Dexter Ramos <blitzde...@gmail.com> added the comment:
Thank you Mr. Erick Smith. Now I know. I also tried to find the hard way like
this:
--------finding nemo-------------
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4] --->index
[4, 1, 2, 5, 7, 4, 2, 8, 9, 5, 3, 2, 4, 6] --->original list
/ --->started at index 0 with value of
4
[1, 2, 5, 7, 2, 8, 9, 5, 3, 2, 6, 4] --->1st iteration, all 4's are removed
then appended so the index adjusted
/ --->next at index 1 with a value of 2
(whereas value 1 was skipped which had index 1 originally; this is not seen on
the output because it has no duplicate)
[1, 5, 7, 8, 9, 5, 3, 6, 4, 2] --->3rd iteration
/ --->next at index 2 with value of 7; value
5 was skipped which had the index 2 originally; cause found!
[1, 5, 8, 9, 5, 3, 6, 4, 2, 7] --->4th ...
...
---------------nemo found-----------------------
Credits to you.
Here is the new working code:
-------code------------------------------------------------
bunchofnumbers = [4, 1, 2, 5, 7, 4, 2, 8, 9, 5, 3, 2, 4, 6]
for eachnumber in bunchofnumbers.copy():
while eachnumber in bunchofnumbers:
bunchofnumbers.remove(eachnumber)
bunchofnumbers.append(eachnumber)
bunchofnumbers.sort()
-------end of code-----------------------------------------
OUTPUT:
[1, 2, 3, 4, 5, 6, 7, 8, 9]
print(bunchofnumbers)
----------
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue41454>
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