Programming Assignment 1

1. 选择题

1. 「C」

2. 「B」

3. 「D」

4.「C」

5.「D」

6.「B 」

7. 「C」

8. 「A」

9. 「A」

10. 「B」

11. 「A」

12. 「C」

13. 「D」

14. 「D」

15. 「C」

16.「A」

17. 「B」

In [13]: for i in s.split():
    ...:     print(i, end=", ")
    ...:
my, name, is, x,

18. 「C」

In [14]: "abcd"[2:]
Out[14]: 'cd'

19. 「B」

20. 「D」

2. Code

This is the Part 2 of your Assignment 1. Together with the Part 1 multiple choice questions on iSpace, it accounts for maximum 10% of the final grade.

这是作业1的第二部分。与iSpace上的第一部分多项选择题一起，它们最多占据期末成绩的10%。

Note

• Write your code after you see # YOUR CODE HERE 「在看到# YOUR CODE HERE后编写你的代码。」
• Read the instruction of each question. You have a limited time to submit: Monday 10 April, 11:00am. Only your last submission counts.「阅读每个问题的说明。你有一个截止提交的有限时间：4月10日星期一上午11:00。只有你最后一次提交的答案会被计入成绩。」
• You should be able to complete the questions within 1 hour (but this is not required now)「你应该能够在1小时内完成这些问题（但现在并非必须）。」
• Copying the solution of other students is forbidden.「禁止抄袭其他学生的答案。」
• For each exercise example, the symbol -> indicates the value the function should return.「对于每个练习示例，符号->表示函数应该返回的值。」
• After the deadline, submission is only possible by email attachment (.ipynb file) to yujiahu@uic.edu.cn. Late submission will be penalized (up to 100%, if late > 48h ).「截止日期后，只能通过电子邮件附件（.ipynb文件）提交给yujiahu@uic.edu.cn。逾期提交将被罚分（如果逾期超过48小时，最高可罚100％）。」
• This assignment will be auto-graded. The auto-grading will check that your answers to the question is correct (or close to be correct). If your submission fails the auto-grade, you will get 0. 「本次作业将进行自动评分。自动评分将检查您对问题的回答是否正确（或接近正确）。如果您的提交未通过自动评分，则将获得0分。」

Question 1

Define the function remove_special(string1) that returns the string1 with the characters ['?', '!'] removed.

定义函数remove_special(string1)，该函数返回删除字符['?', '!']后的string1。

Examples

remove_special("python is such a fun!") -> "python is such a fun"
remove_special("?18?? UIC!") -> "18 UIC"

def remove_special(string1):
    # YOUR CODE HERE

Answer

def remove_special(string1):
    # YOUR CODE HERE
    return string1.replace("?", "").replace("!", "")

def remove_special(string1):
    # YOUR CODE HERE
    return string1.replace("?", "").replace("!", "")


print(remove_special("python is such a fun!"))
print(remove_special("?18?? UIC!"))

Question 2

Given day of the week encoded as 0 = Sun, 1 = Mon, 2 = Tue, ... 6 = Sat, and a boolean vacation indicating if we are on vacation, write the function alarm_clock(day, vacation).

给定一周的某个day，其中0 = 周日，1 = 周一，2 = 周二，... 6 = 周六，以及一个布尔值vacation，表示我们是否在度假，编写函数alarm_clock(day, vacation)。

• If we are not on vacation, 「如果我们不在度假，」

  • on weekdays (Mon-Fri), the function should return a string of the form "7:00"「在工作日（周一至周五），该函数应该返回一个形如"7:00"的字符串。」
  • on weekends (Sat and Sun), the function should return a string of the form "10:00"「在周末（周六和周日），该函数应该返回一个形如"10:00"的字符串。」

• If we are on vacation,「如果我们在度假，」

  • on weekdays (Mon-Fri), the function should return a string of the form "10:00"「在工作日（周一至周五），如果我们在度假，该函数应该返回一个形如"10:00"的字符串。」
  • on weekends (Sat and Sun), the function should return a string of the form "off"「在周末（周六和周日），如果我们在度假，该函数应该返回一个形如"off"的字符串。」

