I have written a code to find out the LCM [Lowest Common Multiple] of a list of numbers but there appears to be an error in my code. The code is given below:
def final_lcm[thelist]:
previous_thelist = thelist
prime_thelist = list[set[thelist] - set[returns_new_thelist[previous_thelist]]
factors = 1
for i in prime_thelist:
factors = factors*i
new_thelist = returns_new_thelist[previous_thelist]
for i in range[1, 10000000000]:
s_empty = []
for j in new_thelist:
if i % j == 0:
s_empty.append[True]
if len[new_thelist] == len[s_empty]:
initial_lcm = i
break
final_lcm = factor*initial_lcm
return final_lcm
def returns_new_thelist[ll]:
if 3 in ll:
ll.remove[3]
for i in ll:
if checks_if_prime[i] == True:
ll.remove[i]
return ll
def checks_if_prime[n]:
if n == 2:
return True
import math
for i in range[math.ceil[0.5*n], 1, -1]:
if n % i == 0:
return False
elif i == 2:
return True
print[final_lcm[[1,2,3,4,5,6,7,8,9]]]
Kindly pardon my poor choice of variables, I request you to see if the logic is correct and that the code is functional.
The syntax error which I am getting is that "factors" is invalid syntax though I don't agree with this. Please tell me where my code is wrong.
asked May 15, 2016 at 11:52
Aradhye AgarwalAradhye Agarwal
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This is the best way that I know of :
from math import gcd
a = [100, 200, 150] #will work for an int array of any length
lcm = 1
for i in a:
lcm = lcm*i//gcd[lcm, i]
print[lcm]
Hope this helps. All queries, contributions and comments are welcome :]
answered Feb 26, 2017 at 19:01
Ananay MitalAnanay Mital
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Works with an arbitrarily long denominator list.
from math import gcd # Python versions 3.5 and above
#from fractions import gcd # Python versions below 3.5
from functools import reduce # Python version 3.x
def lcm[denominators]:
return reduce[lambda a,b: a*b // gcd[a,b], denominators]
Example:
>>> lcm[[100, 200, 300]]
600
answered Apr 13, 2018 at 11:34
TakingItCasualTakingItCasual
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1
As of Python 3.9 lcm[] function has been added in the math
library. It can be called with the following signature:
math.lcm[*integers]
Return the least common multiple of the specified integer arguments. If all arguments are nonzero, then the returned value is the smallest positive integer that is a multiple of all arguments. If any of the arguments is zero, then the returned value is
0.lcm[]without arguments returns1.
Advantages:
- Besides being native,
- Its a one-liner,
- Its fastest,
- Can deal with arbitrarily long list of integers
- And can deal with nearly any kind of exceptions [e.g.
lcm[0,0]] overlooked by custom-built solutions.
answered Jun 17, 2020 at 11:05
In Numpy v1.17 [which is, as of
writing, the non-release development version] there is an lcm function that can be used for two numbers with, e.g.:
import numpy as np
np.lcm[12, 20]
or for multiple numbers with, e.g.:
np.lcm.reduce[[40, 12, 20]]
There's also a gcd function.
answered Feb 8, 2019 at 17:09
Matt PitkinMatt Pitkin
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3
Your solution might be too lengthy ... Try this !
from functools import reduce # need this line if you're using Python3.x
def lcm[a, b]:
if a > b:
greater = a
else:
greater = b
while True:
if greater % a == 0 and greater % b == 0:
lcm = greater
break
greater += 1
return lcm
def get_lcm_for[your_list]:
return reduce[lambda x, y: lcm[x, y], your_list]
ans = get_lcm_for[[1, 2, 3, 4, 5, 6, 7, 8, 9]]
print[ans]
answered May 15, 2016 at 12:11
Jay PatelJay Patel
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2
if you don't want to import anything.
def gcd[n, m]:
if m == 0:
return n
return gcd[m, n % m]
A = [10, 25, 37, 15, 75, 12]
lcm = 1
for i in A:
lcm = lcm * i // gcd[lcm, i]
print[lcm]
answered Apr 22, 2021 at 6:14
SabaSaba
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You're missing a closing parenthesis []] in the third line. Hence the error in line factors.
Moreover in second to last line of your first function, you've named the variable factor instead of factors.
SiHa
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answered May 15, 2016 at 11:57
Jay PatelJay Patel
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To find LCM of given list of numbers
def findDivisor[num]:
# 2,3 are the most common divisor for many numbers hence I go by divisor of 2,3
# if not then by the same number as divisor
if num%2 == 0:
return 2
elif num%3==0:
return 3
return num
def findLCM[lcmArray]:
lcm = 1
while len[lcmArray] > 0:
minOfLCMArray = min[lcmArray]
divisor = findDivisor[minOfLCMArray]
for x in xrange[0, len[lcmArray]]:
Quotient = lcmArray[x]/divisor
Reminder = lcmArray[x]%divisor
if Reminder == 0:
lcmArray[x] = Quotient
lcm*=divisor
minOfLCMArray = min[lcmArray]
if minOfLCMArray == 1:
lcmArray.remove[minOfLCMArray]
return lcm
lcmArray = map[int, raw_input[].split[]]
print findLCM[lcmArray]
answered May 8, 2017 at 10:19
A faster approach without using any math functions would be to calculate GCD and calculate LCM.
def gcd[a,b]:
while b:
a,b = b, a%b
return a
Now find LCM using GCF
def lcm[a,b]:
return a*b // gcd[a,b]
Extend this to work with list as below
LCM = functools.reduce[lambda x, y: lcm[x, y], your_list_to_find_lcm]
answered Jul 26, 2018 at 0:20
YashwanthYashwanth
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I had written a code to find lcm of numbers within a list. The User can input any number of values he wants. I'm attaching the code below, It is simpler than the code you posted. Try checking this... I know your question is to find errors in your code, but try checking this for future purpose.
a = list[map[int,input['enter numbers for the lcm: '].strip[].split[]]]
a.sort[reverse = True]
a
x = a[0]
while 1:
sum = 0
for i in a:
if x%i == 0:
sum += 1
if sum == len[a]:
break
else :
x += 1
print[x,' is the lcm of numbers in the input']
Harsha pps
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answered Jul 22, 2019 at 17:48
c=1
i=0
q=0
j=2;
flag=0;
count=0;
a=input["ente 3 no"]
a=a.split[',']
print[len[a]]
for i in range[len[a]]:
z=int[a[i]]
c=c*z
while[j