A is big: len(a)=10000000
Will python interpreter optimize the op like a[:10]=[1,2,3] to O(1) time ?
Is there any difference between a[:10]=[1,2,3] and a[:3]=[1,2,3]? I mean difference between whether length changes.
There's very much a difference between the two statements:
a[:10] = [1,2,3]
a[:3] = [1,2,3]
The first involves actual deletion of some elements in the list, whereas the second can just change the elements that are already there. You can verify this by executing:
print(len(a))
before and after the operation.
There's a useful web page that shows the various operations on standard Python data structures along with their time complexities. Deleting from a list (which is really an array under the covers) is O(n) as it involves moving all elements beyond the deletion area to fill in the gap that would otherwise be left.
And, in fact, if you look at the list_ass_slice code responsible for list slice assignment, you'll see it has a number of memcpy and memmove operations for modifying the list, for example:
if (d < 0) { /* Delete -d items */
Py_ssize_t tail;
tail = (Py_SIZE(a) - ihigh) * sizeof(PyObject *);
memmove(&item[ihigh+d], &item[ihigh], tail);
if (list_resize(a, Py_SIZE(a) + d) < 0) {
memmove(&item[ihigh], &item[ihigh+d], tail);
memcpy(&item[ilow], recycle, s);
goto Error;
}
item = a->ob_item;
}
Basically, the code first works out the size difference between the slice being copied and the slice it's replacing: d = n - norig. Before copying the individual elements, it inserts some new elements if d > 0, deletes some if d < 0, and does neither if d == 0.
I am wondering what the big-O runtime complexity is for comparing two collections.Counter objects. Here is some code to demonstrate what I mean:
import collections
counter_1 = collections.Counter("abcabcabcabcabcabcdefg")
counter_2 = collections.Counter("xyzxyzxyzabc")
comp = counter_1 == counter_2 # What is the runtime of this comparison statement?
Is the runtime of the equality comparison in the final statement O(1)? Or is it O(num_of_unique_keys_in_largest_counter)? Or is it something else?
For reference, here is the source code for collections.Counter https://github.com/python/cpython/blob/0250de48199552cdaed5a4fe44b3f9cdb5325363/Lib/collections/init.py#L497
I do not see the class implementing an __eq()__ method.
Bonus points: If the answer to this question changes between python2 and python3, I would love to hear the difference?
Counter is a subclass of dict, therefore the big O analysis is the one of dict, with the caveat that Counter objects are specialized to only hold int values (i/e they can not hold collections of values as dicts can); this simplifies the analysis.
Looking at the c code implementation of the equality comparison:
There is an early exit if the number of keys is different. (this does not influence big-O).
Then a loop that iterates over all the keys that exits early if the key is not found, or if the corresponding value is different. (again, this has no bearing on big-O).
if all keys are found, and the corresponding values are all equal, then the dictionaries are declared equal. The lookup and comparisons of each key-value pair is O(1); this operation is repeated at most n times (n being the number of keys)
In all, the time complexity is O(n), with n the number of keys.
This applies to both python 2 and 3.
from dictobject.c
/* Return 1 if dicts equal, 0 if not, -1 if error.
* Gets out as soon as any difference is detected.
* Uses only Py_EQ comparison.
*/
static int
dict_equal(PyDictObject *a, PyDictObject *b)
{
Py_ssize_t i;
if (a->ma_used != b->ma_used)
/* can't be equal if # of entries differ */
return 0;
/* Same # of entries -- check all of 'em. Exit early on any diff. */
for (i = 0; i < a->ma_keys->dk_nentries; i++) {
PyDictKeyEntry *ep = &DK_ENTRIES(a->ma_keys)[i];
PyObject *aval;
if (a->ma_values)
aval = a->ma_values[i];
else
aval = ep->me_value;
if (aval != NULL) {
int cmp;
PyObject *bval;
PyObject *key = ep->me_key;
/* temporarily bump aval's refcount to ensure it stays
alive until we're done with it */
Py_INCREF(aval);
/* ditto for key */
Py_INCREF(key);
/* reuse the known hash value */
b->ma_keys->dk_lookup(b, key, ep->me_hash, &bval);
if (bval == NULL) {
Py_DECREF(key);
Py_DECREF(aval);
if (PyErr_Occurred())
return -1;
return 0;
}
cmp = PyObject_RichCompareBool(aval, bval, Py_EQ);
Py_DECREF(key);
Py_DECREF(aval);
if (cmp <= 0) /* error or not equal */
return cmp;
}
}
return 1;
}
Internally, collections.Counter stores the count as a dictionary (that's why it subclasses dict) so the same rules apply as with comparing dictionaries - namely, it compares each key with each value from both dictionaries to ensure equality. For CPython, that is implemented in dict_equal(), other implementations may vary but, logically, you have to do the each-with-each comparison to ensure equality.
