上机练习:容器类型操作
〉 列表、元组基本操作
+, *, len(), [], in
1 Python 3.7.0 (default, Jun 28 2018, 08:04:48) [MSC v.1912 64 bit (AMD64)] :: Anaconda, Inc. on win32 2 Type "help", "copyright", "credits" or "license" for more information. 3 >>> t1=tuple(range(1,10,2)) 4 >>> t1 5 (1, 3, 5, 7, 9) 6 >>> t2=tuple(range(2,11,2)) 7 >>> t2 8 (2, 4, 6, 8, 10) 9 >>> t1+t2 10 (1, 3, 5, 7, 9, 2, 4, 6, 8, 10) 11 >>> len(t1) 12 5 13 >>> t2[3] 14 8 15 >>> 9 in t1 16 True 17 >>> 7 in t2 18 False 19 >>> L1=list(range(15)) 20 >>> L1 21 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14] 22 >>> L2=L1*3 23 >>> L2 24 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14] 25 >>> 99 in L2 26 False 27 >>> len(L2) 28 45 29 >>> list(t2)+L1 30 [2, 4, 6, 8, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
〉 列表、元组高级操作
mylist=[1,2,3,4,5]
切片:获得[2,3,4],获得[3,4,5],获得[3,2,1],
获得[1,3,5]
mytpl=(1,2,3,4,5)同上操作
t=‘Mike and Tom‘
split拆分、 join合成为‘Mike/and/Tom
>>> mylist=[1,2,3,4,5] >>> list1=mylist[1:4] >>> list1 [2, 3, 4] >>> list2=mylist[2:] >>> list2 [3, 4, 5] >>> list3=mylist[2:0:-1] >>> list3 [3, 2] >>> list3=mylist[2::-1] >>> list3 [3, 2, 1] >>> list4=mylist[-3::-1] >>> list4 [3, 2, 1] >>> list5=mylist[::2] >>> list5 [1, 3, 5] >>> mytpl=tuple(mylist) >>> mytpl (1, 2, 3, 4, 5) >>> tu1=mytpl[1:4] >>> tu1 (2, 3, 4) >>> tu2=mytpl[2::-1] >>> tu2 (3, 2, 1) >>> tu3=mytpl[::2] >>> tu3 (1, 3, 5) >>> t=‘Mike and Tom‘ >>> t_new=t.split(‘ ‘).join("/") Traceback (most recent call last): File "<stdin>", line 1, in <module> AttributeError: ‘list‘ object has no attribute ‘join‘ >>> t_new=t.split(‘ ‘) >>> t_new [‘Mike‘, ‘and‘, ‘Tom‘] >>> t_n=str(t_new) >>> t_n "[‘Mike‘, ‘and‘, ‘Tom‘]" >>> t_n=t.replace(" ","/") >>> t_n ‘Mike/and/Tom‘
总结:split方法中如果不传参,默认以空格分割。list没有join方法,只有str才有。join()中的参数是个序列,可以是list,tuple,set,但最好不要用set,因为set中的元素是无序的。
原文地址:https://www.cnblogs.com/flyingtester/p/12360964.html
时间: 2024-10-28 15:39:56