python 之 itertools模塊
官方:https://yiyibooks.cn/xx/python_352/library/itertools.html
參考:
https://blog.csdn.net/neweastsun/article/details/51965226
https://www.liaoxuefeng.com/wiki/0014316089557264a6b348958f449949df42a6d3a2e542c000/00143200162233153835cfdd1a541a18ddc15059e3ddeec000
python 之 itertools模塊
介紹
itertools是python內置的模塊,使用簡單且功能強大,使用只需簡單一句導入:import itertools
chain()
與其名稱意義一樣,給它一個列表如 lists/tuples/iterables,鏈接在一起;返回iterables對象。
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)
count()
生成無界限序列,count(start=0, step=1) ,示例從100開始,步長為2,循環10,打印對應值;必須手動break,count()會一直循環。
i = 0 for item in itertools.count(100,2): i += 1 if i > 10 : break print(item)
filterfalse ()
Python filterfalse(contintion,data) 叠代過濾條件為false的數據(註意:最終結果僅會展示 為 False 的數據)
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]
compress()
返回我們需要使用的元素,根據b集合中元素真值,返回a集中對應的元素。
print(list(itertools.compress(letters,booleans))) # [‘a‘, ‘c‘, ‘f‘]
starmap()
針對list中的每一項,調用函數功能。starmap(func,list[]) ;
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() 重復
repeat(object[, times]) 重復times次;
repeat(10, 3) --> 10 10 10
dropwhile()
dropwhile(func, seq );當函數f執行返回假時, 開始叠代序列
dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1
takewhile()
takewhile(predicate, iterable);返回序列,當predicate為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的從start開始到stop結束的步長為step的元素的叠代器
for i in islice("abcdef", 0, 4, 2):#a, c print i
矩陣大家族
product()
product(iter1,iter2, ... iterN, [repeat=1]);創建一個叠代器,生成表示item1,item2等中的項目的笛卡爾積的元組,repeat是一個關鍵字參數,指定重復生成序列的次數
# 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)
permutations()
permutations(p[,r]);返回p中任意取r個元素做排列的元組的叠代器
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)
combinations()
combinations(iterable,r);創建一個叠代器,返回iterable中所有長度為r的子序列,返回的子序列中的項按輸入iterable中的順序排序
note:不帶重復
for i in combinations([1, 2, 3], 2): print i (1, 2) (1, 3) (2, 3)
combinations_with_replacement() 同上, 帶重復
例子:
for i in combinations_with_replacement([1, 2, 3], 2): print i (1, 1) (1, 2) (1, 3) (2, 2) (2, 3) (3, 3)
應用示例
# 求質數序列中1,3,5,7,9,11,13,15三個數之和為35的三個數; 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)
python 之 itertools模塊