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+--- Thread: list without repetitions (/thread-15272.html)
list without repetitions - sokadin - Jan-10-2019
Good day dear people, i just need little help with my code. I have to make list without repetitions from the one that has repetitions. I don't know where am i making a mistake. this is code:
liss=[6,7,5,3,6,8,9,6,4,3,2,5,6,8,9,0,1,98]
def lwr(liss):
liss.sort(key=None,reverse=False)
print(liss)
i=1
liss1=[]
for n in liss:
if n!=liss[i]:
liss1.append(n)
i+=1
else:
continue
return print(liss1)
lwr(liss)Thank you.
RE: list without repetitions - nilamo - Jan-10-2019
You generate a new list, but then don't return it.
But also, you only increment
i when a match is found. Which means once there isn't a match, the whole rest of the list would be included.To demonstrate, I added a couple prints, along with a smaller sample size:
def lwr(liss):
liss.sort(key=None, reverse=False)
print(liss)
i=1
liss1=[]
for n in liss:
print(f"Index: {i}, Current Value: {n}, Next Value: {liss[i]}")
if n!=liss[i]:
liss1.append(n)
i+=1
return liss1
liss=[1, 1, 2, 3, 3, 4, 5, 5]
unique = lwr(liss)
print(unique)Output:[1, 1, 2, 3, 3, 4, 5, 5]
Index: 1, Current Value: 1, Next Value: 1
Index: 1, Current Value: 1, Next Value: 1
Index: 1, Current Value: 2, Next Value: 1
Index: 2, Current Value: 3, Next Value: 2
Index: 3, Current Value: 3, Next Value: 3
Index: 3, Current Value: 4, Next Value: 3
Index: 4, Current Value: 5, Next Value: 3
Index: 5, Current Value: 5, Next Value: 4
[2, 3, 4, 5, 5]if n != liss1[-1]