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