如何根据最小值筛选字典?
grades = {"Bob": 46, "Angela":73, "Dave": 94}我想做一本新的字典,它只适用于70岁以上的人。
passing = {"Angela":73, "Dave":94}发布于 2020-06-08 07:14:27
没有那么复杂,只要跟着这个,你就可以走了。我们可以用两种方法来做
grades = {"Bob": 46, "Angela":73, "Dave": 94}
# new dictionary to get what you want
# using dictionary comprehension in python
passing = {key:value for key, value in grades.items() if value > 70}
print(passing)输出
{'Angela': 73, 'Dave': 94}grades = {"Bob": 46, "Angela":73, "Dave": 94}
passing = {}
# traversing through dictionary items
for key, value in grades.items():
# check for passing values
if value > 70: passing[key] = value
print(passing)输出将是相同的。我建议开始学习Dictionary Comprehension,这是一种更方便、更简洁的编码方式。快乐编码:)
发布于 2020-06-08 07:12:17
试着对分数大于70的分数进行筛选:
>>> grades = {"Bob": 46, "Angela":73, "Dave": 94}
>>> {k: v for k, v in grades.items() if v > 70}
{'Angela': 73, 'Dave': 94}发布于 2020-06-08 07:12:34
你可以用字典理解。将新字典存储在同一个变量中
grades = {k:v for k,v in grades.items() if v > 70}将新字典存储在另一个变量中
another = {k:v for k,v in grades.items() if v > 70}输出:
{'Angela': 73, 'Dave': 94}https://stackoverflow.com/questions/62256660
复制相似问题