Recursive approach.
def recurse(n):
print(n)
if n==0:
return 0
return recurse(n-1)
recurse(10)
print(recurse(10))
This is Itrative approach.
def factorial(n):
if n < 0:
return 0
elif n == 0 or n == 1:
return 1
else:
fact = 1
while(n > 1):
fact *= n
n -= 1
return fact
print(factorial(5))
Like LinFelix in the comments said. "Which is better" is a broad question. I have here somesources you can read up on to better understand which is better in your case:
https://www.interviewkickstart.com/learn/difference-between-recursion-and-iteration#:~:text=Recursion%20is%20when%20a%20function,loops%20and%20%22while%22%20loops.
https://www.geeksforgeeks.org/difference-between-recursion-and-iteration/
recursion versus iteration
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I'm trying to figure out how I would use a closure which will find and groups of words that have the same ending letter.
For example: [United, Static, Rapid, Directed]
The return should be ["D":3, "c":1]
If you want to group use:
def list = ['United', 'Static', 'Rapid', 'Directed']
def groupped = list.groupBy{ it[ -1 ] }
assert groupped == [d:['United', 'Rapid', 'Directed'], c:['Static']]
For counting only you can use:
def counted = list.inject( [:].withDefault{ 0 } ){ res, curr ->
res[ curr[ -1 ] ]++
res
}
assert counted == [d:3, c:1]
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Doing some exercises from Think like a CS with Python 3:
Have a task:
Overload the necessary operator(s) that instead of having to write
if t1.after(t2):...
we can use the more convenient
if t1 > t2: ...
How I can do it? Have no ideas.
You need to override the t1.__gt__(t2) method of your class. I would suggest overriding all of the following __gt__ __lt__ __le__ __ge__ special functions.
For example
class Point:
def __init__(self, x = 0, y = 0):
self.x = x
self.y = y
def __lt__(self,other):
self_mag = (self.x ** 2) + (self.y ** 2)
other_mag = (other.x ** 2) + (other.y ** 2)
return self_mag < other_mag
will allow you to write expressions like p1 < p2 but not p1 > p2. But it can be done trivially.
edit: turns out that simply overriding the __eq__ and __lt__ on top of using functools.#total_ordering gives the desired result.
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I'm trying to create a function that receive as argument a number and return an array of 3 numbers max.
I have 3 tokens. 1 unit, 5 unit and 25 unit.
calculateUnit(4) should = [4,0,0]
calculateUnit(7) should = [2,1,0] (because 2 unit of 1 and 1 unit of 5 = 7)
calculateUnit(36) should = [1,2,1] (because 1 unit of 1, 2 unit of 5 and 1 unit of 25 = 36)
I have a basic code and I think I need to use modulo division, I already tried to search here and every other resources I have but I may not use the correct terms.
You can reduce your solution to:
def convertInToken(am):
return [am//25, (am%25)//5, am%5]
This leverages integer-division (3.x upwards, also named floor division) and modulo division.
Floor division returns the full integer that woud have been returned if you did a normal division and floored it.
Modulu division returns the "remainder" of a division.
I managed to do that, but thanks anyway :)
# your code goes here
import math
def convertInToken(am):
result = [];
#need to use 25
if am >= 25:
amount25 = math.floor((am/25))
amount5 = math.floor((am-(amount25*25))/5)
amount1 = math.floor(((am-(amount25*25)-(amount5*5))/1))
result = result+[amount1]
result = result+[amount5]
result = result+[amount25]
#need to use 5
elif am >= 5:
amount5 = math.floor((am/5))
amount1 = math.floor(((am-(amount5*5))/1))
result = result+[amount1]
result = result+[amount5]
result = result+[0]
#need to use 1
elif am < 5:
result = result+[am]
result = result+[0]
result = result+[0]
return result
print(convertInToken(4))
print(convertInToken(7))
print(convertInToken(12))
print(convertInToken(37))
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A positive integer n is said to be perfect if the sum of the factors of n, other than n itself, add up to n. For instance 6 is perfect since the factors of 6 are {1,2,3,6} and 1+2+3=6. Likewise, 28 is perfect because the factors of 28 are {1,2,4,7,14,28} and 1+2+4+7+14=28.
Write a Python function perfect(n) that takes a positive integer argument and returns True if the integer is perfect, and False otherwise.
Here are some examples to show how your function should work.
perfect(6)
True
perfect(12)
False
perfect(28)
True
def perfect(x):
factor_sum = 0
for i in range(1, x-1):
if x % i == 0:
factor_sum = factor_sum + i
if(factor_sum == x):
return True
return False
print perfect(6) #Prints True
print perfect(12) #Prints False
print perfect(28) #Prints True
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I try to print number of combinations of given count of elements. - Yes, there was such topic, but I'm very beginner of Python and I want to understand my errors. Count is true just for x=4. And one more question: why in the end it prints "None"?
x=int(input('Count of elements for combinations: '))
a=x
from math import factorial
def everywithevery(x):
y=x
print ('Graphical representation:')
print (x*'_ ')
while x>0:
print ((x-1)*'* ')
x=x-1
print('Stars count is equal combinations count. Total count is: ',factorial((y-1)));
print(everywithevery(a));
There you can try my script: http://goo.gl/EDFkYM
You print the following line but you didn't return any thing from everywithevery function. Thats why that line prints none
print(everywithevery(a));
Just use
everywithevery(a)