+ 4
I can't believe this question is still in the quiz
"What is the output of this code? x = {1:0, 0:1} y = [1, [4,6,5,4], 2] print(list(set(y[x[0]]))[1]) The answer is 5" What's going on here, you may ask. First of all, by the definition "A set object is an __unordered collection__ of distinct hashable objects. " Second, try changing the code a little bit: x = {1:0, 0:1} y = [1, [-1,6,5,-1], 2] print(list(set(y[x[0]]))[1]) What answer do you expect in this case? The "correct" answer is -1 Or may be you like this combination [-1,-3,0,-1]?
9 Respuestas
+ 3
A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟ This is in principle an incorrect question because there is no order in Set. The answer will depend on the hash implementation.
+ 2
A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟ Did you read properly my question? Second part?
Without knowing the hash implementation, can you predict the behavior of
x = {1:0, 0:1}
y = [1, [-1,6,5,-1], 2]
print(list(set(y[x[0]]))[1])? or may be:
x = {1:0, 0:1}
y = [1, [-1,0,-3,-1], 2]
print(list(set(y[x[0]]))[1])?
+ 2
True, just because the set conversion gave the same order multiple times is not a guarantee that it always will.
So we should not expect the result of the problem to always be the same. And yes, it is not a good problem because it gives the impression that sets are ordered.
more discussion at SO:
https://stackoverflow.com/questions/3812429/is-pythons-set-stable
+ 1
A͢J - S͟o͟l͟o͟H͟e͟l͟p͟e͟r͟ As I said, it depends on the specific hash implementation. You can go to https://github.com/JUMP_LINK__&&__python__&&__JUMP_LINK/cpython/blob/main/Python/pyhash.c for details.
A change in the implementation will not change the set properties, but it will change the answer to this quiz.
0
Alexey Kopyshev
Yes there is no order that's why set([4, 6, 5, 4]) returns [6, 5, 4] so why this question is wrong.
If there is order in set then it should return [4, 5, 6].
Is it not?
0
Alexey Kopyshev
In this case [-1, -3, 0, -1] getting -3. I think it should be 0. There is little confusion.
Can you explain why this behaviour?
0
EXEMPLE :
S=0
for i in range (1,5):
S+=i
print (i, end=" + ")
print ("=", S)
#output
# 1 + 2 + 3 + 4 +=10
I want to remove the last + on the output
- 1
Alexey Kopyshev
x = {1:0, 0:1}
y = [1, [4, 6, 5, 4], 2]
list(set(y[x[0]]))[1])
x[0] = 1
So list(set(y[1]))[1])
Now y[1] = [4, 6, 5, 4]
So list(set([4, 6, 5, 4]))[1]
As set doesn't allow duplicate so
list(6, 5, 4)[1] = 5
- 4
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