Python迭代器模块itertools使用原理解析

import itertools

num = itertools.count(10)
for i in num:
    print(i)
# 10
# 11
# 12
# 13
# 以此类推，无穷无尽

letter = itertools.cycle('ABC')
for i in letter:
    print(i)
# A
# B
# C
# A
# B
# 依次循环，无穷无尽

rp = itertools.repeat('X')
for i in rp:
    print(i)

# X
# X
# X
# 依次类推，无穷无尽

rp2 = itertools.repeat('X', 2) # 限制2次
for i in rp2:
    print(i)

# X
# X

num2 = itertools.takewhile(lambda x: x < 15, num)
list(num2)
# [10,11,12,13,14]

for c in itertools.chain('AB', 'CD'):
    print(c)
# A
# B
# C
# D

for key, group in itertools.groupby('AAABBCCB'):
    print(key, list(group))

# A ['A', 'A', 'A']
# B ['B', 'B', 'B]
# C ['C', 'C']

# 注意这里是区分大小写的，如果要忽略
# 请使用 itertools.groupby('AAABBCCB', lambda c: c.upper())

x = itertools.accumulate(range(5))
print(list(x))
# [0, 1, 3, 6, 10, 15]

a = [1,2,3,4,5]
x1, x2, x3 = itertools.tee(a,3)

# 产生了三个元素和a一样的iter

x = itertools.combinations(range(4), 3)
print(list(x))
# [(0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)]

x = itertools.compress(range(5), (True, False, True, True, False)) # 0,1,2,3,4,5
print(list(x))
# [0, 2, 3]

x = itertools.filterfalse(lambda n: n < 5, (1, 2, 5, 3, 7, 10, 0))
print(list(x))
# [5, 7, 10]

x = itertools.islice(range(10), 0, 9, 2)
print(list(x))
# [0, 2, 4, 6, 8]

x = itertools.product('ABC', range(3))
print(list(x))
# [('A', 0), ('A', 1), ('A', 2), ('B', 0), ('B', 1), ('B', 2), ('C', 0), ('C', 1), ('C', 2)]