Python gcd for list

帅比萌擦擦* 提交于 2020-07-20 07:47:08

问题


I want to calculate gcd for a list of numbers. But I don't know what's wrong with my code.

A = [12, 24, 27, 30, 36]


def Greatest_Common_Divisor(A):
    for c in A:
        while int(c) > 0:
            if int(c) > 12:
                c = int(c) % 12
            else:
                return 12 % int(c)
    print Greatest_Common_Divisor(A)

回答1:


here is the piece of code, that I used:

from fractions import gcd
from functools import reduce
def find_gcd(list):
    x = reduce(gcd, list)
    return x



回答2:


def gcd (a,b):
    if (b == 0):
        return a
    else:
         return gcd (b, a % b)
A = [12, 24, 27, 30, 36]
res = A[0]
for c in A[1::]:
    res = gcd(res , c)
print res

ideone link




回答3:


It's not clear to me why you are using 12 in your function? Do you want to test your algorithm with 12 specifically?

There is built in function that provides a good solution (fraction.gcd()) as referenced in this answer

If you want to develop your own approach, you could do it this way: sort the list and get the minimum number of list (call it min). Loop from 2 to min, you can get the great common divisor of your list.




回答4:


def find_gcd(l):
    def gcd(a, b):
        while b:
            a, b = b, a%b
        return a
    n =1
    f = l[0]
    while n != len(l):
        f = gcd(f,l[n])
        if  f == 1:
            return 1
        else:
            n = n + 1          
    return f

l = [12, 24, 27, 30, 36]
print(find_gcd(l))



回答5:


return exits the function. Inside a for-loop this is normally not intended.




回答6:


As I see your code will simply go in infinite loop. Since you call method Greatest_Common_Divisor recursively but without base case. Align print Greatest_Common_Divisor(A) and "def" in the same column and that problem would be solved. But still what your code does for each number ai, it takes remainder of ai % 12, and then simply print 12 % (ai % 12), and there is no any connection between it and greatestCommonDivisor. Here is simple code for gcd(a,b) which you can use for the whole array:

def gcd (a,b):
    if (b == 0):
        return a
    else:
         return gcd (b, a % b)



回答7:


from functools import reduce
def gcd(a,b):
    if a==0:
        return b 
    else:
        return gcd(b%a,a)
A = [12, 24, 27, 30, 36]
gcdp = reduce(lambda x,y:gcd(x,y),A)
print(gcdp)

I guess this one will clear your doubts.




回答8:


import functools as f
A = [12, 24, 27, 30, 36]
g = lambda a,b:a if b==0 else g(b,a%b)   #Gcd for two numbers
print(f.reduce(lambda x,y:g(x,y),A))     #Calling gcd function throughout the list.

"Lambda" is an anonymous function where 'g' is assigned with GCD of two numbers whenever called.

"Reduce" is a function in "functools" module which is used to perform a particular function to all of the elements in the list. Here reduce() computes the GCD of the complete list A by computing GCD of first two elements, then the GCD of 3rd element with previously computed GCD of first two elements and so on.

Hope this clears your doubt.




回答9:


I used this piece of code:

def gcd(my_list):
    result = my_list[0]
    for x in my_list[1:]:
        if result < x:
            temp = result
            result = x
            x = temp
        while x != 0:
            temp = x
            x = result % x
            result = temp
    return result 



回答10:


gcd of a list input by the user which can be used for any number of input values.

 n = int(input('enter no of numbers: '))
a = list(map(int,input('enter numbers to find gcd: ').strip().split()))[:n]

def gcd(num1,num2):
    x = 1
    while x:
        if max(num1,num2) % min(num1,num2) == 0:
            return min(num1,num2)
            x = 0
        else :
            r = max(num1,num2)%min(num1,num2)
            return gcd(max(num1,num2),r)

while True:
        a[0] = gcd(a[0],a[1])
        a.pop(1)
        if len(set(a))>2:
            a.pop(2)
        if len(set(a)) == 1:
            break
a = set(a)
print(a)



回答11:


If you'd like to use an existing method try `np.gcd.reduce':

import numpy as np

A = [12, 24, 27, 30, 36]

print(np.gcd.reduce(A))

which returns 3



来源:https://stackoverflow.com/questions/29194588/python-gcd-for-list

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