This also means that the complexity is O(N) at its worst (loops through one of the dictionaries, looks if the value is the same in the other). There are no significant changes between Python 2.x and Python 3.x in this regard.
(10 points) Write an O(nlogn) algorithm to find the majority element of a list of items. (Assume that the number of elements is a power of 2). Again, the only operation you can use on items of the list is equality comparison. Hint: solve a problem of size n by solving two sub-problems of size n/2
This was a test question on divide and conquer, for my algorithms class.
Here is the code I wrote in python 3.5.
def majElement(L):
tally = 0
if len(L) == 1:
return 1
for i in range(len(L)):
tally = majElement(L[i:len(L)]) + majElement(L[len(L)/2:])
if tally > (len(L)/2):
print(L[i])
This code results in a stack overflow. Some how I'm not reaching my base case.
How can I stop the infinite recursive calls?
Not sure about the divide and conquer approach, but majority element in an array/list can be identified in O(nlogn) time.
This can be done using a binary search tree using structure
struct tree
{
int element;
int count;
}BST;
Algorithm:
Insert elements in BST one by one and if already present then increment the count of the node.
At any stage, if count of a node becomes more than n/2 then return.
Now, the worst case complexity can be O(n^2) in case of skewed BST. So, use a self balancing BST to ensure O(nlogn) time.
If you want to do it using Divide and Conquer approach,
Algorithm :
Divide array into two parts L and R.
int m1 = Majority(L); int m2 = Majority(R);
if m1 is majority return it.
if m2 is majority retun it.
Otherwise return "no majority element".
Code :
i=0;j=arr.length;
int majority(int *A, int i, int j)
{
int m1= majority(A, i, j/2-1);
int m2= majority(A, j/2+1, j);
int count = 0;
for(int i=0; i<j; i++)
if(A[i] == m1)
count++;
if(count > j/2)
return m1;
count = 0;
for(int i=0; i<j; i++)
if(A[i] == m2)
count++;
if(count > j/2)
return m2;
}
return -1;
}
Your first problem is that you are only returning a value in the base case; the rest of the time, you print a value and return None.
Second, the first recursive call is made when i == 0, which means L[i:len(L)] is the same as L, so you aren't acutually reducing the size of the problem. At the very least, you want for i in range(1, len(L)), but I suspect you aren't properly decomposing the problem into subproblems. (You shouldn't have to make as many recursive calls as you are proposing.)
Not sure how your algorithm works,
mine does the implementation of finding majority element using divide and conquer technique. It divides the array into two slices and checks for potential candidate. It is then checked that whether it is majority element or not.
def divideAndConquer(array): #Function to solve by divide and conquer
if(len(array)==1):
return array[0]
middle = len(array)//2
left = array[:middle] #First half
right = array[middle:] #Second half
cLeft = divideAndConquer(left) #candidate for left side
cRight = divideAndConquer(right) #candidate for right side
if cLeft == cRight: #if both have the same candidate, it is the one
return cLeft
if array.count(cLeft)>middle: #if not, check in left
return cLeft
if array.count(cRight)>middle: #else check in right
return cRight
return "No majority element" #no majority element
It is my understanding that the range() function, which is actually an object type in Python 3, generates its contents on the fly, similar to a generator.
This being the case, I would have expected the following line to take an inordinate amount of time because, in order to determine whether 1 quadrillion is in the range, a quadrillion values would have to be generated:
1_000_000_000_000_000 in range(1_000_000_000_000_001)
Furthermore: it seems that no matter how many zeroes I add on, the calculation more or less takes the same amount of time (basically instantaneous).
I have also tried things like this, but the calculation is still almost instant:
# count by tens
1_000_000_000_000_000_000_000 in range(0,1_000_000_000_000_000_000_001,10)
If I try to implement my own range function, the result is not so nice!
def my_crappy_range(N):
i = 0
while i < N:
yield i
i += 1
return
What is the range() object doing under the hood that makes it so fast?
Martijn Pieters's answer was chosen for its completeness, but also see abarnert's first answer for a good discussion of what it means for range to be a full-fledged sequence in Python 3, and some information/warning regarding potential inconsistency for __contains__ function optimization across Python implementations. abarnert's other answer goes into some more detail and provides links for those interested in the history behind the optimization in Python 3 (and lack of optimization of xrange in Python 2). Answers by poke and by wim provide the relevant C source code and explanations for those who are interested.
