数据的迭代(1.1)

Posted 2021-02-04 wangjiale1024

列表

1. filter + lambda函数

   >>> import random
   >>> l1 = [random.randint(-10, 10) for i in range(10)]
   >>> print(l1)
   [3, 8, -5, 4, -4, -5, -7, -8, -6, -7]
   >>> print(list(filter(lambda x: x>0, l1)))
   [3, 8, 4]

1. 迭代器

  >>> print([x for x in l1 if x>0])
  [3, 8, 4]

集合

1. 迭代器

   >>> s1 = set(random.randint(-10, 10) for i in range(10))
   >>> print(s1)
   {-10, -5, -1, 0, 3, 8, 9}
   >>> print({x for x in s1 if x >0})
   {8, 9, 3}

字典

• 产生随机字典

      # 产生随机字典
      letters = ‘abcdefghijklmnopqrstuvwxyz‘
      l2 = []
      l1 = [0 for i in range(len(letters))]
      d1 = {}
      for got_num in range(len(letters)):
          temp_int = random.randint(got_num+1, len(letters)) - got_num # 表示第temp_int个0，从1开始
          current_num,i  = 0, -1
          while current_num < temp_int:
              i += 1
              current_num += 1 if l1[i] == 0 else 0
          # i 就是我们要找的位置
          l2.append(i)
          l1[i] = 1
  
      d1 = {letters[x]: random.randint(-10, 10) for x in l2}
      # 打印一下
      print(d1)

• d1的迭代

  >>> d2 = {k:v for k,v in d1.items() if v>0}
  >>> print(d2)
      {‘a‘: 9,
       ‘b‘: 3,
       ‘c‘: 5,
       ‘d‘: 10,
       ‘f‘: 3,
       ‘i‘: 8,
       ‘j‘: 2,
       ‘n‘: 5,
       ‘v‘: 1,
       ‘w‘: 10,
       ‘x‘: 3}