详解python itertools功能

letters = ['a', 'b', 'c', 'd', 'e', 'f']
booleans = [1, 0, 1, 0, 0, 1]
   print(list(itertools.chain(letters,booleans)))
#   ['a', 'b', 'c', 'd', 'e', 'f', 1, 0, 1, 0, 0, 1]
 
  print(tuple(itertools.chain(letters,letters[3:])))
#   ('a', 'b', 'c', 'd', 'e', 'f', 'd', 'e', 'f')
 
  print(set(itertools.chain(letters,letters[3:])))
#   {'a', 'd', 'b', 'e', 'c', 'f'}
    
  print(list(itertools.chain(letters,letters[3:])))
#   ['a', 'b', 'c', 'd', 'e', 'f', 'd', 'e', 'f']
 
  for item in list(itertools.chain(letters,booleans)):
    print(item)

  i = 0
  for item in itertools.count(100,2):
    i += 1
    if i > 10 : break
    
    print(item) 
 
filterfalse ()
   Python filterfalse(contintion,data) 迭代过滤条件为false的数据。如果条件为空，返回data中为false的项；
booleans = [1, 0, 1, 0, 0, 1]
numbers = [23, 20, 44, 32, 7, 12]
 
print(list(itertools.filterfalse(None,booleans)))
#   [0, 0, 0]
print(list(itertools.filterfalse(lambda x : x < 20,numbers)))
#  [23, 20, 44, 32]

print(list(itertools.compress(letters,booleans)))
# ['a', 'c', 'f']

starmap(pow, [(2,5), (3,2), (10,3)]) --> 32 9 1000
 
>>> from itertools import *
>>> x = starmap(max,[[5,14,5],[2,34,6],[3,5,2]])
>>> for i in x:
>>> print (i)
14
34
5
repeat()
repeat(object[, times]) 重复times次；
repeat(10, 3) --> 10 10 10
dropwhile()
dropwhile(func, seq );当函数f执行返回假时, 开始迭代序列
dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1
takewhile()
takewhile(predicate, iterable)；返回序列，当predicate为true是截止。
takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4
islice()
islice(seq[, start], stop[, step]);返回序列seq的从start开始到stop结束的步长为step的元素的迭代器
for i in islice("abcdef", 0, 4, 2):#a, c
  print i

# product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
  # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
for i in product([1, 2, 3], [4, 5], [6, 7]):
  print i
(1, 4, 6)
(1, 4, 7)
(1, 5, 6)
(1, 5, 7)
(2, 4, 6)
(2, 4, 7)
(2, 5, 6)
(2, 5, 7)
(3, 4, 6)
(3, 4, 7)
(3, 5, 6)
(3, 5, 7)

for i in permutations([1, 2, 3], 3):
  print i
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)

for i in combinations([1, 2, 3], 2):
  print i
(1, 2)
(1, 3)
(2, 3)
combinations_with_replacement()

for i in combinations_with_replacement([1, 2, 3], 2):
  print i
(1, 1)
(1, 2)
(1, 3)
(2, 2)
(2, 3)
(3, 3)

def get_three_data(data_list,amount):
  for data in list(itertools.combinations(data_list, 3)):
    if sum(data) == amount:
      print(data)
#(7, 13, 15)
#(9, 11, 15)