Third part of PythonMaterial.

list

""" Show list operations """
 
import sys
 
print "Size of a list:", sys.getsizeof([]), sys.getsizeof([1, 2, 3, 4, 5]),
print sys.getsizeof([1, 2, 3, 4, 'this is much bigger'])
print
print "Constants:", [1, 2, 3, 'a']
 
l = [1, 2, 2, 3]
print "Operators:", l + ['a', 'b'], l * 2, l[1]
print "Slicing:", l[1:3], l[2:], l[-1:], l[:-2], [1, 2, 3, 4, 5, 6, 7][::2]
print "Functions:", len(l), max(l), min(l), l.index(2), l.count(2)
 
del l[2]
print "Removed 3:", l
l += [3, 4, 5]
print "Added some:", l
l.append(6)
print "Added more:", l
l.extend([7, 8, 9])
print "Even more:", l
del l[:2]
print "Removed start:", l
del l[-4:]
print "Removed tail:", l
print "Boolean:", l < [4], l == [3, 3, 4, 5]
 
print "Loop",
for v in l:
    print v * 2,
print
 
l[:2] = [1, 2, 3]
print "Overwrite:", l,
l.reverse()
print l
 
l.reverse()
l.insert(3, 16)
print "Insert:", l
l.sort()
print "Sorted:", l, sorted("Me and my house alone!".split(), key=str.lower)
 
l.remove(16)
print "Removed:", l
 
print "Pop:", l.pop(), l, l.pop(0), l

• A vector like (1, 5) is an immutable list.

Assignment
1 Write a function that returns a list with every element multiplied with a given value. The first argument is the list the second the multiplier. Try it out with some test cases.
2 When the first argument was a variable the contents should be unaltered after the function. Try it with the first argument a vector that cannot be altered.
3 Try the function with a string as the first argument. And with a numerical list and another list in the second argument.
4 Write a function that returns the 4 highest values in a list ordered with the lowest of them first till the highest last.
5 Remove the key=str.lower argument from the example code and run it again. The ordering is different now.

string

Strings have a second life as list of single characters.

""" Show list operations on strings """
 
l = "abcdef"
print "Operators:", l[1], l[-3:], l[::2]
 
print "Iterate string:",
for c in "abcdef":
    print c * 2,
print
 
print "Zipping:",
for c in zip(xrange(5), "abcdefgh"):
    print c,
print
 
print "Mutable strings:",
l = []
l += "abc"
print l,
l += "efg"
l[2] += "d"
print l, "".join(l)
 
print "Sorting:",
l = []
l += "edabc"
l += ["fg"]
l.sort()
print l,
l.sort(reverse=True)
print l

Assignment
1 Write a function that writes a string with each character on a separate line.
2 Show a '!' after each capital letter and '#' after each numeric token and '.' after each space.
3 Write a function that get's a string and when it is enclosed with " or ' quotes return the normal string.
4 Alter to function to replace two " tokens with a single one or when it was enclosed in ' replace two tokens with a single one.
5 Return an integer value when the argument was not enclosed in quotes and contains only numeric tokens or starts with a '-' token. Also show the type of the result.
6 Return a float value when the string also contained a single 'e' or '.' or '+'/'-' after the 'e' token.

set

""" Show set operations """
 
s = set([1, 2, 5])
t = set([1, 3, 5])
print "Constants:", s, t, "length", len(s)
print "Operators:",
print s | t,  # s.union(t)
print s & t,  # s.intersection(t)
print s - t,  # s.difference(t)
print s ^ t  # s.symmetric_difference(t)
print "Tests:", 2 in s, 5 not in s,
print s.isdisjoint([4, 6, 2]),  # True is there are no elements in common
print s <= (s & frozenset(xrange(4))), s.issubset(xrange(6)),
print s | t > set(xrange(4)) - set(xrange(1))  # (s | t).issuperset(o)
 
s |= frozenset([6, 7])  # s.update([6,7])
s &= frozenset(xrange(7))  # s.intersection_update(xrange(7))
s ^= frozenset(xrange(9))  # s.symmetric_difference_update(xrange(9))
s -= frozenset([8])  # s.difference_update([8])
print "Change operators:", s
 
s.add(5)
s.remove(0)
s.discard(0)
print "Set mutations:", s, s.pop(), s
s.clear()

A lot of the operators on sets also have set method alternatives.

Assignment
1 Fill a set with names of persons. Add a name twice and show that this name is only once in the set.
2 Make a second set with names and show a list of persons in both sets and persons only in the second set.
3 Show the persons that are only in one set and not in both.
4 Remove from the set the person with the highest and the lowest sorted name.

dict

""" Show usage of the dict type """
 
a = {1: 'this', 2: 'is'}
a[3] = "no"
a[4] = "list"
b = dict(another=3, value=4)
c = dict([(3, 'different'), (4, 'value')])
d = dict.fromkeys(xrange(1, 5), "a")
print "Constants:", a, "length", len(a), b, c, d
 
print "Keys:",
for v in a:  # a.iter() or a.iterkeys() or a.keys()
    print str(v) + ':' + a[v],
print
 
print "Values:",
for v in a.itervalues():  # a.values() is possible but less efficient
    print v,
print
 
a[3] = "a"
print "Items:",
for k, v in a.iteritems():  # a.items()
    print str(k) + ':' + v,
print
 
del a[1]
print "Removed:", a, a.get(1, "something")
 
print "Tests:", 'another' in b, 1 not in a
print "Dict mutation:", a.pop(3), a.popitem(), a.setdefault(4, ""), a
a.clear()
 
b.update(added=5, too=6)
b.update([('another', 2)])
print "Update:", b, sorted(b)

Assignment
1 Create a dictionary with persons and a number. For example the estimated number of times you met.
2 Write a function that add one to a person in the dictionary when already there or adds the persons when 1 when not in the dictionary.
3 Show the persons on the dictionary but skip the persons you met less then 5 times. Now show them alphabetically ordered.
4 Show the list ordered by the number of times you met with the highest times first.