举一反三
- ls = [3.5, "Python", [10,"LIST"], 3.6],ls[2][-1][1]的运行结果是:
- 已知列表ls = [7,"Python", [8,"LIST"], 9],则ls[2][-1][3]的运行结果是( ) A: [8,"LIST"] B: "T" C: "LIST" D: "Python"
- 获取列表list=[1,2,3,4]第一个元素正确的方法是()。 A: list[-1] B: list[1] C: list[0] D: list[[1]]
- ls = [3.5, "Python", [10, "LIST"], 3.6]<br/>ls[2][–1][1]的运行结果是 A: I B: P C: Y D: L
- 补全代码,使输出结果为True Truea = 1b = "2"list = [1,2,3,4,5]print(a _____ list,b ____ list)
内容
- 0
dict={'a':1, 'b':2, 'c':3} str=list(dict.keys())[list(dict.values()).index(2)] print(str) 输出结果是( ) A: a B: C: c D: 2
- 1
Python语句pint(type([]))的输出结果是() 未知类型:{'options': ["[ class 'tuple‘]", "[class'dict']", "[ class 'set]", "[class 'list']"], 'type': 102}
- 2
ls = [3.5, "Python", [10, "LIST"], 3.6],ls[2][ –1][1]的运行结果是 A: L B: I C: P D: Y
- 3
59ls=[3.5,"Python",[10,"LIST"],3.6],ls[2][–1][1]的运行结果是 A: I B: Y C: L D: P
- 4
59ls=[3.5,"Python",[10,"LIST"],3.6],ls[2][–1][1]的运行结果是 A: I B: Y C: L D: P