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python - 给定字典中的键，提取 x 个先前的键值对

问题描述

我有一本这样的字典：

 dict_test ={1: 111, 2: 2222, 3:333, 4:4444, 5:5555, 6:6666,
        7: 777, 8: 8888, 9:9999, 10:100010101}

并想创建一个字典的子集，例如给定键 8 的前四个值。因此，生成的预期字典将如下所示：

 dict_new ={4:4444, 5:5555, 6:6666, 7: 777, 8: 8888}

我尝试在下面编写一个更通用的函数，在那里我可以更普遍地确定我应该回顾多少以前的值。

def get_x_prev_entries(dictionary: dict, key: str, prev: int):
   if key in dictionary:
      token = object()
      keys = [token]*(prev*-1) + sorted(dictionary) + [token]*diff
      print('keys' + str(keys))
      new_dict = []
      newkeys = []
      new_prev= prev
      # extract all keys that are between 0 and the specified difference
      while new_prev is not 0:
          new_prev -= 1
          if len(newkeys) == 0:
              newkeys= newkeys
          else:
              newkeys = newkeys.append(keys[keys.index(key)-new_diff])
              print(newkeys)
          print(new_diff)

      new_dict = {k:v for k, v in dictionary.items() if k in newkeys}
      return new_dict
  else:
      print('Key not found')

所以要创建我想要的字典，我最好输入

get_x_prev_entries(dict_test, 8, 4)

但此时我只返回一个空字典。任何意见，将不胜感激。谢谢！

标签： pythondictionary

解决方案

使用命令字典

from collections import OrderedDict as od

dict_test ={1: 111, 2: 2222, 3:333, 4:4444, 5:5555, 6:6666, 7: 777, 8: 8888, 9:9999, 10:100010101}

od_dict = od(dict_test)

def get_previous_keys(od_dict, prev=4, given=8):
    if given not in od_dict:
        return
    k, v = [], []
    for i in range(given-prev, given):
        k.append(i)
        v.append(od_dict[i])
    return dict(zip(k,v))

print(get_previous_keys(od_dict))

{4: 4444, 5: 5555, 6: 6666, 7: 777}

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