listsal2 = [1,2,2,3,3,4,5,6,7,8]
listsal3 = []
counter = 0
for i in listsal2:
item = listsal2.count(i)
if item > 1:
counter = item
while counter > 1:
listsal3.append(i)
counter = counter - 1
print (listsal3)
I've been working on a mode function but for some reason it keeps the last couple numbers in the list and the more items in the list, the more that don't get removed.
EDIT: just realized I forgot the 2nd part of the code which is now in
EDIT2: code is shrunken down and easier to read
EDIT3: changed code so the duplicate numbers go into a new list but it has the multiple amounts of the list item
Thanks to all for the help I think I've got it now
In your case, even after you account for i=2 once, you go through the list again for 2, since it exists multiple times. This is why 2 and 3 end up in listsal3 twice. Instead, what you want to do is only go through the list once for each unique item.
listsal2 = [1,2,2,3,3, 4]
newlist = set(listsal2)
listsal3 = []
counter = 0
for i in newlist:
item = listsal2.count(i)
if item > 1:
counter = item
print counter
while counter > 1:
listsal3.append(i)
counter = counter - 1
print listsal2, listsal3
To get the unique items from your list, convert it to a set.
Another way to do it: Simply keep a list containing the counts of each unique element, take the maximum of it, and trace back to the element that it corresponds to.
newset = set(listsal2)
newlist = list(set)
counts = []
for item in newlist:
counts.append(listsal2.count(item))
maxcount = max(counts)
max_occurring_item = newlist[counts.index(maxcount)]