I'm looking for the median of three, using this for a pivot in a QuickSort. I would not like to import any statistics library because I believe it creates a bit of overhead which I would like to reduce as much as possible.
def median(num_list):
if (num_list[0] > num_list[len(num_list) - 1]) and (num_list[0] < num_list[int(len(num_list)//2)]):
return num_list[0]
elif (num_list[int(len(num_list)//2)] > num_list[len(num_list) - 1]) and (num_list[0] > num_list[int(len(num_list)//2)]):
return num_list[int(len(num_list)//2)]
else:
return num_list[len(num_list) - 1]
this seems to be returning the last else statement every time, I'm stumped...
In Quicksort you do not usually want just to know the median of three, you want to arrange the three values so the smallest is in one spot, the median in another, and the maximum in yet another. But if you really just want the median of three, here are two ways, plus another that rearranges.
Here's a short way to find the median of a, b, and c.
return a + b + c - min(a, b, c) - max(a, b, c)
If you want only comparisons, and to get what may be the quickest code, realize that three comparisons may need to be executed but you want to try for only two. (Two comparisons can handle four cases, but there are six arrangements of three objects.) Try
if a < b:
if b < c:
return b
elif a < c:
return c
else:
return a
else:
if a < c:
return a
elif b < c:
return c
else:
return b
If you want to rearrange the values so a <= b <= c,
if a > b:
a, b = b, a
if b > c:
b, c = c, b
if a > b
a, b = b, a
return b
Let Python do the work for you. Sort the three elements, then return the middle one.
def median(num_list):
return sorted([num_list[0], num_list[len(num_list) // 2], num_list[-1]])[1]
Using min and max:
>>> numlist = [21, 12, 16]
>>> a, b, c = numlist
>>> max(min(a,b), min(b,c), min(a,c))
16
>>>
Going out on a limb - I have a functional streak so here is the itertools equivalent, even though it means importing a module
>>> import itertools
>>> numlist = [21, 12, 16]
>>> z = itertools.combinations(numlist, 2)
>>> y = itertools.imap(min, z)
>>> max(y)
16
Related
The rating for Alice's challenge is the triplet a = (a[0], a[1], a[2]), and the rating for Bob's challenge is the triplet b = (b[0], b[1], b[2]).
The task is to find their comparison points by comparing a[0] with b[0], a[1] with b[1], and a[2] with b[2].
If a[i] > b[i], then Alice is awarded 1 point.
If a[i] < b[i], then Bob is awarded 1 point.
If a[i] = b[i], then neither person receives a point.
Below is the function body:
# The function is expected to return an INTEGER_ARRAY.
# The function accepts following parameters:
# 1. INTEGER_ARRAY a
# 2. INTEGER_ARRAY b
#
def compareTriplets(a, b):
# Write your code here
c=0
d=0
for i in range(0,len(a)-1):
if(a[i]>b[i]):
c+=1
elif(a[i]<b[i]):
d+=1
return array.array("i",[c,d])
First of all, not sure why you're iterating with range(0,len(a)-1) - you are skipping the last element. It should be range(len(a)).
Second, it is more "pythonic" in these cases to use the zip function instead of using indexes. So you could do:
def compareTriplets(alice, bob):
alice_score = 0
bob_score = 0
for a, b in zip(alice, bob):
if a > b:
alice_score += 1
elif a < b:
bob_score += 1
return array.array("i", [alice_score, bob_score])
If you're looking for shorter code (note that it doesn't mean better or faster code!), you could use the sum function:
def compareTriplets(alice, bob):
alice_score = sum(a > b for a, b in zip(alice, bob))
bob_score = sum(a < b for a, b in zip(alice, bob))
return array.array("i", [alice_score, bob_score])
As you can see this loops over the zip twice. For two tuples of three elements this is insignificant, but it is important to take in count that once you're dealing with big data, this approach is not ideal.
Here is the comprehension for ya:
def compareTriplets(a, b):
return [sum(tot) for tot in zip(*((aa > bb, aa < bb) for aa, bb in zip(a, b)))]
I have three variables called a, b and c, each of these can assume a different value defined in a range. I'd like to create a function that tests every possible variable value and gives me their best combination for the output 'f'.
a = list(range(1, 10, 2))
b = list(range(5, 8, 1))
c = list(range(1, 3, 1))
def all_combinations (a, b, c):
#something
f = a + (b * a) - (c*(a ^ b))
return BEST a, b, c for my f
it's possible to do it ? what is the best way to do it?