The Python 3 range() object doesn't produce numbers immediately; it is a smart sequence object that produces numbers on demand. All it contains is your start, stop and step values, then as you iterate over the object the next integer is calculated each iteration.
The object also implements the object.__contains__ hook, and calculates if your number is part of its range. Calculating is a (near) constant time operation *. There is never a need to scan through all possible integers in the range.
From the range() object documentation:
The advantage of the range type over a regular list or tuple is that a range object will always take the same (small) amount of memory, no matter the size of the range it represents (as it only stores the start, stop and step values, calculating individual items and subranges as needed).
So at a minimum, your range() object would do:
class my_range:
def __init__(self, start, stop=None, step=1, /):
if stop is None:
start, stop = 0, start
self.start, self.stop, self.step = start, stop, step
if step < 0:
lo, hi, step = stop, start, -step
else:
lo, hi = start, stop
self.length = 0 if lo > hi else ((hi - lo - 1) // step) + 1
def __iter__(self):
current = self.start
if self.step < 0:
while current > self.stop:
yield current
current += self.step
else:
while current < self.stop:
yield current
current += self.step
def __len__(self):
return self.length
def __getitem__(self, i):
if i < 0:
i += self.length
if 0 <= i < self.length:
return self.start + i * self.step
raise IndexError('my_range object index out of range')
def __contains__(self, num):
if self.step < 0:
if not (self.stop < num <= self.start):
return False
else:
if not (self.start <= num < self.stop):
return False
return (num - self.start) % self.step == 0
This is still missing several things that a real range() supports (such as the .index() or .count() methods, hashing, equality testing, or slicing), but should give you an idea.
I also simplified the __contains__ implementation to only focus on integer tests; if you give a real range() object a non-integer value (including subclasses of int), a slow scan is initiated to see if there is a match, just as if you use a containment test against a list of all the contained values. This was done to continue to support other numeric types that just happen to support equality testing with integers but are not expected to support integer arithmetic as well. See the original Python issue that implemented the containment test.
* Near constant time because Python integers are unbounded and so math operations also grow in time as N grows, making this a O(log N) operation. Since it’s all executed in optimised C code and Python stores integer values in 30-bit chunks, you’d run out of memory before you saw any performance impact due to the size of the integers involved here.
The fundamental misunderstanding here is in thinking that range is a generator. It's not. In fact, it's not any kind of iterator.
You can tell this pretty easily:
>>> a = range(5)
>>> print(list(a))
[0, 1, 2, 3, 4]
>>> print(list(a))
[0, 1, 2, 3, 4]
If it were a generator, iterating it once would exhaust it:
>>> b = my_crappy_range(5)
>>> print(list(b))
[0, 1, 2, 3, 4]
>>> print(list(b))
[]
What range actually is, is a sequence, just like a list. You can even test this:
>>> import collections.abc
>>> isinstance(a, collections.abc.Sequence)
True
This means it has to follow all the rules of being a sequence:
>>> a[3] # indexable
3
>>> len(a) # sized
5
>>> 3 in a # membership
True
>>> reversed(a) # reversible
<range_iterator at 0x101cd2360>
>>> a.index(3) # implements 'index'
3
>>> a.count(3) # implements 'count'
1
The difference between a range and a list is that a range is a lazy or dynamic sequence; it doesn't remember all of its values, it just remembers its start, stop, and step, and creates the values on demand on __getitem__.
(As a side note, if you print(iter(a)), you'll notice that range uses the same listiterator type as list. How does that work? A listiterator doesn't use anything special about list except for the fact that it provides a C implementation of __getitem__, so it works fine for range too.)
Now, there's nothing that says that Sequence.__contains__ has to be constant time—in fact, for obvious examples of sequences like list, it isn't. But there's nothing that says it can't be. And it's easier to implement range.__contains__ to just check it mathematically ((val - start) % step, but with some extra complexity to deal with negative steps) than to actually generate and test all the values, so why shouldn't it do it the better way?
But there doesn't seem to be anything in the language that guarantees this will happen. As Ashwini Chaudhari points out, if you give it a non-integral value, instead of converting to integer and doing the mathematical test, it will fall back to iterating all the values and comparing them one by one. And just because CPython 3.2+ and PyPy 3.x versions happen to contain this optimization, and it's an obvious good idea and easy to do, there's no reason that IronPython or NewKickAssPython 3.x couldn't leave it out. (And in fact, CPython 3.0-3.1 didn't include it.)
