0
to get sum of the sub lists
Hi there, I tried to perform a code which divide the given list into three. And then get sum of each part into a new list (without build in). For example; given list = [1,2,3,4,5,6,7,8,9] output=[6,15,24] My code: def y(x): t=len(x) z=0 f=0 sum=[] if t%3!=0: return "can not be divided into 3 parts" else: for i in range(z,t,int(t/3)): z=i r=x[z:z+int(t/3)] for j in r: f=+j sum.append (f) return sum y([1,2,3,4,5,6]) i could divide the list into three part. However i can not get the true result for sum of each part. Can anyone help me?
7 odpowiedzi
+ 11
Hercule Poirot , nice code. Abhay , you have been faster than me, but i found the same issues. So I have no choice but to improve something else.
The code is working with 3 slices in a loop, then calculating the sum of each slice. I thought it would be nice to shorten this by using the sum function:
lst = [1,2,3,4,5,6,7,8,9]
def y(x):
t=len(x)
sum_=[]
if t%3!=0:
return "can not be divided into 3 parts"
else:
for i in range(0,t,int(t/3)):
sum_.append (sum(x[i:i+int(t/3)]))
return sum_
print(y(lst))
+ 3
here it is,
def y(x):
t=len(x)
z=0
f=0
sum=[]
if t%3!=0:
return "can not be divided into 3 parts"
else:
for i in range(z,t,int(t/3)):
z=i
r=x[z:z+int(t/3)]
print(r)
for j in r:
print(j)
f+=j
sum.append(f)
f=0
return sum
print(y([1,2,3,4,5,6]))
+ 3
Hercule Poirot for j in r:
Loops over first two numbers and add to f so f has a value of 3 now ,if you won't set f to 0 again it will be keep adding new values from for j in r: loop to 3 so instead of 3,7 you will get 3,10 and so on
+ 3
ty a lot for your great help. i really appricate it. :)
+ 2
ty a lot Abhay. I m now in phyton and also coding. could you explain what iş the role of f=0 in the code?
+ 2
Lothar nice approach and seems like we don't even need sum variable as well and can use list comprehension instead ,
lst = [1,2,3,4,5,6,7,8,9]
def y(x):
t=len(x)
if t%3!=0:
return "can not be divided into 3 parts"
else:
return([sum(x[i:i+int(t/3)]) for i in range(0,t,int(t/3))])
print(y(lst))
+ 1
ty Lothar for your help. :)