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单选题 What will be the output of the following Python code? s="a@b@c@d" a=list(s.partition("@")) print(a) b=list(s.split("@")) print(b)
学科：Python程序设计（双语）
时间：2023-11-24 00:22:56
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题目1单选题
What will be the output of the following Python code? a=[13,56,17] a.append([87]) a.extend([45,67]) print(a)A. [13, 56, 17, 87, 45, 67]B. [13, 56, 17, [87], [45, 67]]C. [13, 56, 17, 87,[ 45, 67]]D. [13, 56, 17, [87], 45, 67]
题目2单选题
The value of ls after the assign statement ls=[m+n for m in 'AB' for n in 'CD'] isA. ['AC', 'AD', 'BC', 'BD']B. ABCDC. 错误D. AABBCCDD
题目3单选题
What will be the output of the following Python code? >>>list("a#b#c#d".split('#'))A. ['a#b#c#d']B. ['a b c d']C. ['abcd']D. ['a', 'b', 'c', 'd']
题目4单选题
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], which output of the following choices is correctA. >>> numbers[: -1] [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]B. >>> numbers[-2:] [9, 10]C. >>> numbers[0::3] [1, 3, 5, 7, 9]D. >>> numbers[0: 2] [1, 2, 3]
题目5单选题
ls=[1,2,3,4,5],which of the following choices could NOT remove the first element from the listA. ls.remove(1)B. ls.pop(1)C. ls = ls[1:]D. del ls[0]
题目6单选题
What will be the output of the following Python code? >>> s=["pune", "mumbai", "delhi"] >>> [(w.upper(), len(w)) for w in s]A. [('PUNE', 4), ('MUMBAI', 6), ('DELHI', 5)]B. ErrorC. ['PUNE', 4, 'MUMBAI', 6,'DELHI', 5]D. [PUNE, 4, MUMBAI, 6, DELHI, 5]
题目7单选题
What will be the output of the following Python code? num = ['One', 'Two', 'Three'] for i, x in enumerate(num): print('{}: {}'.format(i, x),end=" ")A. Exception is thrownB. 0: One 1: Two 2: ThreeC. One Two ThreeD. 1: 2: 3
题目8单选题
What will be the output of the following Python code snippet? ['hello', 'morning'][bool('')]A. errorB. no outputC. morningD. hello
题目9单选题
What will be the output of the following Python code? >>> l=["good", "oh!", "excellent!", "#450"] >>> [n for n in l if n.isalpha() or n.isdigit()]A. ['oh!', 'excellent!', '#450']B. ['good']C. ['good', '#450']D. ['good', 'oh', 'excellent', '450' ]
题目10单选题
What will be the output of the following Python code? s="a@b@c@d" a=list(s.partition("@")) print(a) b=list(s.split("@")) print(b)A. ['a', '@', 'b@c@d'] ['a', 'b', 'c', 'd']B. ['a', '@', 'b@c@d'] ['a', '@', 'b', '@', 'c', '@', 'd']C. ['a', 'b', 'c', 'd'] ['a', 'b', 'c', 'd']D. ['a', '@', 'b', '@', 'c', '@', 'd'] ['a', 'b', 'c', 'd']
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