You can use itertools.product() to get all the possible combinations of a, b, and c.
Then calculate your formula for each unique combination of a b c, keep track of the result, and if the result is better than the previous best, save the current values of a b c.
import itertools
def all_combinations (alist, blist, clist):
best_a = 0
best_b = 0
best_c = 0
best_f = 0
for a,b,c in itertools.product(alist, blist, clist):
f = a + (b * a) - (c*(a ^ b))
if f > best_f: # use your own definition of "better"
best_a = a
best_b = b
best_c = c
best_f = f
return best_a, best_b, best_c
First of all, you said I have three variables called a, b and c, each of these can assume a different value defined in a range. Note that the variables in your code are actually equal to three lists of integers, not three integers.
The naive algorithm to test all possible combinations is 3 nested for loops. Here I assume that by "best" you mean "maximum value":
def all_combinations (list1, list2, list3):
best_f, best_a, best_b, best_c = None, None, None, None
for a in list1:
for b in list2:
for c in list3:
f = a + (b * a) - (c*(a ^ b))
# here you have to define what f being "better" than best_f means:
if not f or f > best_f:
best_f = f
best_a = a
best_b = b
best_c = c
return best_a, best_b, best_c
If you're sure those are the only values you want to test, then the following will work. Otherwise you might want to look into scipy.optimize.
from itertools import product
import numpy as np
parameters = list(product(a, b, c))
results = [my_fun(*x) for x in parameters]
print(parameters[np.argmax(results)])
obviously replace np.argmax with np.argmin if you want to minimize the function
So I'm writing a program in Python to get the GCD of any amount of numbers.
def GCD(numbers):
if numbers[-1] == 0:
return numbers[0]
# i'm stuck here, this is wrong
for i in range(len(numbers)-1):
print GCD([numbers[i+1], numbers[i] % numbers[i+1]])
print GCD(30, 40, 36)
The function takes a list of numbers.
This should print 2. However, I don't understand how to use the the algorithm recursively so it can handle multiple numbers. Can someone explain?
updated, still not working:
def GCD(numbers):
if numbers[-1] == 0:
return numbers[0]
gcd = 0
for i in range(len(numbers)):
gcd = GCD([numbers[i+1], numbers[i] % numbers[i+1]])
gcdtemp = GCD([gcd, numbers[i+2]])
gcd = gcdtemp
return gcd
Ok, solved it
def GCD(a, b):
if b == 0:
return a
else:
return GCD(b, a % b)
and then use reduce, like
reduce(GCD, (30, 40, 36))
Since GCD is associative, GCD(a,b,c,d) is the same as GCD(GCD(GCD(a,b),c),d). In this case, Python's reduce function would be a good candidate for reducing the cases for which len(numbers) > 2 to a simple 2-number comparison. The code would look something like this:
if len(numbers) > 2:
return reduce(lambda x,y: GCD([x,y]), numbers)
Reduce applies the given function to each element in the list, so that something like
gcd = reduce(lambda x,y:GCD([x,y]),[a,b,c,d])
is the same as doing
gcd = GCD(a,b)
gcd = GCD(gcd,c)
gcd = GCD(gcd,d)
Now the only thing left is to code for when len(numbers) <= 2. Passing only two arguments to GCD in reduce ensures that your function recurses at most once (since len(numbers) > 2 only in the original call), which has the additional benefit of never overflowing the stack.
You can use reduce:
>>> from fractions import gcd
>>> reduce(gcd,(30,40,60))
10
which is equivalent to;
>>> lis = (30,40,60,70)
>>> res = gcd(*lis[:2]) #get the gcd of first two numbers
>>> for x in lis[2:]: #now iterate over the list starting from the 3rd element
... res = gcd(res,x)
>>> res
10
help on reduce:
>>> reduce?
Type: builtin_function_or_method
reduce(function, sequence[, initial]) -> value
Apply a function of two arguments cumulatively to the items of a sequence,
from left to right, so as to reduce the sequence to a single value.
For example, reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates
((((1+2)+3)+4)+5). If initial is present, it is placed before the items
of the sequence in the calculation, and serves as a default when the
sequence is empty.
