Part IV, Functions

See Test Your Knowledge: Part IV Exercises in Chapter 20 for the exercises.

 

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  1. The basics. There’s not much to this one, but notice that using print (and hence your function) is technically a polymorphic operation, which does the right thing for each type of object:

    % python
    >>> def func(x): print(x)
    ...
    >>> func("spam")
    spam
    >>> func(42)
    42
    >>> func([1, 2, 3])
    [1, 2, 3]
    >>> func({'food': 'spam'})
    {'food': 'spam'}

  2. Arguments. Here’s a sample solution. Remember that you have to use print to see results in the test calls because a file isn’t the same as code typed interactively; Python doesn’t normally echo the results of expression statements in files:

    def adder(x, y):
        return x + y

    print(adder(2, 3))
    print(adder('spam', 'eggs'))
    print(adder(['a', 'b'], ['c', 'd']))

    % python mod.py
    5
    spameggs
    ['a', 'b', 'c', 'd']

  3. varargs. Two alternative adder functions are shown in the following file, adders.py. The hard part here is figuring out how to initialize an accumulator to an empty value of whatever type is passed in. The first solution uses manual type testing to look for an integer, and an empty slice of the first argument (assumed to be a sequence) if the argument is determined not to be an integer. The second solution uses the first argument to initialize and scan items 2 and beyond, much like one of the min function variants shown in Chapter 18.
    The second solution is better. Both of these assume all arguments are of the same type, and neither works on dictionaries (as we saw in Part II, + doesn’t work on mixed types or dictionaries). You could add a type test and special code to allow dictionaries, too, but that’s extra credit.

    def adder1(*args):
        print('adder1', end=' ')
        if type(args[0]) == type(0):              # Integer?
             sum = 0                              # Init to zero
        else:                                     # else sequence:
             sum = args[0][:0]                    # Use empty slice of arg1
        for arg in args:
            sum = sum + arg
        return sum

    def adder2(*args):
        print('adder2', end=' ')
        sum = args[0]                             # Init to arg1
        for next in args[1:]:
            sum += next                           # Add items 2..N
        return sum

    for func in (adder1, adder2):
        print(func(2, 3, 4))
        print(func('spam', 'eggs', 'toast'))
        print(func(['a', 'b'], ['c', 'd'], ['e', 'f']))

    % python adders.py
    adder1 9
    adder1 spameggstoast
    adder1 ['a', 'b', 'c', 'd', 'e', 'f']
    adder2 9
    adder2 spameggstoast
    adder2 ['a', 'b', 'c', 'd', 'e', 'f']

  4. Keywords. Here is my solution to the first and second parts of this exercise (coded in the file mod.py). To iterate over keyword arguments, use the **args form in the function header and use a loop (e.g., for x in args.keys(): use args[x]), or use args.values() to make this the same as summing *args positionals:

    def adder(good=1, bad=2, ugly=3):
        return good + bad + ugly

    print(adder())
    print(adder(5))
    print(adder(5, 6))
    print(adder(5, 6, 7))
    print(adder(ugly=7, good=6, bad=5))

    % python mod.py
    6
    10
    14
    18
    18


    # Second part solutions

    def adder1(*args):                  # Sum any number of positional args
        tot = args[0]
        for arg in args[1:]:
            tot += arg
        return tot

    def adder2(**args):                 # Sum any number of keyword args
        argskeys = list(args.keys())    # list needed in 3.0!
        tot = args[argskeys[0]]
        for key in argskeys[1:]:
            tot += args[key]
        return tot

    def adder3(**args):                 # Same, but convert to list of values
        args = list(args.values())      # list needed to index in 3.0!
        tot = args[0]
        for arg in args[1:]:
            tot += arg
        return tot

    def adder4(**args):                 # Same, but reuse positional version
        return adder1(*args.values())

    print(adder1(1, 2, 3),       adder1('aa', 'bb', 'cc'))
    print(adder2(a=1, b=2, c=3), adder2(a='aa', b='bb', c='cc'))
    print(adder3(a=1, b=2, c=3), adder3(a='aa', b='bb', c='cc'))
    print(adder4(a=1, b=2, c=3), adder4(a='aa', b='bb', c='cc'))

