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Why do these list operations (methods: clear / extend / reverse / append / sort / remove) return None, rather than the resulting list?
(6 answers)
Closed 7 months ago.
I've been able to verify that the findUniqueWords does result in a sorted list. However, it does not return the list. Why?
def findUniqueWords(theList):
newList = []
words = []
# Read a line at a time
for item in theList:
# Remove any punctuation from the line
cleaned = cleanUp(item)
# Split the line into separate words
words = cleaned.split()
# Evaluate each word
for word in words:
# Count each unique word
if word not in newList:
newList.append(word)
answer = newList.sort()
return answer
list.sort sorts the list in place, i.e. it doesn't return a new list. Just write
newList.sort()
return newList
The problem is here:
answer = newList.sort()
sort does not return the sorted list; rather, it sorts the list in place.
Use:
answer = sorted(newList)
Here is an email from Guido van Rossum in Python's dev list explaining why he choose not to return self on operations that affects the object and don't return a new one.
This comes from a coding style (popular in various other languages, I
believe especially Lisp revels in it) where a series of side effects
on a single object can be chained like this:
x.compress().chop(y).sort(z)
which would be the same as
x.compress()
x.chop(y)
x.sort(z)
I find the chaining form a threat to readability; it requires that the
reader must be intimately familiar with each of the methods. The
second form makes it clear that each of these calls acts on the same
object, and so even if you don't know the class and its methods very
well, you can understand that the second and third call are applied to
x (and that all calls are made for their side-effects), and not to
something else.
I'd like to reserve chaining for operations that return new values,
like string processing operations:
y = x.rstrip("\n").split(":").lower()
Python has two kinds of sorts: a sort method (or "member function") and a sort function. The sort method operates on the contents of the object named -- think of it as an action that the object is taking to re-order itself. The sort function is an operation over the data represented by an object and returns a new object with the same contents in a sorted order.
Given a list of integers named l the list itself will be reordered if we call l.sort():
>>> l = [1, 5, 2341, 467, 213, 123]
>>> l.sort()
>>> l
[1, 5, 123, 213, 467, 2341]
This method has no return value. But what if we try to assign the result of l.sort()?
>>> l = [1, 5, 2341, 467, 213, 123]
>>> r = l.sort()
>>> print(r)
None
r now equals actually nothing. This is one of those weird, somewhat annoying details that a programmer is likely to forget about after a period of absence from Python (which is why I am writing this, so I don't forget again).
The function sorted(), on the other hand, will not do anything to the contents of l, but will return a new, sorted list with the same contents as l:
>>> l = [1, 5, 2341, 467, 213, 123]
>>> r = sorted(l)
>>> l
[1, 5, 2341, 467, 213, 123]
>>> r
[1, 5, 123, 213, 467, 2341]
Be aware that the returned value is not a deep copy, so be cautious about side-effecty operations over elements contained within the list as usual:
>>> spam = [8, 2, 4, 7]
>>> eggs = [3, 1, 4, 5]
>>> l = [spam, eggs]
>>> r = sorted(l)
>>> l
[[8, 2, 4, 7], [3, 1, 4, 5]]
>>> r
[[3, 1, 4, 5], [8, 2, 4, 7]]
>>> spam.sort()
>>> eggs.sort()
>>> l
[[2, 4, 7, 8], [1, 3, 4, 5]]
>>> r
[[1, 3, 4, 5], [2, 4, 7, 8]]
Python habitually returns None from functions and methods that mutate the data, such as list.sort, list.append, and random.shuffle, with the idea being that it hints to the fact that it was mutating.
If you want to take an iterable and return a new, sorted list of its items, use the sorted builtin function.
To understand why it does not return the list:
sort() doesn't return any value while the sort() method just sorts the elements of a given list in a specific order - ascending or descending without returning any value.
So problem is with answer = newList.sort() where answer is none.
Instead you can just do return newList.sort().
The syntax of the sort() method is:
list.sort(key=..., reverse=...)
Alternatively, you can also use Python's in-built function sorted() for the same purpose.
sorted(list, key=..., reverse=...)
Note: The simplest difference between sort() and sorted() is: sort() doesn't return any value while, sorted() returns an iterable list.
So in your case answer = sorted(newList).
A small piece of wisdom which I didn't see in other answers:
All methods for mutable objects in python (like lists) which modify the list return None. So, for lists this also includes list.append(), list.reverse(), etc. That's why the syntax should be
myList.sort()
Meanwhile, methods for any immutable object (like strings) must be assigned like so:
myString = myString.strip()
you can use sorted() method if you want it to return the sorted list.
