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>>> amounts = [120.15, 764.05, 823.14]
>>> for checknum, amount in izip(count(1200), amounts):
... print 'Check %d is for $%.2f' % (checknum, amount)
...
Check 1200 is for $120.15
Check 1201 is for $764.05
Check 1202 is for $823.14
>>> import operator
>>> for cube in imap(operator.pow, xrange(1,4), repeat(3)):
... print cube
...
1
8
27
>>> reportlines = ['EuroPython', 'Roster', '', 'alex', '', 'laura', '', 'martin', '', 'walter', '', 'samuele']
>>> for name in islice(reportlines, 3, None, 2):
... print name.title()
...
Alex
Laura
Martin
Walter
Samuele
>>> from operator import itemgetter
>>> d = dict(a=1, b=2, c=1, d=2, e=1, f=2, g=3)
>>> di = sorted(sorted(d.iteritems()), key=itemgetter(1))
>>> for k, g in groupby(di, itemgetter(1)):
... print k, map(itemgetter(0), g)
...
1 ['a', 'c', 'e']
2 ['b', 'd', 'f']
3 ['g']
# Find runs of consecutive numbers using groupby. The key to the solution
# is differencing with a range so that consecutive numbers all appear in
# same group.
>>> data = [ 1, 4,5,6, 10, 15,16,17,18, 22, 25,26,27,28]
>>> for k, g in groupby(enumerate(data), lambda (i,x):i-x):
... print map(operator.itemgetter(1), g)
...
[1]
[4, 5, 6]
[10]
[15, 16, 17, 18]
[22]
[25, 26, 27, 28]
>>> def take(n, seq):
... return list(islice(seq, n))
>>> def enumerate(iterable):
... return izip(count(), iterable)
>>> def tabulate(function):
... "Return function(0), function(1), ..."
... return imap(function, count())
>>> def iteritems(mapping):
... return izip(mapping.iterkeys(), mapping.itervalues())
>>> def nth(iterable, n):
... "Returns the nth item"
... return list(islice(iterable, n, n+1))
>>> def all(seq, pred=bool):
... "Returns True if pred(x) is True for every element in the iterable"
... for elem in ifilterfalse(pred, seq):
... return False
... return True
>>> def any(seq, pred=bool):
... "Returns True if pred(x) is True for at least one element in the iterable"
... for elem in ifilter(pred, seq):
... return True
... return False
>>> def no(seq, pred=bool):
... "Returns True if pred(x) is False for every element in the iterable"
... for elem in ifilter(pred, seq):
... return False
... return True
>>> def quantify(seq, pred=bool):
... "Count how many times the predicate is True in the sequence"
... return sum(imap(pred, seq))
>>> def padnone(seq):
... "Returns the sequence elements and then returns None indefinitely"
... return chain(seq, repeat(None))
>>> def ncycles(seq, n):
... "Returns the sequence elements n times"
... return chain(*repeat(seq, n))
>>> def dotproduct(vec1, vec2):
... return sum(imap(operator.mul, vec1, vec2))
>>> def flatten(listOfLists):
... return list(chain(*listOfLists))
>>> def repeatfunc(func, times=None, *args):
... "Repeat calls to func with specified arguments."
... " Example: repeatfunc(random.random)"
... if times is None:
... return starmap(func, repeat(args))
... else:
... return starmap(func, repeat(args, times))
>>> def pairwise(iterable):
... "s -> (s0,s1), (s1,s2), (s2, s3), ..."
... a, b = tee(iterable)
... try:
... b.next()
... except StopIteration:
... pass
... return izip(a, b)
This is not part of the examples but it tests to make sure the definitions
perform as purported.
>>> take(10, count())
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> list(enumerate('abc'))
[(0, 'a'), (1, 'b'), (2, 'c')]
>>> list(islice(tabulate(lambda x: 2*x), 4))
[0, 2, 4, 6]
>>> nth('abcde', 3)
['d']
>>> all([2, 4, 6, 8], lambda x: x%2==0)
True
>>> all([2, 3, 6, 8], lambda x: x%2==0)
False
>>> any([2, 4, 6, 8], lambda x: x%2==0)
True
>>> any([1, 3, 5, 9], lambda x: x%2==0,)
False
>>> no([1, 3, 5, 9], lambda x: x%2==0)
True
>>> no([1, 2, 5, 9], lambda x: x%2==0)
False
>>> quantify(xrange(99), lambda x: x%2==0)
50
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> flatten(a)
[1, 2, 3, 4, 5, 6]
>>> list(repeatfunc(pow, 5, 2, 3))
[8, 8, 8, 8, 8]
>>> import random
>>> take(5, imap(int, repeatfunc(random.random)))
[0, 0, 0, 0, 0]
>>> list(pairwise('abcd'))
[('a', 'b'), ('b', 'c'), ('c', 'd')]
>>> list(pairwise([]))
[]
>>> list(pairwise('a'))
[]
>>> list(islice(padnone('abc'), 0, 6))
['a', 'b', 'c', None, None, None]
>>> list(ncycles('abc', 3))
['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b', 'c']
>>> dotproduct([1,2,3], [4,5,6])
32
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