Python 3.9 introduced multiple arguments version of math.gcd, so you can use:
import math
math.gcd(30, 40, 36)
3.5 <= Python <= 3.8.x:
import functools
import math
functools.reduce(math.gcd, (30, 40, 36))
3 <= Python < 3.5:
import fractions
import functools
functools.reduce(fractions.gcd, (30, 40, 36))
A solution to finding out the LCM of more than two numbers in PYTHON is as follow:
#finding LCM (Least Common Multiple) of a series of numbers
def GCD(a, b):
#Gives greatest common divisor using Euclid's Algorithm.
while b:
a, b = b, a % b
return a
def LCM(a, b):
#gives lowest common multiple of two numbers
return a * b // GCD(a, b)
def LCMM(*args):
#gives LCM of a list of numbers passed as argument
return reduce(LCM, args)
Here I've added +1 in the last argument of range() function because the function itself starts from zero (0) to n-1. Click the hyperlink to know more about range() function :
print ("LCM of numbers (1 to 5) : " + str(LCMM(*range(1, 5+1))))
print ("LCM of numbers (1 to 10) : " + str(LCMM(*range(1, 10+1))))
print (reduce(LCMM,(1,2,3,4,5)))
those who are new to python can read more about reduce() function by the given link.
The GCD operator is commutative and associative. This means that
gcd(a,b,c) = gcd(gcd(a,b),c) = gcd(a,gcd(b,c))
So once you know how to do it for 2 numbers, you can do it for any number
To do it for two numbers, you simply need to implement Euclid's formula, which is simply:
// Ensure a >= b >= 1, flip a and b if necessary
while b > 0
t = a % b
a = b
b = t
end
return a
Define that function as, say euclid(a,b). Then, you can define gcd(nums) as:
if (len(nums) == 1)
return nums[1]
else
return euclid(nums[1], gcd(nums[:2]))
This uses the associative property of gcd() to compute the answer
Try calling the GCD() as follows,
i = 0
temp = numbers[i]
for i in range(len(numbers)-1):
temp = GCD(numbers[i+1], temp)
My way of solving it in Python. Hope it helps.
def find_gcd(arr):
if len(arr) <= 1:
return arr
else:
for i in range(len(arr)-1):
a = arr[i]
b = arr[i+1]
while b:
a, b = b, a%b
arr[i+1] = a
return a
def main(array):
print(find_gcd(array))
main(array=[8, 18, 22, 24]) # 2
main(array=[8, 24]) # 8
main(array=[5]) # [5]
main(array=[]) # []
Some dynamics how I understand it:
ex.[8, 18] -> [18, 8] -> [8, 2] -> [2, 0]
18 = 8x + 2 = (2y)x + 2 = 2z where z = xy + 1
ex.[18, 22] -> [22, 18] -> [18, 4] -> [4, 2] -> [2, 0]
22 = 18w + 4 = (4x+2)w + 4 = ((2y)x + 2)w + 2 = 2z
As of python 3.9 beta 4, it has got built-in support for finding gcd over a list of numbers.
Python 3.9.0b4 (v3.9.0b4:69dec9c8d2, Jul 2 2020, 18:41:53)
[Clang 6.0 (clang-600.0.57)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import math
>>> A = [30, 40, 36]
>>> print(math.gcd(*A))
2
One of the issues is that many of the calculations only work with numbers greater than 1. I modified the solution found here so that it accepts numbers smaller than 1. Basically, we can re scale the array using the minimum value and then use that to calculate the GCD of numbers smaller than 1.
# GCD of more than two (or array) numbers - alows folating point numbers
# Function implements the Euclidian algorithm to find H.C.F. of two number
def find_gcd(x, y):
while(y):
x, y = y, x % y
return x
# Driver Code
l_org = [60e-6, 20e-6, 30e-6]
min_val = min(l_org)
l = [item/min_val for item in l_org]
num1 = l[0]
num2 = l[1]
gcd = find_gcd(num1, num2)
for i in range(2, len(l)):
gcd = find_gcd(gcd, l[i])
gcd = gcd * min_val
print(gcd)
HERE IS A SIMPLE METHOD TO FIND GCD OF 2 NUMBERS
a = int(input("Enter the value of first number:"))
b = int(input("Enter the value of second number:"))
c,d = a,b
while a!=0:
b,a=a,b%a
print("GCD of ",c,"and",d,"is",b)
As You said you need a program who would take any amount of numbers
and print those numbers' HCF.