If range actually were a generator, like my_crappy_range, then it wouldn't make sense to test __contains__ this way, or at least the way it makes sense wouldn't be obvious. If you'd already iterated the first 3 values, is 1 still in the generator? Should testing for 1 cause it to iterate and consume all the values up to 1 (or up to the first value >= 1)?
Use the source, Luke!
In CPython, range(...).__contains__ (a method wrapper) will eventually delegate to a simple calculation which checks if the value can possibly be in the range. The reason for the speed here is we're using mathematical reasoning about the bounds, rather than a direct iteration of the range object. To explain the logic used:
Check that the number is between start and stop, and
Check that the stride value doesn't "step over" our number.
For example, 994 is in range(4, 1000, 2) because:
4 <= 994 < 1000, and
(994 - 4) % 2 == 0.
The full C code is included below, which is a bit more verbose because of memory management and reference counting details, but the basic idea is there:
static int
range_contains_long(rangeobject *r, PyObject *ob)
{
int cmp1, cmp2, cmp3;
PyObject *tmp1 = NULL;
PyObject *tmp2 = NULL;
PyObject *zero = NULL;
int result = -1;
zero = PyLong_FromLong(0);
if (zero == NULL) /* MemoryError in int(0) */
goto end;
/* Check if the value can possibly be in the range. */
cmp1 = PyObject_RichCompareBool(r->step, zero, Py_GT);
if (cmp1 == -1)
goto end;
if (cmp1 == 1) { /* positive steps: start <= ob < stop */
cmp2 = PyObject_RichCompareBool(r->start, ob, Py_LE);
cmp3 = PyObject_RichCompareBool(ob, r->stop, Py_LT);
}
else { /* negative steps: stop < ob <= start */
cmp2 = PyObject_RichCompareBool(ob, r->start, Py_LE);
cmp3 = PyObject_RichCompareBool(r->stop, ob, Py_LT);
}
if (cmp2 == -1 || cmp3 == -1) /* TypeError */
goto end;
if (cmp2 == 0 || cmp3 == 0) { /* ob outside of range */
result = 0;
goto end;
}
/* Check that the stride does not invalidate ob's membership. */
tmp1 = PyNumber_Subtract(ob, r->start);
if (tmp1 == NULL)
goto end;
tmp2 = PyNumber_Remainder(tmp1, r->step);
if (tmp2 == NULL)
goto end;
/* result = ((int(ob) - start) % step) == 0 */
result = PyObject_RichCompareBool(tmp2, zero, Py_EQ);
end:
Py_XDECREF(tmp1);
Py_XDECREF(tmp2);
Py_XDECREF(zero);
return result;
}
static int
range_contains(rangeobject *r, PyObject *ob)
{
if (PyLong_CheckExact(ob) || PyBool_Check(ob))
return range_contains_long(r, ob);
return (int)_PySequence_IterSearch((PyObject*)r, ob,
PY_ITERSEARCH_CONTAINS);
}
The "meat" of the idea is mentioned in the comment lines:
/* positive steps: start <= ob < stop */
/* negative steps: stop < ob <= start */
/* result = ((int(ob) - start) % step) == 0 */
As a final note - look at the range_contains function at the bottom of the code snippet. If the exact type check fails then we don't use the clever algorithm described, instead falling back to a dumb iteration search of the range using _PySequence_IterSearch! You can check this behaviour in the interpreter (I'm using v3.5.0 here):
>>> x, r = 1000000000000000, range(1000000000000001)
>>> class MyInt(int):
... pass
...
>>> x_ = MyInt(x)
>>> x in r # calculates immediately :)
True
>>> x_ in r # iterates for ages.. :(
^\Quit (core dumped)
To add to Martijn’s answer, this is the relevant part of the source (in C, as the range object is written in native code):
static int
range_contains(rangeobject *r, PyObject *ob)
{
if (PyLong_CheckExact(ob) || PyBool_Check(ob))
return range_contains_long(r, ob);
return (int)_PySequence_IterSearch((PyObject*)r, ob,
PY_ITERSEARCH_CONTAINS);
}
So for PyLong objects (which is int in Python 3), it will use the range_contains_long function to determine the result. And that function essentially checks if ob is in the specified range (although it looks a bit more complex in C).
If it’s not an int object, it falls back to iterating until it finds the value (or not).
