3.6. Lists¶
The designers of Python had many choices to make when they implemented the list data structure. Each of these choices could have an impact on how fast list operations perform. To help them make the right choices they looked at the ways that people would most commonly use the list data structure and they optimized their implementation of a list so that the most common operations were very fast. Of course they also tried to make the less common operations fast, but when a tradeoff had to be made the performance of a less common operation was often sacrificed in favor of the more common operation.
Two common operations are indexing and assigning to an index position. Both of these operations take the same amount of time no matter how large the list becomes. When an operation like this is independent of the size of the list they are \(O(1)\).
Another very common programming task is to grow a list. There are two ways to create a longer list. You can use the append method or the concatenation operator. The append method is \(O(1)\). However, the concatenation operator is \(O(k)\) where \(k\) is the size of the list that is being concatenated. This is important for you to know because it can help you make your own programs more efficient by choosing the right tool for the job.
Let’s look at four different ways we might generate a list of n
numbers starting with 0. First we’ll try a for
loop and create the
list by concatenation, then we’ll use append rather than concatenation.
Next, we’ll try creating the list using list comprehension and finally,
and perhaps the most obvious way, using the range function wrapped by a
call to the list constructor. Listing 3 shows the code for
making our list four different ways.
Listing 3
def test1():
l = []
for i in range(1000):
l = l + [i]
def test2():
l = []
for i in range(1000):
l.append(i)
def test3():
l = [i for i in range(1000)]
def test4():
l = list(range(1000))
To capture the time it takes for each of our functions to execute we
will use Python’s timeit
module. The timeit
module is designed
to allow Python developers to make cross-platform timing measurements by
running functions in a consistent environment and using timing
mechanisms that are as similar as possible across operating systems.
To use timeit
you create a Timer
object whose parameters are two
Python statements. The first parameter is a Python statement that you
want to time; the second parameter is a statement that will run once to
set up the test. The timeit
module will then time how long it takes
to execute the statement some number of times. By default timeit
will try to run the statement one million times. When its done it
returns the time as a floating point value representing the total number
of seconds. However, since it executes the statement a million times you
can read the result as the number of microseconds to execute the test
one time. You can also pass timeit
a named parameter called
number
that allows you to specify how many times the test statement
is executed. The following session shows how long it takes to run each
of our test functions 1000 times.
t1 = Timer("test1()", "from __main__ import test1")
print("concat ",t1.timeit(number=1000), "milliseconds")
t2 = Timer("test2()", "from __main__ import test2")
print("append ",t2.timeit(number=1000), "milliseconds")
t3 = Timer("test3()", "from __main__ import test3")
print("comprehension ",t3.timeit(number=1000), "milliseconds")
t4 = Timer("test4()", "from __main__ import test4")
print("list range ",t4.timeit(number=1000), "milliseconds")
concat 6.54352807999 milliseconds
append 0.306292057037 milliseconds
comprehension 0.147661924362 milliseconds
list range 0.0655000209808 milliseconds
In the experiment above the statement that we are timing is the function
call to test1()
, test2()
, and so on. The setup statement may
look very strange to you, so let’s consider it in more detail. You are
probably very familiar with the from
, import
statement, but this
is usually used at the beginning of a Python program file. In this case
the statement from __main__ import test1
imports the function
test1
from the __main__
namespace into the namespace that
timeit
sets up for the timing experiment. The timeit
module does
this because it wants to run the timing tests in an environment that is
uncluttered by any stray variables you may have created, that may
interfere with your function’s performance in some unforeseen way.
From the experiment above it is clear that the append operation at 0.30
milliseconds is much faster than concatenation at 6.54 milliseconds. In
the above experiment we also show the times for two additional methods
for creating a list; using the list constructor with a call to range
and a list comprehension. It is interesting to note that the list
comprehension is twice as fast as a for
loop with an append
operation.
One final observation about this little experiment is that all of the times that you see above include some overhead for actually calling the test function, but we can assume that the function call overhead is identical in all four cases so we still get a meaningful comparison of the operations. So it would not be accurate to say that the concatenation operation takes 6.54 milliseconds but rather the concatenation test function takes 6.54 milliseconds. As an exercise you could test the time it takes to call an empty function and subtract that from the numbers above.