In this code you give numbers separated with space and click enter to get GCD
num =list(map(int,input().split())) #TAKES INPUT
def print_factors(x): #MAKES LIST OF LISTS OF COMMON FACTROS OF INPUT
list = [ i for i in range(1, x + 1) if x % i == 0 ]
return list
p = [print_factors(numbers) for numbers in num]
result = set(p[0])
for s in p[1:]: #MAKES THE SET OF COMMON VALUES IN LIST OF LISTS
result.intersection_update(s)
for values in result:
values = values*values #MULTIPLY ALL COMMON FACTORS TO FIND GCD
values = values//(list(result)[-1])
print('HCF',values)
Hope it helped
UPDATED
a = int(input("Give a value: "))
b = int(input("Give a value: "))
c = int(input("Give a value: "))
def middle(a, b ,c) :
m = min(a,b,c)
M = max(a,b,c)
return a+b+c-m-M
This is where im at. It takes my numbers into the data. How would I get it to display the middle one?! Sorry I'm so terrible at this. Way in over my head on this intro course. #CommuSoft #Zorg #paxdiablo and everyone else
Like others mentioned, you're missing a colon, but for simplicity sake:
def middle(a, b, c):
return sorted([a, b, c])[1]
You should put a colon (:) on the first line (def) as well.
This works for the online python environment:
def input(a, b, c) :
if a <= b <= c or c <= b <= a :
return b
elif b <= a <= c or c <= a <= b :
return a
else:
return c
Furthermore it is more advisable to make use of min and max I guess. Min and max are sometimes directly supported by a CPU and there are implementations that avoid branching (if-then-else's):
def input(a, b, c) :
m = min(a,b,c)
M = max(a,b,c)
return a+b+c-m-M
or:
def input(a, b, c) :
return min(max(a,b),max(b,c),max(a,c))
The last one is also numerically stable.
In most cases if-then-else clauses should be avoided. They reduce the amount of pipelining although in interpreted languages this might not increase performance.
Based on the comments, I guess you want to write an interactive program. This can be done like:
def middle(a, b, c) : #defining a method
return min(max(a,b),max(b,c),max(a,c))
a = int(input("Give a value: "))
b = int(input("Give b value: "))
c = int(input("Give c value: "))
print("The requested value is ")
print(middle(a,b,c)) #calling a method
Defining a method will never result in Python using that method. The a, b and c in the def block are not the a, b and c in the rest of your program. These are "other variables that happen to have the same name". In order to call a method. You write the methods name and between brackets the parameters with which you wish to call your method.
Post your full syntax error (or any other full traceback) whenever you're having trouble.
And your def line needs a colon.
You could do:
def middle(a,b,c):
s={a,b,c}
s-={min(s),max(s)}
return s.pop()
What this does:
Create a set of the unduplicated values:
>>> a,b,c=1,2,3
>>> s={a,b,c}
>>> s
{1, 2, 3}
Remove the max and the min, leaving the middle:
>>> s-={min(s), max(s)}
>>> s
{2}
Pop the only remaining value:
>>> s.pop()
2
I am trying to write a program in Python, but I am stuck in this piece of code:
def function():
a=[3,4,5,2,4]
b=1
c=0
for x in range(5):
if a[x-1]>b:
c=c+1
return c
print(function())
It gives me value 1 instead of 5. Actually the function I am trying to write is a little bit more complicated, but the problem is actually the same, it doesn't give me the right result.
def result():
r=[0]*len(y)
a=2
b=an_integer
while b>0:
for x in range(len(y)) :
if y[x-1] > 1/a and b>0:
r[x-1]=r[x-1]+1
b=b-1
a=a+1
return r
print(result())
v is a list of values smaller than 1 and b has an integer as value. If some values x in v are bigger than 1/a then the values x in r should get 1 bigger, then it should repeat a=a+1 until b becomes 0. I want this function to give a result of the type for ex. [7,6,5,4,3] where the sum of the elements in this list is equal to b.
Sometimes it gives me the right value, sometimes not and when the elements in v are equal for example v=[0.33333,0.33333,0.33333] it gets stuck and doesn't give me a result.
I don't know what I am doing wrong !
Your return statements are incorrectly indented. You want to return after the loop ends, not inside the loop.
def function():
a = [3, 4, 5, 2, 4]
b = 1
c = 0
for x in range(5):
if a[x-1] > b:
c = c + 1
return c
Also, a couple of optimizations to the code:
def function(a, b):
c = 0
for x in a:
if x > b:
c += 1
return c
or further:
def function(a, b):
return sum(x > b for x in a)
return; only inside the fun in the end it.
and name the Variable v