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How to sort iterable with Python without using stable sort?

So, I have an iterable in Input like this:
[4, 6, 2, 2, 6, 4, 4, 4]
And I want to sort it based on decreased frequency order. So that the result will be this:
[4, 4, 4, 4, 6, 6, 2, 2]
So what happened here is that, when an element has the same frequency of another one, they will be in the same order (6 appeared first so the 6 goes before the 2).
I tried to implement this mechanism using the sorted function but I have a big problem.
def frequency_sort(items):
return sorted(items, key=lambda elem: sum([True for i in items if i == elem]), reverse=True)
I know this short way is difficult to read but it just sort the array using the key parameter to extract the frequency of a number. But, the output is this:
[4, 4, 4, 4, 6, 2, 2, 6]
As you can see the output is a little different from what it should be. And that happened (I think) because sorted() is a function that does a "stable sort" i.e. a sort that will keep the order as it is if there are same keys.
So what is happening here is like a strong stable sort. I want more like a soft-sort that will take into account the order but will put the same elements next to each other.

You could use collections.Counter and use most_common that returns in descending order of frequency:
from collections import Counter
def frequency_sorted(lst):
counts = Counter(lst)
return [k for k, v in counts.most_common() for _ in range(v)]
result = frequency_sorted([4, 6, 2, 2, 6, 4, 4, 4])
print(result)
Output
[4, 4, 4, 4, 6, 6, 2, 2]
From the documentation on most_common:
Return a list of the n most common elements and their counts from the
most common to the least. If n is omitted or None, most_common()
returns all elements in the counter. Elements with equal counts are
ordered in the order first encountered

Using Regex in Python 3 [duplicate]

How can I check if a list has any duplicates and return a new list without duplicates?

The common approach to get a unique collection of items is to use a set. Sets are unordered collections of distinct objects. To create a set from any iterable, you can simply pass it to the built-in set() function. If you later need a real list again, you can similarly pass the set to the list() function.
The following example should cover whatever you are trying to do:
>>> t = [1, 2, 3, 1, 2, 3, 5, 6, 7, 8]
>>> list(set(t))
[1, 2, 3, 5, 6, 7, 8]
>>> s = [1, 2, 3]
>>> list(set(t) - set(s))
[8, 5, 6, 7]
As you can see from the example result, the original order is not maintained. As mentioned above, sets themselves are unordered collections, so the order is lost. When converting a set back to a list, an arbitrary order is created.
Maintaining order
If order is important to you, then you will have to use a different mechanism. A very common solution for this is to rely on OrderedDict to keep the order of keys during insertion:
>>> from collections import OrderedDict
>>> list(OrderedDict.fromkeys(t))
[1, 2, 3, 5, 6, 7, 8]
Starting with Python 3.7, the built-in dictionary is guaranteed to maintain the insertion order as well, so you can also use that directly if you are on Python 3.7 or later (or CPython 3.6):
>>> list(dict.fromkeys(t))
[1, 2, 3, 5, 6, 7, 8]
Note that this may have some overhead of creating a dictionary first, and then creating a list from it. If you don’t actually need to preserve the order, you’re often better off using a set, especially because it gives you a lot more operations to work with. Check out this question for more details and alternative ways to preserve the order when removing duplicates.
Finally note that both the set as well as the OrderedDict/dict solutions require your items to be hashable. This usually means that they have to be immutable. If you have to deal with items that are not hashable (e.g. list objects), then you will have to use a slow approach in which you will basically have to compare every item with every other item in a nested loop.

In Python 2.7, the new way of removing duplicates from an iterable while keeping it in the original order is:
>>> from collections import OrderedDict
>>> list(OrderedDict.fromkeys('abracadabra'))
['a', 'b', 'r', 'c', 'd']
In Python 3.5, the OrderedDict has a C implementation. My timings show that this is now both the fastest and shortest of the various approaches for Python 3.5.
In Python 3.6, the regular dict became both ordered and compact. (This feature is holds for CPython and PyPy but may not present in other implementations). That gives us a new fastest way of deduping while retaining order:
>>> list(dict.fromkeys('abracadabra'))
['a', 'b', 'r', 'c', 'd']
In Python 3.7, the regular dict is guaranteed to both ordered across all implementations. So, the shortest and fastest solution is:
>>> list(dict.fromkeys('abracadabra'))
['a', 'b', 'r', 'c', 'd']

It's a one-liner: list(set(source_list)) will do the trick.
A set is something that can't possibly have duplicates.
Update: an order-preserving approach is two lines:
from collections import OrderedDict
OrderedDict((x, True) for x in source_list).keys()
Here we use the fact that OrderedDict remembers the insertion order of keys, and does not change it when a value at a particular key is updated. We insert True as values, but we could insert anything, values are just not used. (set works a lot like a dict with ignored values, too.)

