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By Root 1442 0

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'))

(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}

See #5.

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,)

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')