如何访问具有多个值的字典中的键？答案

【问题标题】：How to access keys in a dictionary with multiple values?如何访问具有多个值的字典中的键？
【发布时间】：2021-06-14 12:12:28
【问题描述】：

这是返回值中包含“10”的字典键的代码。

word_freq = {'is': [1, 3, 4, 8, 10],
         'at': [3, 10, 15, 7, 9],
         'test': [5, 3, 7, 8, 1],
         'this': [2, 3, 5, 6, 11],
         'why': [10, 3, 9, 8, 12]}
# Check if a value exist in dictionary with multiple value
value = 10
# Get list of keys that contains the given value
  list_of_keys = [key
            for key, list_of_values in word_freq.items()
            if value in list_of_values]
if list_of_keys:
   print(list_of_keys)
else:
   print('Value does not exist in the dictionary')

输出：['is', 'at', 'why']

我想修改它以查找多个值，例如“10”和“3”，并返回值中包含这些数字的键。

请提出解决方案

【问题讨论】：

看or命令。

标签： python list dictionary key-value

【解决方案1】：

您可以根据您的用例使用all 或any：

values = [10, 3, 9]
# Get list of keys that contains the given value
list_of_keys = [key
            for key, list_of_values in word_freq.items()
            if all(val in list_of_values for val in values)]

list_of_keys = [key
            for key, list_of_values in word_freq.items()
            if any(val in list_of_values for val in values)]

【讨论】：

【解决方案2】：

将值用作数组而不是单个数字。

word_fre[q = {'is': [1, 3, 4, 8, 10],
         'at': [3, 10, 15, 7, 9],
         'test': [5, 3, 7, 8, 1],
         'this': [2, 3, 5, 6, 11],
         'why': [10, 3, 9, 8, 12]}
# Check if a value exist in dictionary with multiple value
value = [10]
# Get list of keys that contains the given value
list_of_keys = []
for i in value:
    for key, list_of_values in word_freq.items():
        if i in list_of_values:
            list_of_keys.append(key)
    
if list_of_keys:
   print(set(list_of_keys))#this gives all the values only once
else:
   print('Value does not exist in the dictionary')

【讨论】：

【解决方案3】：

使用intersection():

word_freq = {'is': [1, 3, 4, 8, 10],
         'at': [3, 10, 15, 7, 9],
         'test': [5, 3, 7, 8, 1],
         'this': [2, 3, 5, 6, 11],
         'why': [10, 3, 9, 8, 12]}
# Check if a value exist in dictionary with multiple value
values = {10, 3}
# Get list of keys that contains the given value
  list_of_keys = [key
            for key, list_of_values in word_freq.items()
            if values.intersection(set(list_of_values))]
if list_of_keys:
   print(list_of_keys)
else:
   print('Values does not exist in the dictionary')

【讨论】：

【解决方案4】：

这似乎工作正常。根据需求进行微调。

word_freq = {'is': [1, 3, 4, 8, 10],
         'at': [3, 10, 15, 7, 9],
         'test': [5, 3, 7, 8, 9],
         'this': [2, 3, 5, 6, 11],
         'why': [10, 3, 9, 8, 12]}

value = [10, 1]

out = []

for check_val in value :
  print("\nChecking for value : ", check_val)
  for key in word_freq.keys() :
    print("Checking for key : " , key)
    if check_val in word_freq[key] :    
      out.append(key)
  print(out)

print("Final keyset : ", set(out))

输出相同：

【讨论】：