>>> t = [1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> t
[1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> s = []
>>> for i in t:
if i not in s:
s.append(i)
>>> s
[1, 2, 3, 5, 6, 7, 8]

If you don't care about the order, just do this:
def remove_duplicates(l):
return list(set(l))
A set is guaranteed to not have duplicates.

To make a new list retaining the order of first elements of duplicates in L:
newlist = [ii for n,ii in enumerate(L) if ii not in L[:n]]
For example: if L = [1, 2, 2, 3, 4, 2, 4, 3, 5], then newlist will be [1, 2, 3, 4, 5]
This checks each new element has not appeared previously in the list before adding it.
Also it does not need imports.

There are also solutions using Pandas and Numpy. They both return numpy array so you have to use the function .tolist() if you want a list.
t=['a','a','b','b','b','c','c','c']
t2= ['c','c','b','b','b','a','a','a']
Pandas solution
Using Pandas function unique():
import pandas as pd
pd.unique(t).tolist()
>>>['a','b','c']
pd.unique(t2).tolist()
>>>['c','b','a']
Numpy solution
Using numpy function unique().
import numpy as np
np.unique(t).tolist()
>>>['a','b','c']
np.unique(t2).tolist()
>>>['a','b','c']
Note that numpy.unique() also sort the values. So the list t2 is returned sorted. If you want to have the order preserved use as in this answer:
_, idx = np.unique(t2, return_index=True)
t2[np.sort(idx)].tolist()
>>>['c','b','a']
The solution is not so elegant compared to the others, however, compared to pandas.unique(), numpy.unique() allows you also to check if nested arrays are unique along one selected axis.

In this answer, there will be two sections: Two unique solutions, and a graph of speed for specific solutions.
Removing Duplicate Items
Most of these answers only remove duplicate items which are hashable, but this question doesn't imply it doesn't just need hashable items, meaning I'll offer some solutions which don't require hashable items.
collections.Counter is a powerful tool in the standard library which could be perfect for this. There's only one other solution which even has Counter in it. However, that solution is also limited to hashable keys.
To allow unhashable keys in Counter, I made a Container class, which will try to get the object's default hash function, but if it fails, it will try its identity function. It also defines an eq and a hash method. This should be enough to allow unhashable items in our solution. Unhashable objects will be treated as if they are hashable. However, this hash function uses identity for unhashable objects, meaning two equal objects that are both unhashable won't work. I suggest you override this, and changing it to use the hash of an equivalent mutable type (like using hash(tuple(my_list)) if my_list is a list).
I also made two solutions. Another solution which keeps the order of the items, using a subclass of both OrderedDict and Counter which is named 'OrderedCounter'. Now, here are the functions:
from collections import OrderedDict, Counter
class Container:
def __init__(self, obj):
self.obj = obj
def __eq__(self, obj):
return self.obj == obj
def __hash__(self):
try:
return hash(self.obj)
except:
return id(self.obj)
class OrderedCounter(Counter, OrderedDict):
'Counter that remembers the order elements are first encountered'
def __repr__(self):
return '%s(%r)' % (self.__class__.__name__, OrderedDict(self))
def __reduce__(self):
return self.__class__, (OrderedDict(self),)
def remd(sequence):
cnt = Counter()
for x in sequence:
cnt[Container(x)] += 1
return [item.obj for item in cnt]
def oremd(sequence):
cnt = OrderedCounter()
for x in sequence:
cnt[Container(x)] += 1
return [item.obj for item in cnt]
remd is non-ordered sorting, while oremd is ordered sorting. You can clearly tell which one is faster, but I'll explain anyways. The non-ordered sorting is slightly faster, since it doesn't store the order of the items.
Now, I also wanted to show the speed comparisons of each answer. So, I'll do that now.
Which Function is the Fastest?
For removing duplicates, I gathered 10 functions from a few answers. I calculated the speed of each function and put it into a graph using matplotlib.pyplot.
I divided this into three rounds of graphing. A hashable is any object which can be hashed, an unhashable is any object which cannot be hashed. An ordered sequence is a sequence which preserves order, an unordered sequence does not preserve order. Now, here are a few more terms:
Unordered Hashable was for any method which removed duplicates, which didn't necessarily have to keep the order. It didn't have to work for unhashables, but it could.
Ordered Hashable was for any method which kept the order of the items in the list, but it didn't have to work for unhashables, but it could.
Ordered Unhashable was any method which kept the order of the items in the list, and worked for unhashables.
On the y-axis is the amount of seconds it took.
On the x-axis is the number the function was applied to.
I generated sequences for unordered hashables and ordered hashables with the following comprehension: [list(range(x)) + list(range(x)) for x in range(0, 1000, 10)]
For ordered unhashables: [[list(range(y)) + list(range(y)) for y in range(x)] for x in range(0, 1000, 10)]
Note there is a step in the range because without it, this would've taken 10x as long. Also because in my personal opinion, I thought it might've looked a little easier to read.
Also note the keys on the legend are what I tried to guess as the most vital parts of the implementation of the function. As for what function does the worst or best? The graph speaks for itself.
With that settled, here are the graphs.
Unordered Hashables
(Zoomed in)
Ordered Hashables
(Zoomed in)
Ordered Unhashables
(Zoomed in)