The whole logic could be translated to pseudo-Python like this:
def range_contains (rangeObj, obj):
if isinstance(obj, int):
return range_contains_long(rangeObj, obj)
# default logic by iterating
return any(obj == x for x in rangeObj)
def range_contains_long (r, num):
if r.step > 0:
# positive step: r.start <= num < r.stop
cmp2 = r.start <= num
cmp3 = num < r.stop
else:
# negative step: r.start >= num > r.stop
cmp2 = num <= r.start
cmp3 = r.stop < num
# outside of the range boundaries
if not cmp2 or not cmp3:
return False
# num must be on a valid step inside the boundaries
return (num - r.start) % r.step == 0
If you're wondering why this optimization was added to range.__contains__, and why it wasn't added to xrange.__contains__ in 2.7:
First, as Ashwini Chaudhary discovered, issue 1766304 was opened explicitly to optimize [x]range.__contains__. A patch for this was accepted and checked in for 3.2, but not backported to 2.7 because "xrange has behaved like this for such a long time that I don't see what it buys us to commit the patch this late." (2.7 was nearly out at that point.)
Meanwhile:
Originally, xrange was a not-quite-sequence object. As the 3.1 docs say:
Range objects have very little behavior: they only support indexing, iteration, and the len function.
This wasn't quite true; an xrange object actually supported a few other things that come automatically with indexing and len,* including __contains__ (via linear search). But nobody thought it was worth making them full sequences at the time.
Then, as part of implementing the Abstract Base Classes PEP, it was important to figure out which builtin types should be marked as implementing which ABCs, and xrange/range claimed to implement collections.Sequence, even though it still only handled the same "very little behavior". Nobody noticed that problem until issue 9213. The patch for that issue not only added index and count to 3.2's range, it also re-worked the optimized __contains__ (which shares the same math with index, and is directly used by count).** This change went in for 3.2 as well, and was not backported to 2.x, because "it's a bugfix that adds new methods". (At this point, 2.7 was already past rc status.)
So, there were two chances to get this optimization backported to 2.7, but they were both rejected.
* In fact, you even get iteration for free with indexing alone, but in 2.3 xrange objects got a custom iterator.
** The first version actually reimplemented it, and got the details wrong—e.g., it would give you MyIntSubclass(2) in range(5) == False. But Daniel Stutzbach's updated version of the patch restored most of the previous code, including the fallback to the generic, slow _PySequence_IterSearch that pre-3.2 range.__contains__ was implicitly using when the optimization doesn't apply.
The other answers explained it well already, but I'd like to offer another experiment illustrating the nature of range objects:
>>> r = range(5)
>>> for i in r:
print(i, 2 in r, list(r))
0 True [0, 1, 2, 3, 4]
1 True [0, 1, 2, 3, 4]
2 True [0, 1, 2, 3, 4]
3 True [0, 1, 2, 3, 4]
4 True [0, 1, 2, 3, 4]
As you can see, a range object is an object that remembers its range and can be used many times (even while iterating over it), not just a one-time generator.
It's all about a lazy approach to the evaluation and some extra optimization of range.
Values in ranges don't need to be computed until real use, or even further due to extra optimization.
By the way, your integer is not such big, consider sys.maxsize
sys.maxsize in range(sys.maxsize) is pretty fast
due to optimization - it's easy to compare given integer just with min and max of range.
but:
Decimal(sys.maxsize) in range(sys.maxsize) is pretty slow.
(in this case, there is no optimization in range, so if python receives unexpected Decimal, python will compare all numbers)
You should be aware of an implementation detail but should not be relied upon, because this may change in the future.
TL;DR
The object returned by range() is actually a range object. This object implements the iterator interface so you can iterate over its values sequentially, just like a generator, list, or tuple.
But it also implements the __contains__ interface which is actually what gets called when an object appears on the right-hand side of the in operator. The __contains__() method returns a bool of whether or not the item on the left-hand side of the in is in the object. Since range objects know their bounds and stride, this is very easy to implement in O(1).
Due to optimization, it is very easy to compare given integers just with min and max range.
The reason that the range() function is so fast in Python3 is that here we use mathematical reasoning for the bounds, rather than a direct iteration of the range object.
So for explaining the logic here:
Check whether the number is between the start and stop.
Check whether the step precision value doesn't go over our number.
Take an example, 997 is in range(4, 1000, 3) because:
4 <= 997 < 1000, and (997 - 4) % 3 == 0.
Try x-1 in (i for i in range(x)) for large x values, which uses a generator comprehension to avoid invoking the range.__contains__ optimisation.
TLDR;
the range is an arithmetic series so it can very easily calculate whether the object is there. It could even get the index of it if it were list like really quickly.
__contains__ method compares directly with the start and end of the range