Examples

alarm_clock(3, False) -> "7:00"
alarm_clock(6, False) -> "10:00"
alarm_clock(1, True) -> "10:00"
alarm_clock(6, True) -> "off"

Answer

def alarm_clock(day, vacation):
    # YOUR CODE HERE
    if vacation:
        if 1 <= day <= 5:
            return "10:00"
        else:
            return "off"
    else:
        if 1 <= day <= 5:
            return "7:00"
        else:
            return "10:00"


print(alarm_clock(3, False))
print(alarm_clock(6, False))
print(alarm_clock(1, True))
print(alarm_clock(6, True))

Question 3

Define the function is_prime(num) to check if given positive integer is a prime number or not. The function returns a boolean True if a number is divisible only by 1 and itself, and returns a boolean False if it is divisible by any other number than 1 or itself. 1 is considered as a prime number. 2 is also a prime number.

定义函数is_prime(num)，用于检查给定的正整数是否是质数。如果一个数仅能被1和它本身整除，该函数返回布尔值True，否则，如果它可以被1或它本身以外的其他数整除，该函数返回布尔值False。1被认为是质数，2也是质数。

Examples

is_prime(7) -> True
is_prime(100) -> False
is_prime(1) -> True
is_prime(2) -> True

Answer

def is_prime(num):
    # YOUR CODE HERE
    if num < 1:
        return False
    for i in range(2, num):
        if num % i == 0:
            return False
    return True


print(is_prime(7))
print(is_prime(100))
print(is_prime(1))
print(is_prime(2))

Question 4

Define the function second_highest(list_arg) that returns the second highest number in the list argument list_arg. The input list has at least two unique numbers.

定义函数second_highest(list_arg)，该函数接受一个列表参数list_arg，并返回该列表中第二大的数字。输入列表至少包含两个不同的数字。

Examples

second_highest([1,2,3,4,5]) -> 4
second_highest([1,2,3,5,5]) -> 3

Answer

def second_highest(list_arg):
    # YOUR CODE HERE
    s = set(list_arg)
    numbers = sorted(s, reverse=True)
    return numbers[1]


def second_highest(list_arg):
    unique_numbers = sorted(set(list_arg), reverse=True)
    return unique_numbers[1]


# 测试用例
print(second_highest([1, 2, 3, 4, 5]))  # 输出: 4
print(second_highest([1, 2, 3, 5, 5]))  # 输出: 3

Question 5

Define the function intersect_list(list1, list2) that returns a list whose element are common elements between list1 and list2 without duplicates. You may want to convert list1 and list2 to set first.

定义一个函数 intersect_list(list1, list2)，返回一个列表，其中的元素是 list1 和 list2 之间的公共元素，且不包含重复元素。你可能需要先将 list1 和 list2 转换为集合。

Examples

list1 = [1, 2, 3, 4, 5]
list2 = [5, 6, 7, 8, 9]
intersect_list(list1, list2) -> [5]

list3 = [1, 2, 3, 4, 5, 5, 6, 7]
list4 = [5, 5, 6, 7, 8, 9]
intersect_list(list3, list4) -> [5, 6, 7]

Answer

def intersect_list(list1, list2):
    # YOUR CODE HERE
    set1 = set(list1)
    set2 = set(list2)
    intersection = set1.intersection(set2)
    return list(intersection)


def intersect_list(list1, list2):
    # YOUR CODE HERE
    set1 = set(list1)
    set2 = set(list2)
    intersection = set1 & set2
    return list(intersection)


# Test cases
list1 = [1, 2, 3, 4, 5]
list2 = [5, 6, 7, 8, 9]
print(intersect_list(list1, list2))  # Output: [5]

list3 = [1, 2, 3, 4, 5, 5, 6, 7]
list4 = [5, 5, 6, 7, 8, 9]
print(intersect_list(list3, list4))  # Output: [5, 6, 7]

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