python itertools库详解（让你的代码更pythonic）
itertools
api:
Itertool functions
itertools.accumulate(iterable[, func ])
def accumulate(iterable, func=operator.add):
'Return running totals'
# accumulate([1,2,3,4,5]) --> 1 3 6 10 15
# accumulate([1,2,3,4,5], operator.mul) --> 1 2 6 24 120
it = iter(iterable)
try:
total = next(it)
except StopIteration:
return
yield total
for element in it:
total = func(total, element)
yield total
eg:
>>> data = [3, 4, 6, 2, 1, 9, 0, 7, 5, 8]
>>> list(accumulate(data, operator.mul)) # running product
[3, 12, 72, 144, 144, 1296, 0, 0, 0, 0]
>>> list(accumulate(data, max)) # running maximum
[3, 4, 6, 6, 6, 9, 9, 9, 9, 9]
# Amortize a 5% loan of 1000 with 4 annual payments of 90
>>> cashflows = [1000, -90, -90, -90, -90]
>>> list(accumulate(cashflows, lambda bal, pmt: bal*1.05 + pmt))
[1000, 960.0, 918.0, 873.9000000000001, 827.5950000000001]itertools.chain 连接多个列表或者迭代器
>>> x = itertools.chain(range(3), range(4), [3,2,1])
>>> print(list(x))
[0, 1, 2, 0, 1, 2, 3, 3, 2, 1]itertools.combinations 求列表或生成器中指定数目的元素不重复的所有组合
源码：
def combinations(iterable, r):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = list(range(r))
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)
def combinations(iterable, r):
pool = tuple(iterable)
n = len(pool)
for indices in permutations(range(n), r):
if sorted(indices) == list(indices):
yield tuple(pool[i] for i in indices)
>>> x = itertools.combinations(range(4), 3)
>>> print(list(x))
[(0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)]itertools.combinations_with_replacement(iterable, r) 允许重复元素的组合
>>> x = itertools.combinations_with_replacement('ABC', 2)
>>> print(list(x))
[('A', 'A'), ('A', 'B'), ('A', 'C'), ('B', 'B'), ('B', 'C'), ('C', 'C')]itertools.compress(data, selectors) 按照真值表筛选元素
def compress(data, selectors):
# compress('ABCDEF', [1,0,1,0,1,1]) --> A C E F
return (d for d, s in zip(data, selectors) if s)
>>> list(itertools.compress(s,[1,0,1,0,1,1]))
['a', 'c', 'e', 'f']itertools.count(start=0, step=1)
def count(start=0, step=1):
# count(10) --> 10 11 12 13 14 ...
# count(2.5, 0.5) -> 2.5 3.0 3.5 ...
n = start
while True:
yield n
n += step
>>> x = itertools.count(start=20, step=-1)
>>> print(list(itertools.islice(x, 0, 10, 1)))
[20, 19, 18, 17, 16, 15, 14, 13, 12, 11]itertools.cycle(iterable) 循环指定的列表和迭代器
def cycle(iterable):
# cycle('ABCD') --> A B C D A B C D A B C D ...
saved = []
for element in iterable:
yield element
saved.append(element)
while saved:
for element in saved:
yield element
>>> x = itertools.cycle('ABC')
>>> print(list(itertools.islice(x, 0, 10, 1)))
['A', 'B', 'C', 'A', 'B', 'C', 'A', 'B', 'C', 'A']itertools.dropwhile(predicate, iterable) 根据真值表舍弃第一个不满足条件前面的值
def dropwhile(predicate, iterable):
# dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1
iterable = iter(iterable)
for x in iterable:
if not predicate(x):
yield x
break
for x in iterable:
yield x
>>> x = itertools.dropwhile(lambda e: e < 5, range(10))
>>> print(list(x))
[5, 6, 7, 8, 9]itertools.filterfalse 保留对应真值为False的元素
def filterfalse(predicate, iterable):
# filterfalse(lambda x: x%2, range(10)) --> 0 2 4 6 8
if predicate is None:
predicate = bool
for x in iterable:
if not predicate(x):
yield x
>>> x = itertools.filterfalse(lambda e: e < 5, (1, 5, 3, 6, 9, 4))
>>> print(list(x))
[5, 6, 9]itertools.groupby(iterable, key=None) 按照分组函数的值对元素进行分组
class groupby:
# [k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B
# [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D
def __init__(self, iterable, key=None):
if key is None:
key = lambda x: x
self.keyfunc = key
self.it = iter(iterable)
self.tgtkey = self.currkey = self.currvalue = object()
def __iter__(self):
return self
def __next__(self):
self.id = object()
while self.currkey == self.tgtkey:
self.currvalue = next(self.it) # Exit on StopIteration
self.currkey = self.keyfunc(self.currvalue)
self.tgtkey = self.currkey
return (self.currkey, self._grouper(self.tgtkey, self.id))
def _grouper(self, tgtkey, id):
while self.id is id and self.currkey == tgtkey:
yield self.currvalue
try:
self.currvalue = next(self.it)
except StopIteration:
return
self.currkey = self.keyfunc(self.currvalue)
>>> x = itertools.groupby(range(10), lambda x: x < 5 or x > 8)
>>> for condition, numbers in x:
print(condition, list(numbers))
True [0, 1, 2, 3, 4]
False [5, 6, 7, 8]
True [9]itertools.islice(iterable, stop)
itertools.islice(iterable, start, stop[, step ])
def islice(iterable, *args):
# islice('ABCDEFG', 2) --> A B
# islice('ABCDEFG', 2, 4) --> C D
# islice('ABCDEFG', 2, None) --> C D E F G
# islice('ABCDEFG', 0, None, 2) --> A C E G
s = slice(*args)
it = iter(range(s.start or 0, s.stop or sys.maxsize, s.step or 1))
try:
nexti = next(it)
except StopIteration:
return
for i, element in enumerate(iterable):
if i == nexti:
yield element
nexti = next(it)
>>> x = itertools.islice(range(10), 0, 9, 2)
>list(x)
>[0, 2, 4, 6, 8]itertools.permutations(iterable, r=None),输出输入序列的全排列
The number of items returned is n! / (n-r)! when 0 <= r <= n or zero when r > n.