It's more convenient.
l1 = []
n = int(input())
for i in range(n):
user = int(input())
l1.append(user)
sorted(l1,reverse=True)
list.sort() method modifies the list in-place and returns None.
if you still want to use sort you can do this.
l1 = []
n = int(input())
for i in range(n):
user = int(input())
l1.append(user)
l1.sort(reverse=True)
print(l1)
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
I have a list say l = [1,5,8,-3,6,8,-3,2,-4,6,8]. Im trying to split it into sublists of positive integers i.e. the above list would give me [[1,5,8],[6,8],[2],[6,8]]. I've tried the following:
l = [1,5,8,-3,6,8,-3,2,-4,6,8]
index = 0
def sublist(somelist):
a = []
for i in somelist:
if i > 0:
a.append(i)
else:
global index
index += somelist.index(i)
break
return a
print sublist(l)
With this I can get the 1st sublist ( [1,5,8] ) and the index number of the 1st negative integer at 3. Now if I run my function again and pass it l[index+1:], I cant get the next sublist and assume that index will be updated to show 6. However i cant, for the life of me cant figure out how to run the function in a loop or what condition to use so that I can keep running my function and giving it l[index+1:] where index is the updated, most recently encountered position of a negative integer. Any help will be greatly appreciated
You need to keep track of two levels of list here - the large list that holds the sublists, and the sublists themselves. Start a large list, start a sublist, and keep appending to the current sublist while i is non-negative (which includes positive numbers and 0, by the way). When i is negative, append the current sublist to the large list and start a new sublist. Also note that you should handle cases where the first element is negative or the last element isn't negative.
l = [1,5,8,-3,6,8,-3,2,-4,6,8]
def sublist(somelist):
result = []
a = []
for i in somelist:
if i > 0:
a.append(i)
else:
if a: # make sure a has something in it
result.append(a)
a = []
if a: # if a is still accumulating elements
result.append(a)
return result
The result:
>>> sublist(l)
[[1, 5, 8], [6, 8], [2], [6, 8]]
Since somelist never changes, rerunning index will always get index of the first instance of an element, not the one you just reached. I'd suggest looking at enumerate to get the index and element as you loop, so no calls to index are necessary.
That said, you could use the included batteries to solve this as a one-liner, using itertools.groupby:
from itertools import groupby
def sublist(somelist):
return [list(g) for k, g in groupby(somelist, key=(0).__le__) if k]
Still worth working through your code to understand it, but the above is going to be fast and fairly simple.
This code makes use of concepts found at this URL:
Python list comprehension- "pop" result from original list?
Applying an interesting concept found here to your problem, the following are some alternatives to what others have posted for this question so far. Both use list comprehensions and are commented to explain the purpose of the second option versus the first. Did this experiment for me as part of my learning curve, but hoping it may help you and others on this thread as well:
What's nice about these is that if your input list is very very large, you won't have to double your memory expenditure to get the job done. You build one up as you shrink the other down.
This code was tested on Python 2.7 and Python 3.6:
o1 = [1,5,8,-3,6,9,-4,2,-5,6,7,-7, 999, -43, -1, 888]
# modified version of poster's list
o1b = [1,5,8,-3,6,8,-3,2,-4,6,8] # poster's list
o2 = [x for x in (o1.pop() for i in range(len(o1))) \
if (lambda x: True if x < 0 else o1.insert(0, x))(x)]
o2b = [x for x in (o1b.pop() for i in range(len(o1b))) \
if (lambda x: True if x < 0 else o1b.insert(0, x))(x)]
print(o1)
print(o2)
print("")
print(o1b)
print(o2b)
It produces result sets like this (on iPython Jupyter Notebooks):
[1, 5, 8, 6, 9, 2, 6, 7, 999, 888]
[-1, -43, -7, -5, -4, -3]
[1, 5, 8, 6, 8, 2, 6, 8]
[-4, -3, -3]
Here is another version that also uses list comprehensions as the work horse, but functionalizes the code in way that is more read-able (I think) and easier to test with different numeric lists. Some will probably prefer the original code since it is shorter:
p1 = [1,5,8,-3,6,9,-4,2,-5,6,7,-7, 999, -43, -1, 888]
# modified version of poster's list
p1b = [1,5,8,-3,6,8,-3,2,-4,6,8] # poster's list
def lst_mut_byNeg_mod(x, pLst): # list mutation by neg nums module
# this function only make sense in context of usage in
# split_pos_negs_in_list()
if x < 0: return True
else:
pLst.insert(0,x)
return False
def split_pos_negs_in_list(pLst):
pLngth = len(pLst) # reduces nesting of ((()))
return [x for x in (pLst.pop() for i in range(pLngth)) \
if lst_mut_byNeg_mod(x, pLst)]
p2 = split_pos_negs_in_list(p1)
print(p1)
print(p2)
print("")
p2b = split_pos_negs_in_list(p1b)
print(p1b)
print(p2b)
Final Thoughts:
Link provided earlier had a number of ideas in the comment thread:
It recommends a Google search for the "python bloom filter library" - this sounds promising from a performance standpoint but I have not yet looked into it
There is a post on that thread with 554 up-voted, and yet it has at least 4 comments explaining what might be faulty with it. When exploring options, it may be advisable to scan the comment trail and not just review what gets the most votes. There are many options proposed for situations like this.
Just for fun you can use re too for a one liner.
l = [1,5,8,-3,6,8,-3,2,-4,6,8]
print map(lambda x: map(int,x.split(",")), re.findall(r"(?<=[,\[])\s*\d+(?:,\s*\d+)*(?=,\s*-\d+|\])", str(l)))
Output:[[1, 5, 8], [6, 8], [2], [6, 8]]