Very late answer.
If you don't care about the list order, you can use *arg expansion with set uniqueness to remove dupes, i.e.:
l = [*{*l}]
Python3 Demo

A colleague have sent the accepted answer as part of his code to me for a codereview today.
While I certainly admire the elegance of the answer in question, I am not happy with the performance.
I have tried this solution (I use set to reduce lookup time)
def ordered_set(in_list):
out_list = []
added = set()
for val in in_list:
if not val in added:
out_list.append(val)
added.add(val)
return out_list
To compare efficiency, I used a random sample of 100 integers - 62 were unique
from random import randint
x = [randint(0,100) for _ in xrange(100)]
In [131]: len(set(x))
Out[131]: 62
Here are the results of the measurements
In [129]: %timeit list(OrderedDict.fromkeys(x))
10000 loops, best of 3: 86.4 us per loop
In [130]: %timeit ordered_set(x)
100000 loops, best of 3: 15.1 us per loop
Well, what happens if set is removed from the solution?
def ordered_set(inlist):
out_list = []
for val in inlist:
if not val in out_list:
out_list.append(val)
return out_list
The result is not as bad as with the OrderedDict, but still more than 3 times of the original solution
In [136]: %timeit ordered_set(x)
10000 loops, best of 3: 52.6 us per loop

Another way of doing:
>>> seq = [1,2,3,'a', 'a', 1,2]
>> dict.fromkeys(seq).keys()
['a', 1, 2, 3]

Simple and easy:
myList = [1, 2, 3, 1, 2, 5, 6, 7, 8]
cleanlist = []
[cleanlist.append(x) for x in myList if x not in cleanlist]
Output:
>>> cleanlist
[1, 2, 3, 5, 6, 7, 8]

I had a dict in my list, so I could not use the above approach. I got the error:
TypeError: unhashable type:
So if you care about order and/or some items are unhashable. Then you might find this useful:
def make_unique(original_list):
unique_list = []
[unique_list.append(obj) for obj in original_list if obj not in unique_list]
return unique_list
Some may consider list comprehension with a side effect to not be a good solution. Here's an alternative:
def make_unique(original_list):
unique_list = []
map(lambda x: unique_list.append(x) if (x not in unique_list) else False, original_list)
return unique_list

All the order-preserving approaches I've seen here so far either use naive comparison (with O(n^2) time-complexity at best) or heavy-weight OrderedDicts/set+list combinations that are limited to hashable inputs. Here is a hash-independent O(nlogn) solution:
Update added the key argument, documentation and Python 3 compatibility.
# from functools import reduce <-- add this import on Python 3
def uniq(iterable, key=lambda x: x):
"""
Remove duplicates from an iterable. Preserves order.
:type iterable: Iterable[Ord => A]
:param iterable: an iterable of objects of any orderable type
:type key: Callable[A] -> (Ord => B)
:param key: optional argument; by default an item (A) is discarded
if another item (B), such that A == B, has already been encountered and taken.
If you provide a key, this condition changes to key(A) == key(B); the callable
must return orderable objects.
"""
# Enumerate the list to restore order lately; reduce the sorted list; restore order
def append_unique(acc, item):
return acc if key(acc[-1][1]) == key(item[1]) else acc.append(item) or acc
srt_enum = sorted(enumerate(iterable), key=lambda item: key(item[1]))
return [item[1] for item in sorted(reduce(append_unique, srt_enum, [srt_enum[0]]))]

