Hello can someone explain to me why this code doesnt work out
import random
def randomlist(nums):
emptylist = []
for j in range(0, 2**len(nums)):
emptylist[j].append(random.shuffle(nums))
print(emptylist)
randomlist([1,2,3]) I want to get a list(in that case is the emptylist) that contains as sublists(the [1,2,3] in that case)shuffled n times
Even tho i use this for loop when it prints the result it prints the list but all the sublists are shuffled in the same way
The first observation is line 5.
"Append" decides for itself where to put the next item.
You don't need to force the index upon the emptylist.
Paul
(Nov-12-2020, 04:22 PM)DPaul Wrote: [ -> ]The first observation is line 5.
"Append" decides for itself where to put the next item.
You don't need to force the index upon the emptylist.
Paul
Yes you are right about this.I didnt notice it to be honest but my problem is that even tho i have this for loop every sublist has the same order
(Nov-12-2020, 04:09 PM)giorgosmarga Wrote: [ -> ]Hello can someone explain to me why this code doesnt work out
What does mean 'doesnt work out'? Don't give expected results?
Python does everything correct:
>>> emptylist = []
>>> emptylist[0].append('whatever')
/.../
IndexError: list index out of rangerandom.shuffle documentation: "Shuffle list x in place, and return None.". Python obediently returns None-s to emptylist.
And now something totally different: it is maybe also worth to know that functions without
return or
yield return
None.
(Nov-12-2020, 04:42 PM)perfringo Wrote: [ -> ] (Nov-12-2020, 04:09 PM)giorgosmarga Wrote: [ -> ]Hello can someone explain to me why this code doesnt work out
What does mean 'doesnt work out'? Don't give expected results?
Python does everything correct:
>>> emptylist = []
>>> emptylist[0].append('whatever')
/.../
IndexError: list index out of rangerandom.shuffle documentation: "Shuffle list x in place, and return None.". Python obediently returns None-s to emptylist.
And now something totally different: it is maybe also worth to know that functions without return or yield return None.
You are right i didnt give much details.I updated my post
(Nov-12-2020, 04:53 PM)giorgosmarga Wrote: [ -> ]You are right i didnt give much details.I updated my post
Never change original post, explain and clarify in subsequent comments. Otherwise thread doesn't make sense for readers.
I don't get how this code can produce: "every sublist has the same order". There are no sublists, there are None-s:
>>> emptylist = []
>>> emptylist.append(random.shuffle([1, 2, 3]))
>>> emptylist
[None]
(Nov-12-2020, 05:09 PM)giorgosmarga Wrote: [ -> ] (Nov-12-2020, 05:03 PM)perfringo Wrote: [ -> ]Never change original post, explain and clarify in subsequent comments. Otherwise thread doesn't make sense for readers.
I don't get how this code can produce: "every sublist has the same order". There are no sublists, there are None-s:
>>> emptylist = []
>>> emptylist.append(random.shuffle([1, 2, 3]))
>>> emptylist
[None]
def randomlist(nums):
emptylist = []
for i in range(2**len(nums)):
emptylist.append(nums)
for sublist in emptylist:
random.shuffle(sublist)
return emptylist
print(randomlist([1,2,3]))Because i am confused shouldnt that code return the emptylist with 8 shuffled differenlty sublists?
(Nov-12-2020, 05:12 PM)giorgosmarga Wrote: [ -> ]Because i am confused shouldnt that code return the emptylist with 8 shuffled differenlty sublists?
No, it shouldn't. There is only one list and you shuffle it 8 times.
Maybe following code can explain:
>>> nums = [1, 2, 3]
>>> emptylist = []
>>> for i in range(3):
... emptylist.append(nums) # you are adding same list n times
...
>>> emptylist
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]
>>> for item in emptylist:
... print(id(item))
...
140256603923232 # same object 3 times
140256603923232 # or more correctly three references to same object
140256603923232
>>> emptylist[0][0] = 'no way' # changing item in emptylist
>>> emptylist # changes all items because they reference same object
[['no way', 2, 3], ['no way', 2, 3], ['no way', 2, 3]]
>>> nums[1] = "it's happening" # it's getting worse; you change original list
>>> emptylist # and as this is same object it is reflected in emptylist
[['no way', "it's happening", 3],
['no way', "it's happening", 3],
['no way', "it's happening", 3]]
(Nov-12-2020, 05:36 PM)perfringo Wrote: [ -> ] (Nov-12-2020, 05:12 PM)giorgosmarga Wrote: [ -> ]Because i am confused shouldnt that code return the emptylist with 8 shuffled differenlty sublists?
No, it shouldn't. There is only one list and you shuffle it 8 times.
Maybe following code can explain:
>>> nums = [1, 2, 3]
>>> emptylist = []
>>> for i in range(3):
... emptylist.append(nums) # you are adding same list n times
...
>>> emptylist
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]
>>> for item in emptylist:
... print(id(item))
...
140256603923232 # same object 3 times
140256603923232 # or more correctly three references to same object
140256603923232
>>> emptylist[0][0] = 'no way' # changing item in emptylist
>>> emptylist # changes all items because they reference same object
[['no way', 2, 3], ['no way', 2, 3], ['no way', 2, 3]]
>>> nums[1] = "it's happening" # it's getting worse; you change original list
>>> emptylist # and as this is same object it is reflected in emptylist
[['no way', "it's happening", 3],
['no way', "it's happening", 3],
['no way', "it's happening", 3]]
Ohh i got it thank you very much.But is there a way to do what i want to do?