Function | ||||
Infinite iterators (无限迭代) |
count() | cycle() | repeat() | |
Combinatoric generators (排列组合) |
product() | permutations() | combinations() | |
有限迭代 | accumulate(p[, func]) | chain(p, q, ...) | chain.from_iterable([p, q, ...]) | compress(data, selectors) |
dropwhile(pred, seq) | groupby(iterable[, keyfunc]) | filterfalse(pred, seq) | islice(seq, [start,] stop [, step]) | |
starmap(fun, seq) | tee(it, n=2) | takewhile(pred, seq) | zip_longest(p, q, ...) |
itertools.tee(it, n=2) 复制独立的迭代器(默认复制两次)
1 import itertools 2 num = [1, 2, 3, 4] 3 for i, item in enumerate(itertools.tee(num, 3)): # 3是指复制三次,默认为2次 4 print('第{}个迭代对象'.format(i)) 5 for it in item: 6 print(it, end=' ') 7 print('')
itertools.accumulate(iter[, func, initial]) 对iter中对象进行指定函数的迭代运算
1 for item in itertools.accumulate(num): # 默认迭代相加 2 print(item, end=' ') 3 print('') 4 for item in itertools.accumulate(num, func=operator.mul): # 指定迭代函数 5 print(item, end=' ') 6 print('') 7 for item in itertools.accumulate(num, func=operator.mul, initial=2): # 指定迭代初始值 8 print(item, end=' ') 9 print('')
itertools.chain ( *iterables ) # 一次性遍历多个迭代器
1 # itertools. chain ( *iterables ) 2 for item in itertools.chain(itertools.tee(num, 3)): 3 for i in item: 4 print(i, end=' ') 5 print(''
itertools.chain.from_iterable ( *iterables ) # 一次性遍历多个迭代器
1 # itertools. chain.from_iterable ( *iterables ) 2 for item in itertools.chain.from_iterable(itertools.tee(num, 3)): 3 print(item, end=' ') 4 print('')
itertools. combinations ( iterable, r ) # 组合,从iterable对象中抽取r个对象的所有情况
1 # itertools. combinations ( iterable, r ) 2 for item in itertools.combinations (num, 3): 3 print(item, end=' ') 4 print('')
itertools.combinations_with_replacement( iterable, r ) # 组合,从iterable对象中抽取r个对象的所有情况,
抽取r次,每次随机抽取,会出现重复对象,适合独立重复试验
1 # itertools. combinations_with_replacement ( iterable, r ) 2 for item in itertools.combinations_with_replacement(num, 2): 3 print(item, end=' ') 4 print('')
itertools.compress(data, selectors) 返回selectors中计算结果为True的相应位置中的data数据
1 # itertools.compress ( data, selectors ) 2 for item in itertools.compress('ABCDEF', [1, 0, 1, 0, 1, 1]): 3 print(item, end=' ') 4 print()
itertools.count( start = 0, step = 1 ) 从0开始,生成步长为1的无限序列
1 # itertools. count ( start=0, step=1 ) 2 for item in itertools.count(start=3, step=3): 3 print(item, end=' ') 4 if item > 20: 5 break 6 print('')
itertools.cycle(iterable) 对iterable对象中的元素进行无限循环
1 # itertools. cycle ( iterable ) 2 for item in itertools.cycle(num): 3 print(item, end=' ') 4 print()
itertools.dropwhile(predicate, iterable) 遍历iterable,删除满足predicate条件的元素,直到出现不满足条件的元素,输出剩余元素
1 # itertools. dropwhile ( predicate, iterable ) 2 for item in itertools.dropwhile(lambda x: x < 5, [1, 4, 6, 3, 0]): 3 print(item, end=' ') 4 print()
itertools.filterfalse ( predicate, iterable ) 遍历iterable,返回满足predicate条件的为False的所有元素
1 # itertools.filterfalse ( predicate, iterable ) 2 for item in itertools.filterfalse(lambda x: x % 2, range(10)): 3 print(item, end=' ') 4 print()
itertools. groupby ( iterable, key=None ) 对iterable做分组处理,但是该处分组不同于sql中,连续输入同一个字符算作一组,重新打断输入则为另外一组
1 # itertools. groupby ( iterable, key=None ) 2 for k, g in itertools.groupby('AAAABBBCCDAABBB'): 3 print(k, list(g)) 4 for k, g in itertools.groupby('AAAABBBCCDAABBB', key=lambda x: x == 'A'): 5 print(k, list(g))
itertools.islice( iterable, start, stop [, step ] ) 对literable进行切片处理
itertools.islice ( iterable, stop )
1 # itertools.islice ( iterable, stop ) 2 # itertools.islice ( iterable, start, stop [, step ] ) 3 for item in itertools.islice('ABCDEFG', 2): 4 print(item, end=' ') 5 print('') 6 for item in itertools.islice('ABCDEFG', 2, 4): 7 print(item, end=' ') 8 print('') 9 for item in itertools.islice('ABCDEFG', 2, None, 2): 10 print(item, end=' ') 11 print('')
itertools.permutations(iterable, r=None) 排列,抽取r个iterable对象进行排列,当r=None时,对所有元素进行排列
1 # itertools. permutations ( iterable, r=None ) 2 for item in itertools.permutations('ABC', 2): 3 print(item, end=' ') 4 print('')
itertools.product(*iterable, repeat=1) 计算笛卡尔积,repeat为iterable所有对象的重复次数
1 # itertools. product ( *iterables, repeat=1 ) 2 for item in itertools.product('AB', 'xy', repeat=2): 3 print(item) 4 print('')
# itertools. repeat ( object [, times ] )
repeat(10, 3) --> 10 10 10
# itertools. starmap ( function, iterable )
starmap(pow, [(2,5), (3,2), (10,3)]) --> 32 9 1000
# itertools. takewhile ( predicate, iterable ) 遇到不满足条件的即停止运行
takewhile(lambda x: x<5, [1,4,6,3,5]) --> 1 4
# itertools. zip_longest ( *iterables, fillvalue=None )
zip_longest('ABCD', 'xy', fillvalue='-') --> Ax By C- D-