If you want to preserve the order, and not use any external modules here is an easy way to do this:
>>> t = [1, 9, 2, 3, 4, 5, 3, 6, 7, 5, 8, 9]
>>> list(dict.fromkeys(t))
[1, 9, 2, 3, 4, 5, 6, 7, 8]
Note: This method preserves the order of appearance, so, as seen above, nine will come after one because it was the first time it appeared. This however, is the same result as you would get with doing
from collections import OrderedDict
ulist=list(OrderedDict.fromkeys(l))
but it is much shorter, and runs faster.
This works because each time the fromkeys function tries to create a new key, if the value already exists it will simply overwrite it. This wont affect the dictionary at all however, as fromkeys creates a dictionary where all keys have the value None, so effectively it eliminates all duplicates this way.

I've compared the various suggestions with perfplot. It turns out that, if the input array doesn't have duplicate elements, all methods are more or less equally fast, independently of whether the input data is a Python list or a NumPy array.
If the input array is large, but contains just one unique element, then the set, dict and np.unique methods are costant-time if the input data is a list. If it's a NumPy array, np.unique is about 10 times faster than the other alternatives.
It's somewhat surprising to me that those are not constant-time operations, too.
Code to reproduce the plots:
import perfplot
import numpy as np
import matplotlib.pyplot as plt
def setup_list(n):
# return list(np.random.permutation(np.arange(n)))
return [0] * n
def setup_np_array(n):
# return np.random.permutation(np.arange(n))
return np.zeros(n, dtype=int)
def list_set(data):
return list(set(data))
def numpy_unique(data):
return np.unique(data)
def list_dict(data):
return list(dict.fromkeys(data))
b = perfplot.bench(
setup=[
setup_list,
setup_list,
setup_list,
setup_np_array,
setup_np_array,
setup_np_array,
],
kernels=[list_set, numpy_unique, list_dict, list_set, numpy_unique, list_dict],
labels=[
"list(set(lst))",
"np.unique(lst)",
"list(dict(lst))",
"list(set(arr))",
"np.unique(arr)",
"list(dict(arr))",
],
n_range=[2 ** k for k in range(23)],
xlabel="len(array)",
equality_check=None,
)
# plt.title("input array = [0, 1, 2,..., n]")
plt.title("input array = [0, 0,..., 0]")
b.save("out.png")
b.show()

You could also do this:
>>> t = [1, 2, 3, 3, 2, 4, 5, 6]
>>> s = [x for i, x in enumerate(t) if i == t.index(x)]
>>> s
[1, 2, 3, 4, 5, 6]
The reason that above works is that index method returns only the first index of an element. Duplicate elements have higher indices. Refer to here:
list.index(x[, start[, end]])
Return zero-based index in the list of
the first item whose value is x. Raises a ValueError if there is no
such item.

Best approach of removing duplicates from a list is using set() function, available in python, again converting that set into list
In [2]: some_list = ['a','a','v','v','v','c','c','d']
In [3]: list(set(some_list))
Out[3]: ['a', 'c', 'd', 'v']

You can use set to remove duplicates:
mylist = list(set(mylist))
But note the results will be unordered. If that's an issue:
mylist.sort()

Try using sets:
import sets
t = sets.Set(['a', 'b', 'c', 'd'])
t1 = sets.Set(['a', 'b', 'c'])
print t | t1
print t - t1

One more better approach could be,
import pandas as pd
myList = [1, 2, 3, 1, 2, 5, 6, 7, 8]
cleanList = pd.Series(myList).drop_duplicates().tolist()
print(cleanList)
#> [1, 2, 3, 5, 6, 7, 8]
and the order remains preserved.

This one cares about the order without too much hassle (OrderdDict & others). Probably not the most Pythonic way, nor shortest way, but does the trick:
def remove_duplicates(item_list):
''' Removes duplicate items from a list '''
singles_list = []
for element in item_list:
if element not in singles_list:
singles_list.append(element)
return singles_list

Reduce variant with ordering preserve:
Assume that we have list:
l = [5, 6, 6, 1, 1, 2, 2, 3, 4]
Reduce variant (unefficient):
>>> reduce(lambda r, v: v in r and r or r + [v], l, [])
[5, 6, 1, 2, 3, 4]
5 x faster but more sophisticated
>>> reduce(lambda r, v: v in r[1] and r or (r[0].append(v) or r[1].add(v)) or r, l, ([], set()))[0]
[5, 6, 1, 2, 3, 4]
Explanation:
default = (list(), set())
# user list to keep order
# use set to make lookup faster
def reducer(result, item):
if item not in result[1]:
result[0].append(item)
result[1].add(item)
return result
reduce(reducer, l, default)[0]