Now that we have seen how performance can be measured concretely you can
look at Table 2 to see the Big-O efficiency of all the
basic list operations. After thinking carefully about
Table 2, you may be wondering about the two different times
for pop
. When pop
is called on the end of the list it takes
\(O(1)\) but when pop is called on the first element in the list
or anywhere in the middle it is \(O(n)\). The reason for this lies
in how Python chooses to implement lists. When an item is taken from the
front of the list, in Python’s implementation, all the other elements in
the list are shifted one position closer to the beginning. This may seem
silly to you now, but if you look at Table 2 you will see
that this implementation also allows the index operation to be
\(O(1)\). This is a tradeoff that the Python implementors thought
was a good one.
Operation |
Big-O Efficiency |
---|---|
index [] |
O(1) |
index assignment |
O(1) |
append |
O(1) |
pop() |
O(1) |
pop(i) |
O(n) |
insert(i,item) |
O(n) |
del operator |
O(n) |
iteration |
O(n) |
contains (in) |
O(n) |
get slice [x:y] |
O(k) |
del slice |
O(n) |
set slice |
O(n+k) |
reverse |
O(n) |
concatenate |
O(k) |
sort |
O(n log n) |
multiply |
O(nk) |
As a way of demonstrating this difference in performance let’s do
another experiment using the timeit
module. Our goal is to be able
to verify the performance of the pop
operation on a list of a known
size when the program pops from the end of the list, and again when the
program pops from the beginning of the list. We will also want to
measure this time for lists of different sizes. What we would expect to
see is that the time required to pop from the end of the list will stay
constant even as the list grows in size, while the time to pop from the
beginning of the list will continue to increase as the list grows.
Listing 4 shows one attempt to measure the difference between the two uses of pop. As you can see from this first example, popping from the end takes 0.0003 milliseconds, whereas popping from the beginning takes 4.82 milliseconds. For a list of two million elements this is a factor of 16,000.
There are a couple of things to notice about Listing 4. The
first is the statement from __main__ import x
. Although we did not
define a function we do want to be able to use the list object x in our
test. This approach allows us to time just the single pop
statement
and get the most accurate measure of the time for that single operation.
Because the timer repeats 1000 times it is also important to point out
that the list is decreasing in size by 1 each time through the loop. But
since the initial list is two million elements in size we only reduce
the overall size by \(0.05\%\)
Listing 4
popzero = timeit.Timer("x.pop(0)",
"from __main__ import x")
popend = timeit.Timer("x.pop()",
"from __main__ import x")
x = list(range(2000000))
popzero.timeit(number=1000)
4.8213560581207275
x = list(range(2000000))
popend.timeit(number=1000)
0.0003161430358886719
While our first test does show that pop(0)
is indeed slower than
pop()
, it does not validate the claim that pop(0)
is
\(O(n)\) while pop()
is \(O(1)\). To validate that claim
we need to look at the performance of both calls over a range of list
sizes. Listing 5 implements this test.
Listing 5
popzero = Timer("x.pop(0)",
"from __main__ import x")
popend = Timer("x.pop()",
"from __main__ import x")
print("pop(0) pop()")
for i in range(1000000,100000001,1000000):
x = list(range(i))
pt = popend.timeit(number=1000)
x = list(range(i))
pz = popzero.timeit(number=1000)
print("%15.5f, %15.5f" %(pz,pt))
Figure 3 shows the results of our experiment. You can see
that as the list gets longer and longer the time it takes to pop(0)
also increases while the time for pop
stays very flat. This is
exactly what we would expect to see for a \(O(n)\) and
\(O(1)\) algorithm.
Some sources of error in our little experiment include the fact that there are other processes running on the computer as we measure that may slow down our code, so even though we try to minimize other things happening on the computer there is bound to be some variation in time. That is why the loop runs the test one thousand times in the first place to statistically gather enough information to make the measurement reliable.