Python 迭代器工具包

__iter__()

__next__()

yield

APL, Haskell, and SML

itertools

itertools

1. 无限迭代

for...in...

count(start, [step])

cycle(p)

repeat(elem [,n])

map

zip

from itertools import cycle, count, repeat
print(count.__doc__)

count(start=0, step=1) --> count object

Return a count object whose .__next__() method returns consecutive values.
Equivalent to:

def count(firstval=0, step=1):
x = firstval
while 1:
yield x
x += step

counter = count()
print(next(counter))
print(next(counter))
print(list(map(lambda x, y: x+y, range(10), counter)))

odd_counter = map(lambda x: 'Odd#{}'.format(x), count(1, 2))
print(next(odd_counter))
print(next(odd_counter))

0
1
[2, 4, 6, 8, 10, 12, 14, 16, 18, 20]
Odd#1
Odd#3

print(cycle.__doc__)

cycle(iterable) --> cycle object

Return elements from the iterable until it is exhausted.
Then repeat the sequence indefinitely.

cyc = cycle(range(5))
print(list(zip(range(6), cyc)))
print(next(cyc))
print(next(cyc))

[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 0)]
1
2

print(repeat.__doc__)

repeat(object [,times]) -> create an iterator which returns the object
for the specified number of times.  If not specified, returns the object
endlessly.

print(list(repeat('Py', 3)))
rep = repeat('p')
print(list(zip(rep, 'y'*3)))

['Py', 'Py', 'Py']
[('p', 'y'), ('p', 'y'), ('p', 'y')]

2. 整合两序列迭代

zip

zip

accumulate()

chain()

chain.from_iterable()

compress()

dropwhile()

filterfalse()

takewhile()

groupby()

islice()

starmap()

tee()

zip_longest()

print(method.__doc__)

itertools

from itertools import cycle, compress, islice, takewhile, count

# 这三个方法（如果使用恰当）可以限定无限迭代
# print(compress.__doc__)
print(list(compress(cycle('PY'), [1, 0, 1, 0])))

# 像操作列表 l[start:stop:step] 一样操作其它序列
# print(islice.__doc__)
print(list(islice(cycle('PY'), 0, 2)))

# 限制版的 filter
# print(takewhile.__doc__)
print(list(takewhile(lambda x: x < 5, count())))

['P', 'P']
['P', 'Y']
[0, 1, 2, 3, 4]

from itertools import groupby
from operator import itemgetter
print(groupby.__doc__)

for k, g in groupby('AABBC'):
print(k, list(g))
db = [dict(name='python', script=True),
dict(name='c', script=False),
dict(name='c++', script=False),
dict(name='ruby', script=True)]
keyfunc = itemgetter('script')

db2 = sorted(db, key=keyfunc) # sorted by `script'
for isScript, langs in groupby(db2, keyfunc):
print(', '.join(map(itemgetter('name'), langs)))

groupby(iterable[, keyfunc]) -> create an iterator which returns
(key, sub-iterator) grouped by each value of key(value).

A ['A', 'A']
B ['B', 'B']
C ['C']
c, c++
python, ruby

from itertools import zip_longest

# 内置函数 zip 以较短序列为基准进行合并，
# zip_longest 则以最长序列为基准，并提供补足参数 fillvalue
# Python 2.7 中名为 izip_longest

print(list(zip_longest('ABCD', '123', fillvalue=0)))

[('A', '1'), ('B', '2'), ('C', '3'), ('D', 0)]

3. 组合生成器

product(*iterables, repeat=1)

permutations(iterable, r=None)

combinations(iterable, r)

combinations_with_replacement(iterable, r)

from itertools import product, permutations, combinations, combinations_with_replacement
print(list(product(range(2), range(2))))
print(list(product('AB', repeat=2)))

[(0, 0), (0, 1), (1, 0), (1, 1)]
[('A', 'A'), ('A', 'B'), ('B', 'A'), ('B', 'B')]

print(list(combinations_with_replacement('AB', 2)))

[('A', 'A'), ('A', 'B'), ('B', 'B')]

# 赛马问题：4匹马前2名的排列组合（A^4_2）
print(list(permutations('ABCDE', 2)))

[('A', 'B'), ('A', 'C'), ('A', 'D'), ('A', 'E'), ('B', 'A'), ('B', 'C'), ('B', 'D'), ('B', 'E'), ('C', 'A'), ('C', 'B'), ('C', 'D'), ('C', 'E'), ('D', 'A'), ('D', 'B'), ('D', 'C'), ('D', 'E'), ('E', 'A'), ('E', 'B'), ('E', 'C'), ('E', 'D')]

# 彩球问题：4种颜色的球任意抽出2个的颜色组合（C^4_2）
print(list(combinations('ABCD', 2)))

[('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D')]

总结