There are many other answers suggesting different ways to do this, but they're all batch operations, and some of them throw away the original order. That might be okay depending on what you need, but if you want to iterate over the values in the order of the first instance of each value, and you want to remove the duplicates on-the-fly versus all at once, you could use this generator:
def uniqify(iterable):
seen = set()
for item in iterable:
if item not in seen:
seen.add(item)
yield item
This returns a generator/iterator, so you can use it anywhere that you can use an iterator.
for unique_item in uniqify([1, 2, 3, 4, 3, 2, 4, 5, 6, 7, 6, 8, 8]):
print(unique_item, end=' ')
print()
Output:
1 2 3 4 5 6 7 8
If you do want a list, you can do this:
unique_list = list(uniqify([1, 2, 3, 4, 3, 2, 4, 5, 6, 7, 6, 8, 8]))
print(unique_list)
Output:
[1, 2, 3, 4, 5, 6, 7, 8]

You can use the following function:
def rem_dupes(dup_list):
yooneeks = []
for elem in dup_list:
if elem not in yooneeks:
yooneeks.append(elem)
return yooneeks
Example:
my_list = ['this','is','a','list','with','dupicates','in', 'the', 'list']
Usage:
rem_dupes(my_list)
['this', 'is', 'a', 'list', 'with', 'dupicates', 'in', 'the']

Using set :
a = [0,1,2,3,4,3,3,4]
a = list(set(a))
print a
Using unique :
import numpy as np
a = [0,1,2,3,4,3,3,4]
a = np.unique(a).tolist()
print a

Without using set
data=[1, 2, 3, 1, 2, 5, 6, 7, 8]
uni_data=[]
for dat in data:
if dat not in uni_data:
uni_data.append(dat)
print(uni_data)

The Magic of Python Built-in type
In python, it is very easy to process the complicated cases like this and only by python's built-in type.
Let me show you how to do !
Method 1: General Case
The way (1 line code) to remove duplicated element in list and still keep sorting order
line = [1, 2, 3, 1, 2, 5, 6, 7, 8]
new_line = sorted(set(line), key=line.index) # remove duplicated element
print(new_line)
You will get the result
[1, 2, 3, 5, 6, 7, 8]
Method 2: Special Case
TypeError: unhashable type: 'list'
The special case to process unhashable (3 line codes)
line=[['16.4966155686595', '-27.59776154691', '52.3786295521147']
,['16.4966155686595', '-27.59776154691', '52.3786295521147']
,['17.6508629295574', '-27.143305738671', '47.534955022564']
,['17.6508629295574', '-27.143305738671', '47.534955022564']
,['18.8051102904552', '-26.688849930432', '42.6912804930134']
,['18.8051102904552', '-26.688849930432', '42.6912804930134']
,['19.5504702331098', '-26.205884452727', '37.7709192714727']
,['19.5504702331098', '-26.205884452727', '37.7709192714727']
,['20.2929416861422', '-25.722717575124', '32.8500163147157']
,['20.2929416861422', '-25.722717575124', '32.8500163147157']]
tuple_line = [tuple(pt) for pt in line] # convert list of list into list of tuple
tuple_new_line = sorted(set(tuple_line),key=tuple_line.index) # remove duplicated element
new_line = [list(t) for t in tuple_new_line] # convert list of tuple into list of list
print (new_line)
You will get the result :
[
['16.4966155686595', '-27.59776154691', '52.3786295521147'],
['17.6508629295574', '-27.143305738671', '47.534955022564'],
['18.8051102904552', '-26.688849930432', '42.6912804930134'],
['19.5504702331098', '-26.205884452727', '37.7709192714727'],
['20.2929416861422', '-25.722717575124', '32.8500163147157']
]
Because tuple is hashable and you can convert data between list and tuple easily

below code is simple for removing duplicate in list
def remove_duplicates(x):
a = []
for i in x:
if i not in a:
a.append(i)
return a
print remove_duplicates([1,2,2,3,3,4])
it returns [1,2,3,4]