  5. (and 6.) Here are my solutions to exercises 5 and 6 (file dicts.py). These are just coding exercises, though, because Python 1.5 added the dictionary methods D.copy() and D1.update(D2) to handle things like copying and adding (merging) dictionaries. (See Python’s library manual or O’Reilly’s Python Pocket Reference for more details.) X[:] doesn’t work for dictionaries, as they’re not sequences (see Chapter 8 for details). Also, remember that if you assign (e = d) rather than copying, you generate a reference to a shared dictionary object; changing d changes e, too:

    def copyDict(old):
        new = {}
        for key in old.keys():
            new[key] = old[key]
        return new

    def addDict(d1, d2):
        new = {}
        for key in d1.keys():
            new[key] = d1[key]
        for key in d2.keys():
            new[key] = d2[key]
        return new

    % python
    >>> from dicts import *
    >>> d = {1: 1, 2: 2}
    >>> e = copyDict(d)
    >>> d[2] = '?'
    >>> d
    {1: 1, 2: '?'}
    >>> e
    {1: 1, 2: 2}

    >>> x = {1: 1}
    >>> y = {2: 2}
    >>> z = addDict(x, y)
    >>> z
    {1: 1, 2: 2}

  6. See #5.
  7. More argument-matching examples. Here is the sort of interaction you should get, along with comments that explain the matching that goes on:

    def f1(a, b): print(a, b)            # Normal args

    def f2(a, *b): print(a, b)           # Positional varargs

    def f3(a, **b): print(a, b)          # Keyword varargs

    def f4(a, *b, **c): print(a, b, c)   # Mixed modes

    def f5(a, b=2, c=3): print(a, b, c)  # Defaults

    def f6(a, b=2, *c): print(a, b, c)   # Defaults and positional varargs


    % python
    >>> f1(1, 2)                         # Matched by position (order matters)
    1 2
    >>> f1(b=2, a=1)                     # Matched by name (order doesn't matter)
    1 2

    >>> f2(1, 2, 3)                      # Extra positionals collected in a tuple
    1 (2, 3)

    >>> f3(1, x=2, y=3)                  # Extra keywords collected in a dictionary
    1 {'x': 2, 'y': 3}

    >>> f4(1, 2, 3, x=2, y=3)            # Extra of both kinds
    1 (2, 3) {'x': 2, 'y': 3}

    >>> f5(1)                            # Both defaults kick in
    1 2 3
    >>> f5(1, 4)                         # Only one default used
    1 4 3

    >>> f6(1)                            # One argument: matches "a"
    1 2 ()
    >>> f6(1, 3, 4)                      # Extra positional collected
    1 3 (4,)

  8. Primes revisited. Here is the primes example, wrapped up in a function and a module (file primes.py) so it can be run multiple times. I added an if test to trap negatives, 0, and 1. I also changed / to // in this edition to make this solution immune to the Python 3.0 / true division changes we studied in Chapter 5, and to enable it to support floating-point numbers (uncomment the from statement and change // to / to see the differences in 2.6):

    #from __future__ import division

    def prime(y):
        if y <= 1:                                       # For some y > 1
            print(y, 'not prime')
        else:
            x = y // 2                                   # 3.0 / fails
            while x > 1:
                if y % x == 0:                           # No remainder?
                    print(y, 'has factor', x)
                    break                                # Skip else
                x -= 1
            else:
                print(y, 'is prime')

    prime(13); prime(13.0)
    prime(15); prime(15.0)
    prime(3);  prime(2)
    prime(1);  prime(-3)

    Here is the module in action; the // operator allows it to work for floating-point numbers too, even though it perhaps should not:

    % python primes.py
    13 is prime
    13.0 is prime
    15 has factor 5
    15.0 has factor 5.0
    3 is prime
    2 is prime
    1 not prime
    -3 not prime

    This function still isn’t very reusable—it could return values, instead of printing—but it’s enough to run experiments. It’s also not a strict mathematical prime (floating points work), and it’s still inefficient. Improvements are left as exercises for more mathematically minded readers. (Hint: a for loop over range(y, 1, −1) may be a bit quicker than the while, but the algorithm is the real bottleneck here.) To time alternatives, use the built-in time module and coding patterns like those used in this general function-call timer (see the library manual for details):

    def timer(reps, func, *args):
        import time
        start = time.clock()
        for i in range(reps):
            func(*args)
        return time.clock() - start

  9. List comprehensions. Here is the sort of code you should write; I may have a preference, but I’m not telling:

    >>> values = [2, 4, 9, 16, 25]
    >>> import math

    >>> res = []
    >>> for x in values: res.append(math.sqrt(x))
    ...
    >>> res
    [1.4142135623730951, 2.0, 3.0, 4.0, 5.0]

    >>> list(map(math.sqrt, values))
    [1.4142135623730951, 2.0, 3.0, 4.0, 5.0]

    >>> [math.sqrt(x) for x in values]
    [1.4142135623730951, 2.0, 3.0, 4.0, 5.0]

  10. Timing tools. Here is some code I wrote to time the three square root options, along with the results in 2.6 and 3.0. The last result of each function is printed to verify that all three do the same work:

    # File mytimer.py (2.6 and 3.0)

    ...same as listed in Chapter 20...

