Third week of Python study notes (2)

Select Sort:

• Time complexity O (n**2)
• There is no way to know whether the current wheel has reached the sorting requirements, but it is possible to know whether the extremum is at the target index position
• Traversal count 1,..., n-1 N (n-1)/2
• Contrast Bubble method: Reduce the number of exchanges, improve efficiency, slightly better performance
• Method three or four actually reduces the average time complexity

Method One:

nums = [1, 2, 6, 7, 8, 9, 3, 4, 5]for i in range(len(nums)):    maxindex = i    for j in range(i + 1, len(nums)):        if nums[j] < nums[maxindex]:            maxindex = j    if i != maxindex:        nums[i], nums[maxindex] = nums[maxindex], nums[i]print(nums)       

Method two (based on optimization of method one):

• Optimization point: Internal loop one-time traversal simultaneously fetch maximum, minimum value at both ends

  nums = [7, 19, 15, 12, 15, 5, 13, 14, 8, 15]  length = len(nums)for i in range(length // 2):maxindex = iminindex = length - i - 1minorigin = length - i - 1for j in range(i + 1, length - i):    if nums[j] > nums[maxindex]:        maxindex = j    if nums[j-1] < nums[minindex]:#正序比较 或者逆序比较都可以 length - j - 1        minindex = j - 1if i != maxindex:    nums[i], nums[maxindex] = nums[maxindex], nums[i]if i == minindex: #如果i就是最小值索引，则执行此if，因为上一行索引i的值与最大值已经进行了交换    minindex = maxindexif minorigin != minindex:    

Method Three (based on the optimization of Method two):

• Optimization point: If the maximum value of one iteration is equal to the minimum value, the remaining element values are the same, that is, sorted

  nums = [7, 19, 15, 12, 15, 5, 13, 14, 8, 15] length = len(nums)for i in range(length // 2):maxindex = iminindex = length - i - 1minorigin = length - i - 1for j in range(i + 1, length - 1):    if nums[j] > nums[maxindex]:        maxindex = j    if nums[length - j - 1] < nums[minindex]:        minindex = length - j - 1if minindex == maxindex: #!    breakif i != maxindex:    nums[i], nums[maxindex] = nums[maxindex], nums[i]if i == minindex:    minindex = maxindexif minorigin != minindex:    nums[minorigin], nums[minindex] = nums[minindex], nums[minorigin]print(nums)

Method Four (based on Method three optimization):

• If the list is the case below, the minimum value does not need to be exchanged

  nums =[1, 1, 1, 1, 1, 1, 1, 1, 2]length = len(nums)for i in range(length // 2):maxindex = iminindex = length - i - 1minorigin = length - i - 1for j in range(i + 1, length - 1):    if nums[j] > nums[maxindex]:        maxindex = j    if nums[length - j - 1] < nums[minindex]:        minindex = length - j - 1if minindex == maxindex: #!    breakif i != maxindex:    nums[i], nums[maxindex] = nums[maxindex], nums[i]if i == minindex:    minindex = maxindexif minorigin != minindex and nums[minorigin] != nums[minindex]: #!    nums[minorigin], nums[minindex] = nums[minindex], nums[minorigin]print(nums)

Problem solving 1. User enters a number

• print each digit and its number of repetitions method one:

  num = input(‘>>>‘)d = {}for c in num:if not d.get(c):    d[c] = 1    continued[c] += 1print(d)

Method Two:

d ={}for c in num:    if c not in d.keys():        d[c] = 1    else:        d[c] += 1print(d)

2. Digital Repeat Statistics

• Randomly generates 100 integers
• range of numbers [-1000, 1000]
• Output all the different numbers in ascending order and the number of repetitions

  import randomn = 100nums = [0] * nfor i in range(n):nums[i] = random.randint(-1000, 1000)print(nums)    t = nums.copy()t.sort()print(t)d = {}for x in nums:if x not in d.keys():    d[x] = 1else:    d[x] += 1print(d)d1 = sorted(d.items())print(d1)

3. String Repetition statistics

• Character ' ABCDEFGHIJKLMNOPQRSTUVWXYZ '
• Randomly select 2 letters to form a string and select 100
• Output all different strings in descending order and the number of repetitions

  import randomalphabet = ‘abcdefghijklmnopqrstuvwxyz‘words = []for _  in range(100):words.append(‘‘.join(random.choice(alphabet) for _ in range(2)))d = {}for x in words:d[x] = d.get(x, 0) + 1print(d)d1 = sorted(d.items(), reverse = True)print(d1)

4. Returns a list of 1-10 squares

[i ** 2 for i in range(1,11)]

5. There is a list of LST = [1,4,9,16,2,5,10,15], which generates a new list that requires the new list element to be a lst adjacent to 2 items and

lst = [1,4,9,16,2,5,10,15][lst[i] + lst[i + 1] for i in range(len(lst) - 1)]

6. Print 99 multiplication table

[print(‘{}*{}={:<{}}{}‘.format(j, i, j * i,2 if j ==1 else 3, ‘‘ if i != j else ‘\n‘),end = ‘‘)for i in range(1,10) for j in range(1,i + 1)]

7. "0001.abadicddws" is the ID format, requires ID format is divided by Dot, the left is 4 bits from 1, the right is a 10-bit random lowercase English letter. Please generate a list of the first 100 IDs in turn

import random[(‘{:>04}‘.format(i) + ‘.‘ + ‘‘.join(‘abcdefghijklmnopqrstuvwxyz‘[random.randint(0,25)] for _ in range(10))) for i in range(1,101)]

Third week of Python study notes (2)