Here's the fastest pythonic solution comaring to others listed in replies.
Using implementation details of short-circuit evaluation allows to use list comprehension, which is fast enough. visited.add(item) always returns None as a result, which is evaluated as False, so the right-side of or would always be the result of such an expression.
Time it yourself
def deduplicate(sequence):
visited = set()
adder = visited.add # get rid of qualification overhead
out = [adder(item) or item for item in sequence if item not in visited]
return out

Dynamically compare list for common occurrences

I want to dynamically use sets to find common entires in lists. If I have a list of list that could contain any number of lists how can I go about find all the common occurrences in all the lists. I have thought about enumerating the list and storing each nested list in its own variable but then I am not sure how to compare all the individual lists.
Example of List:
l = [[1,2,3,4,], [3,6,4,2,1], [6,4,2,6,7,3]]
I want to do something like this but dynamic so it can accept any number of lists:
common = set(l[0]) & set(l[1]) & set(l[2])

Use reduce, with a lambda for
>>> l = [[1,2,3,4,], [3,6,4,2,1], [6,4,2,6,7,3]]
>>> from functools import reduce
>>> common = reduce(lambda l1,l2: set(l1) & set(l2), l)
>>> print(common)
{2, 3, 4}
Or, as a slightly modified version of #tobias_k's solution (as pointed in the comment), you can do it without lambda as
>>> common = reduce(set.intersection, [set(l[0])] + l[1:]))

You can use set.intersection:
set.intersection(*(set(ls) for ls in l)) #evaluates to {2, 3, 4}

You can use reduce from functools package:
from functools import reduce
l = [[1, 2, 3, 4], [3, 6, 4, 2, 1], [6, 4, 2, 6, 7, 3]]
print(reduce(set.intersection, map(set, l)))
Output:
{2, 3, 4}

How to add two lists together, avoid repetitions, and order elements?

I have two lists filled with integers. I wish to add them together such that:
the output list has no duplicate elements,
is in order, and
contains the union of both lists.
Is there any way to do so without creating my own custom function? If not, what would a neat and tidy procedure look like?
For instance:
list1 = [1, 10, 2]
list2 = [3, 4, 10]
Output:
outputlist = [1, 2, 3, 4, 10]

Try this:
combined = [list1, list2]
union = list(set().union(*combined))
This takes advantage of the predefined method (.union()) of set() , which is what you need here.
combined can have as many elements inside it, as the asterisk in *combined means that the union of all of the elements is found.
Also, I list()ed the result but you could leave it as a set().
As #glibdud states in the comments, it's possible that this might produce a sorted list, but it's not guaranteed, so use sorted() to ensure that it's ordered. (like this union = sorted(list(set().union(*combined))))

l1 = [1, 10, 2]
l2 = [3, 4, 10]
sorted(list(set(l1 + l2)))
>>> [1, 2, 3, 4, 10]

Append feature frequency to existing list

I am looking for an fairly efficient way to append the frequency of a feature in a list each item in that list.
For example, given this list:
[['syme', 4, 2], ['said', 4, 2], ['the', 3, 5]]
I would like to append to it the frequency with which the second two items occur in the list. In the list above, this would look something like this:
[['syme', 4, 2, 2], ['said', 4, 2, 2], ['the', 3, 5, 1]]
Where the third number represents how frequently the second two numbers occur as the second two items in the lists. (for example, [4, 2] appears twice as the second two numbers and [3,5] appears once so the first two lists would append a 2 at the end and the third list would append a 1.)
The actual list may have several hundred thousand items so both efficiency AND readable code are both valued here and I would like to maintain the current order of the list.
Thanks in advance!

Probably the most performant method is to use collections.Counter to get the counts based on pairs
counts = Counter(tuple(item[1:]) for item in lst)
then update the list accordingly:
for item in lst:
item.append(counts[tuple(item[1:])])
If the order of the two items doesn't matter, wrap item[1:] with sorted(...) when creating counts and updating lst.

You can use the collections.Counter class:
from collections import Counter
my_list = [['syme', 4, 2], ['said', 4, 2], ['the', 3, 5]]
counts = Counter([(x[1],x[2],) for x in my_list])
for sub_list in my_list:
sub_list.append(counts[(sub_list[1], sub_list[2])])

If order doesn't matter:
import collections
collections.Counter(frozenset((i[1], i[2]))
a_list = [['syme', 4, 2], ['said', 4, 2], ['the', 3, 5]]
counts = Counter(frozenset(l[1], l[2]) for x in a_list)
for l in a_list:
l.append(counts[frozenset(l[1], l[2])])