根据序列位置进行全排列，而不是值，所以如果输入序列有重复值，输出亦会有
源码模拟：
def permutations(iterable, r=None):
# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
# permutations(range(3)) --> 012 021 102 120 201 210
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
if r > n:
return
indices = list(range(n))
cycles = list(range(n, n-r, -1))
yield tuple(pool[i] for i in indices[:r])
while n:
for i in reversed(range(r)):
cycles[i] -= 1
if cycles[i] == 0:
indices[i:] = indices[i+1:] + indices[i:i+1]
cycles[i] = n - i
else:
j = cycles[i]
indices[i], indices[-j] = indices[-j], indices[i]
yield tuple(pool[i] for i in indices[:r])
break
else:
return
eg:
>>> import itertools as iters
>>> x = iters.permutations([0,1,2], 3)
>>> x
0x0000018B993C40A0>
>>> list(x)
[(0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0)]
>>> x = iters.permutations([0,1,1], 3)
>>> x = iters.permutations([0,1,1], 3)
>>> list(x)
[(0, 1, 1), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 0, 1), (1, 1, 0)] itertools.product(*iterables, repeat=1) Cartesian product of input iterables.
product(A, B) returns the same as ((x,y) for x in A for y in B)
def product(*args, repeat=1):
# product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
# product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
pools = [tuple(pool) for pool in args] * repeat
result = [[]]
for pool in pools:
result = [x+[y] for x in result for y in pool]
for prod in result:
yield tuple(prod)
>>> list(iters.product('ABC',repeat =1))
[('A',), ('B',), ('C',)]
>>> list(iters.product('ABC',repeat =2))
[('A', 'A'), ('A', 'B'), ('A', 'C'), ('B', 'A'), ('B', 'B'), ('B', 'C'), ('C', 'A'), ('C', 'B'), ('C', 'C')]
itertools.repeat(object[, times ])
def repeat(object, times=None):
# repeat(10, 3) --> 10 10 10
if times is None:
while True:
yield object
else:
for i in range(times):
yield object
>>> list(map(pow, range(10), repeat(2)))
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
itertools.starmap(function, iterable)
map() and starmap()的区别类似于function(a,b) and function(*c).
def starmap(function, iterable):
# starmap(pow, [(2,5), (3,2), (10,3)]) --> 32 9 1000
for args in iterable:
yield function(*args)
>>> x = itertools.starmap(str.islower, 'aBCDefGhI')
>>> print(list(x))
[True, False, False, False, True, True, False, True, False]itertools.takewhile(predicate, iterable) 保留序列元素直到条件不满足
def takewhile(predicate, iterable):
# takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4
for x in iterable:
if predicate(x):
yield x
else:
breakitertools.tee(iterable, n=2)
一旦迭代器被tee()函数分割，就不要再次使用；list（）其实更快
def tee(iterable, n=2):
it = iter(iterable)
deques = [collections.deque() for i in range(n)]
def gen(mydeque):
while True:
if not mydeque: # when the local deque is empty
try:
newval = next(it) # fetch a new value and
except StopIteration:
return
for d in deques: # load it to all the deques
d.append(newval)
yield mydeque.popleft()
return tuple(gen(d) for d in deques)
>>> x = itertools.tee(range(10), 2)
>>> for letters in x:
... print(list(letters))
...
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
itertools.zip_longest(*iterables, fillvalue=None)
def zip_longest(*args, fillvalue=None):
# zip_longest('ABCD', 'xy', fillvalue='-') --> Ax By C- Diterators
= [iter(it) for it in args]
num_active = len(iterators)
if not num_active:
return
while True:
values = []
for i, it in enumerate(iterators):
try:
value = next(it)
except StopIteration:
num_active -= 1
if not num_active:
return
iterators[i] = repeat(fillvalue)
value = fillvalue
values.append(value)
yield tuple(values)
>>> x = itertools.zip_longest(range(3), range(5))
>>> y = zip(range(3), range(5))
>>> print(list(x))
[(0, 0), (1, 1), (2, 2), (None, 3), (None, 4)]
>>> print(list(y))
[(0, 0), (1, 1), (2, 2)]