3.1 tuple (2,3) and the set { "four", 5,6} a list of synthetic
1 tuple,set,list = (1,2,3),{"four",5,6},[] 2 for i in tuple: 3 list.append(i) 4 for j in set: 5 list.append(j) 6 print(list)
3.2 List [3,7,0,5,1,8] is greater than the element 5 is set to 0, is less than 5 element is set to 1
1 list2 = [3,7,0,5,1,8] 2 print(list2) 3 for i in range(0,len(list2)): 4 if list2[i] >5: 5 list2[i] = 0 6 elif list2[i]<5: 7 list2[i]=1 8 print(list2)
3.3 List [ "mo", "deng", "ge"] and [2,3] is converted to [( "mo", 1), ( "deng", 2), ( "ge", 3) ]
1 # Method a: traversing element method 2 Sl1, Nl1, new_list1 = [ " Mo " , " Deng " , " GE " ], [l, 2,3 ], [] . 3 for I in Sl1: . 4 for J in Nl1 : . 5 IF Sl1.index (I) == Nl1.index (J): . 6 new_list1.append ((I, J)) . 7 Print ( " new_list1 = " , new_list1) . 8 . 9 # standard method the traversal: method two 10 Sl2, Nl2, new_list2 = ["mo","deng","ge"],[1,2,3],[] 11 for a in range(0,len(Sl2)): 12 for b in range(0,len(Nl2)): 13 if a == b: 14 new_list2.append((Sl2[a],Nl2[b])) 15 print("new_list2=",new_list2) 16 17 #方法三：切片组合法 18 Sl3,Nl3=["mo","deng", " GE " ], [l, 2,3 ] . 19 Print ( " new_list3 = " , [(Sl3 [0], Nl3 [0]), (Sl3 [. 1], Nl3 [. 1]), (Sl3 [2 ], Nl3 [2 ])]) 20 is 21 is # four: the subscript opportunistic traversal method 22 is SL4, Nl4, new_list4 = [ " Mo " , " Deng " , " GE " ], [l, 2,3 ], [ ] 23 is for K in Range (0,3 ): 24 new_list4 + = [(SL4 [K], Nl4 [K])] 25 Print ( " new_list4 =" , New_list4)
26 is # run results: 27 " "" 28 new_list1 = [( 'Mo',. 1), ( 'Deng', 2), ( 'GE',. 3)] 29 new_list2 = [( 'Mo', . 1), ( 'Deng', 2), ( 'GE',. 3)] 30 new_list3 = [( 'Mo',. 1), ( 'Deng', 2), ( 'GE',. 3)] 31 is new_list4 = [( 'Mo',. 1), ( 'Deng', 2), ( 'GE',. 3)] 32 "" "
3.4 If a = dict (), so that b = a, performing b.update ({ "x": 1}), a change also, why, how to avoid
Reason: a variable to another variable is equivalent to the value of two variables refer to the same stored address
Solution: re-open space can cancel the association between two variables (what each expression will have value will re-open space, the value depends on the variable name referenced is assigned to it)
1 # Method a: copy () function copies 2 A = {1: " Mo " , 2: " Deng " } . 3 B = a.copy () . 4 b.update ({ " X " : " / " }) . 5 Print (A, B) . 6 . 7 # method two: unpacking assignment method . 8 A = {. 1: " Mo " , 2: " Deng " } . 9 B = dict () 10 b.update (A) . 11 b.update ( { " the X-":"/"}) 12 print(a,b) 13 14 #运行结果： 15 """ 16 {1: 'mo', 2: 'deng'} {1: 'mo', 2: 'deng', 'x': '/'} 17 {1: 'mo', 2: 'deng'} {1: 'mo', 2: 'deng', 'x': '/'} 18 """
3.5 two-dimensional structure of [[ 'a', 1], [ 'b', 2]], and (( 'x', 3), ( 'y', 4)) is converted into the dictionary
1 # two-dimensional structure of [[ "a", "/ "], [ "b", 2]] , and (( "x", 3) , ( "y", 4)) is converted into the dictionary 2 List1, tuple1 = [[ " A " , " / " ], [ " B " , 2]], (( " X " ,. 3), ( " Y " ,. 4 )) . 3 dict1 = dict (List1) . 4 dict2 = dict ( tuple1) . 5 Print (dict1, dict2) . 6 # run results: . 7 "" " . 8 { 'A': '/', 'B':} {2 'X':. 3, 'Y': 4} 9 """
3.6
3.7