April 1999
Beginner
384 pages
11h 15m
English
Coding basic loops. If you work through this exercise, you’ll wind up with code that looks like the following:
>>>S = 'spam'>>>for c in S:...print ord(c)... 115 112 97 109 >>>x = 0>>>for c in S: x = x + ord(c)... >>>x433 >>>x = []>>>for c in S: x.append(ord(c))... >>>x[115, 112, 97, 109] >>>map(ord, S)[115, 112, 97, 109]
Backslash
characters. The example prints the bell character
(\a) 50 times; assuming your machine can handle
it, you’ll get a series of beeps (or one long tone, if your
machine is fast enough). Hey—we warned you.
Sorting dictionaries. Here’s one way to work through this exercise; see Chapter 2 if this doesn’t make sense:
>>>D = {'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6, 'g':7}>>>D{'f': 6, 'c': 3, 'a': 1, 'g': 7, 'e': 5, 'd': 4, 'b': 2} >>> >>>keys = D.keys()>>>keys.sort()>>>for key in keys:...print key, '=>', D[key]... a => 1 b => 2 c => 3 d => 4 e => 5 f => 6 g => 7
Program logic alternatives. Here’s how we coded the solutions; your results may vary a bit.
L = [1, 2, 4, 8, 16, 32, 64]
X = 5
i = 0
while i < len(L):
if 2 ** X == L[i]:
print 'at index', i
break
i = i+1
else:
print X, 'not found'L = [1, 2, 4, 8, 16, 32, 64]
X = 5
for p in L:
if (2 ** X) == p:
print (2 ** X), 'was found at', L.index(p)
break
else:
print X, 'not found'L = [1, 2, 4, 8, 16, 32, 64]
X = 5
if (2 ** X) in L:
print (2 ** X), 'was found at', L.index(2 ** X)
else:
print X, 'not found'X = 5 L = [] for i in range(7): L.append(2 ** i) print L if (2 ** X) ...