How do I remove consecutive duplicates from a list?

问题

How do I remove consecutive duplicates from a list like this in python?

lst = [1,2,2,4,4,4,4,1,3,3,3,5,5,5,5,5]

Having a unique list or set wouldn't solve the problem as there are some repeated values like 1,...,1 in the previous list.

I want the result to be like this:

newlst = [1,2,4,1,3,5]

Would you also please consider the case when I have a list like this [4, 4, 4, 4, 2, 2, 3, 3, 3, 3, 3, 3] and I want the result to be [4,2,3,3] rather than [4,2,3] .

回答1:

itertools.groupby() is your solution.

newlst = [k for k, g in itertools.groupby(lst)]

If you wish to group and limit the group size by the item's value, meaning 8 4's will be [4,4], and 9 3's will be [3,3,3] here are 2 options that does it:

import itertools

def special_groupby(iterable):
    last_element = 0
    count = 0
    state = False
    def key_func(x):
        nonlocal last_element
        nonlocal count
        nonlocal state
        if last_element != x or x >= count:
            last_element = x
            count = 1
            state = not state
        else:
            count += 1
        return state
    return [next(g) for k, g in itertools.groupby(iterable, key=key_func)]

special_groupby(lst)

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return itertools.zip_longest(*args, fillvalue=fillvalue)

newlst = list(itertools.chain.from_iterable(next(zip(*grouper(g, k))) for k, g in itertools.groupby(lst)))

Choose whichever you deem appropriate. Both methods are for numbers > 0.

回答2:

list1 = ['a', 'a', 'a', 'b', 'b' , 'a', 'f', 'c', 'a','a']
temp_list = []


for item in list1:   
   if len(temp_list) == 0:
      temp_list.append(item)

   elif len(temp_list) > 0:
      if  temp_list[-1] != item:
          temp_list.append(item)

print(temp_list)

Fetch each item from the main list(list1).
If the 'temp_list' is empty add that item.
If not , check whether the last item in the temp_list is not same as the item we fetched from 'list1'.
if items are different append into temp_list.

回答3:

If you want to use the itertools method @MaxU suggested, a possible code implementation is:

import itertools as it

lst=[1,2,2,4,4,4,4,1,3,3,3,5,5,5,5,5]

unique_lst = [i[0] for i in it.groupby(lst)]

print(unique_lst)

回答4:

You'd probably want something like this.

lst = [1, 1, 2, 2, 2, 2, 3, 3, 4, 1, 2]
prev_value = None
for number in lst[:]: # the : means we're slicing it, making a copy in other words
    if number == prev_value:
        lst.remove(number)
    else:
        prev_value = number

So, we're going through the list, and if it's the same as the previous number, we remove it from the list, otherwise, we update the previous number.

There may be a more succinct way, but this is the way that looked most apparent to me.

HTH.

回答5:

newlist=[]    
prev=lst[0]
newlist.append(prev)
    for each in lst[:1]: #to skip 1st lst[0]
        if(each!=prev):
            newlist.append(each)  
         prev=each

回答6:

st = ['']
[st.append(a) for a in [1,2,2,4,4,4,4,1,3,3,3,5,5,5,5,5] if a != st[-1]]
print(st[1:])

回答7:

Check if the next element always is not equal to item. If so append.

lst = [1,2,2,4,4,4,4,1,3,3,3,5,5,5,5,5]

new_item = lst[0]
new_list = [lst[0]]
for l in lst:
   if new_item != l:
     new_list.append(l)
     new_item = l

print new_list
print lst

来源：https://stackoverflow.com/questions/39237350/how-do-i-remove-consecutive-duplicates-from-a-list

标签

python

duplicates