    # File timesqrt.py

    import sys, mytimer
    reps = 10000
    repslist = range(reps)              # Pull out range list time for 2.6

    from math import sqrt               # Not math.sqrt: adds attr fetch time
    def mathMod():
        for i in repslist:
            res = sqrt(i)
        return res

    def powCall():
        for i in repslist:
            res = pow(i, .5)
        return res

    def powExpr():
        for i in repslist:
            res = i ** .5
        return res

    print(sys.version)
    for tester in (mytimer.timer, mytimer.best):
        print('<%s>' % tester.__name__)
        for test in (mathMod, powCall, powExpr):
            elapsed, result = tester(test)
            print ('-'*35)
            print ('%s: %.5f => %s' %
                   (test.__name__, elapsed, result))

    Following are the test results for Python 3.0 and 2.6. For both, it looks like the math module is quicker than the ** expression, which is quicker than the pow call; however, you should try this with your code and on your own machine and version of Python. Also, note that Python 3.0 is nearly twice as slow as 2.6 on this test; 3.1 or later might perform better (time this in the future to see for yourself):

    c:\misc> c:\python30\python timesqrt.py
    3.0.1 (r301:69561, Feb 13 2009, 20:04:18) [MSC v.1500 32 bit (Intel)]
    <timer>
    -----------------------------------
    mathMod: 5.33906 => 99.994999875
    -----------------------------------
    powCall: 7.29689 => 99.994999875
    -----------------------------------
    powExpr: 5.95770 => 99.994999875
    <best>
    -----------------------------------
    mathMod: 0.00497 => 99.994999875
    -----------------------------------
    powCall: 0.00671 => 99.994999875
    -----------------------------------
    powExpr: 0.00540 => 99.994999875


    c:\misc> c:\python26\python timesqrt.py
    2.6.1 (r261:67517, Dec  4 2008, 16:51:00) [MSC v.1500 32 bit (Intel)]
    <timer>
    -----------------------------------
    mathMod: 2.61226 => 99.994999875
    -----------------------------------
    powCall: 4.33705 => 99.994999875
    -----------------------------------
    powExpr: 3.12502 => 99.994999875
    <best>
    -----------------------------------
    mathMod: 0.00236 => 99.994999875
    -----------------------------------
    powCall: 0.00402 => 99.994999875
    -----------------------------------
    powExpr: 0.00287 => 99.994999875

    To time the relative speeds of Python 3.0 dictionary comprehensions and equivalent for loops interactively, run a session like the following. It appears that the two are roughly the same in this regard under Python 3.0; unlike list comprehensions, though, manual loops are slightly faster than dictionary comprehensions today (though the difference isn’t exactly earth-shattering—at the end we save half a second when making 50 dictionaries of 1,000,000 items each). Again, rather than taking these results as gospel you should investigate further on your own, on your computer and with your Python:

    c:\misc> c:\python30\python
    >>>
    >>> def dictcomp(I):
    ...     return {i: i for i in range(I)}
    ...
    >>> def dictloop(I):
    ...     new = {}
    ...     for i in range(I): new[i] = i
    ...     return new
    ...
    >>> dictcomp(10)
    {0: 0, 1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9}
    >>> dictloop(10)
    {0: 0, 1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9}
    >>>
    >>> from mytimer import best, timer
    >>> best(dictcomp, 10000)[0]             # 10,000-item dict
    0.0013519874732672577
    >>> best(dictloop, 10000)[0]
    0.001132965223233029
    >>>
    >>> best(dictcomp, 100000)[0]            # 100,000 items: 10 times slower
    0.01816089754424155
    >>> best(dictloop, 100000)[0]
    0.01643484018219965
    >>>
    >>> best(dictcomp, 1000000)[0]           # 1,000,000 items: 10X time
    0.18685105229855026
    >>> best(dictloop, 1000000)[0]           # Time for making one dict
    0.1769041177020938
    >>>
    >>> timer(dictcomp, 1000000, _reps=50)[0]       # 1,000,000-item dict
    10.692516087938543
    >>> timer(dictloop, 1000000, _reps=50)[0]       # Time for making 50
    10.197276050447755