Python中如何避免字典和元组的多重嵌套的方法

一、字典、元组的多重嵌套

例 1：记录全班学生的成绩。

分析：定义一个 SimpleGradebook类，

学生名是字典self._grades的键，成绩是字典self._grades的值。

class SimpleGradebook():
 def __init__(self):
 self._grades = {}
 def add_student(self, name):
 self._grades[name] = []
 def report_grade(self, name, score):
 self._grades[name].append(score)
 def average_grade(self, name):
 grades = self._grades[name]
 return self._grades, sum(grades) / len(grades)

book = SimpleGradebook()
book.add_student('qinlu')
book.report_grade('qinlu', 99)
print(book.average_grade('qinlu'))

({'qinlu': [99]}, 99.0)

字典可能因为功能过多导致结果多重嵌套。

例 2：扩充 SimpleGradebook类，按科目保存成绩。

分析：定义一个 BySubjectGradebook类，字典by_subject嵌套在字典self._grades内。

学生名是字典self._grades的键，科目、成绩是self._grades的值。

科目是字典by_subject的键，成绩是字典by_subject的值。

class BySubjectGradebook():
 """
 report_grade(), average_grade()嵌套了两层的字典
 """
 def __init__(self):
 self._grades = {}
 def add_student(self, name):
 self._grades[name] = {}
 def report_grade(self, name, subject, score):
 by_subject = self._grades[name]
 grade_list = by_subject.setdefault(subject, [])
 grade_list.append(score)
 def average_grade(self, name):
 by_subject = self._grades[name]
 total, count = 0, 0
 for scores in by_subject.values():
 total += sum(scores)
 count += len(scores)
 return self._grades, total / count

book = BySubjectGradebook()
book.add_student('qinlu')
book.report_grade('qinlu', 'Math', 99)
book.report_grade('qinlu', 'Math', 88)
book.report_grade('qinlu', 'Computer', 90)
book.report_grade('qinlu', 'Computer', 80)
print(book.average_grade('qinlu'))

({'qinlu': {'Math': [99, 88], 'Computer': [90, 80]}}, 89.25)

例 3：需求变更，需记录每次成绩占总成绩的权重。（期中、期末考试所占的分量比随堂考大）

class WeightedGradebook():
 def __init__(self):
 self._grades = {}
 def add_student(self, name):
 self._grades[name] = {}
 def report_grade(self, name, subject, score, weight):
 by_subject = self._grades[name]
 grade_list = by_subject.setdefault(subject, [])
 grade_list.append(score, weight)
 def average_grade(self, name):
 by_subject = self._grades[name]
 score_sum, score_count = 0, 0
 for subject, scores in by_subject.items():
 subject_avg, total_weight = 0, 0
 for score, weight in scores:
 #...
 return score_sum / score_count

book = WeightedGradebook()
book.add_student('qinlu')
book.report_grade('qinlu', 'Math', 99, 0.1)
book.report_grade('qinlu', 'Math', 88, 0.6)
book.report_grade('qinlu', 'Computer', 90, 0.1)
book.report_grade('qinlu', 'Computer', 80, 0.6)
print(book.average_grade('qinlu'))

该代码出现字典、元组的多层嵌套，应拆解为类。多层嵌套的代码，很难维护。

二、嵌套结构重构为类

将下面的字典重构为类。

字典by_subject嵌套在字典self._students内。

{'qinlu': {'Math': [(99, 0.1), (88, 0.9)], 'Computer': [(90. 0.1), (80, 0.9)]}}

分析：

• Gradebook()类，学生名是字典self._students的键；科目、成绩、权重是self._grades的值。

• Student()类，科目是字典self._subjects的键；成绩、权重是self._subjects的值。

• Subject()类，成绩是列表self._grades的第一位；权重是列表self._grades的第二位。

从最底层开始重构，即考试成绩。这么简单的信息，没必要写成类。

namedtuple()命名元组。

from collections import namedtuple
#学习中遇到问题没人解答？小编创建了一个Python学习交流群：153708845
Grade = namedtuple('Grade', ('score', 'weight'))
# 科目类，该类包含考试成绩
class Subject():
 def __init__(self):
 self._grades = []
 def report_grade(self, score, weight):
 self._grades.append(Grade(score, weight))
 def average_grade(self):
 total, total_weight = 0, 0
 # print(self._grades)
 for grade in self._grades:
 # print(grade)
 total += grade.score * grade.weight
 total_weight += grade.weight
 return total / total_weight
# 学生类，该类包含学习课程
class Student():
 def __init__(self):
 self._subjects = {}
 def subject(self, name):
 if name not in self._subjects:
 self._subjects[name] = Subject()
 return self._subjects[name]
 def average_grade(self):
 total, count = 0, 0
 for subject in self._subjects.values():
 total += subject.average_grade()
 count += 1
 return total / count
# 成绩册类，包含所有学生考试成绩的容器类，该容器类以学生名字为键，可动态添加学生
class Gradebook():
 def __init__(self):
 self._students = {}
 def student(self, name):
 if name not in self._students:
 self._students[name] = Student()
 return self._students[name]
book = Gradebook()
qin = book.student('qinlu')
math = qin.subject('Math')
math.report_grade(99, 0.1)
math.report_grade(88, 0.9)
print(qin.average_grade())

89.1

虽然代码量是原来的两倍，但更清晰，更易扩展，理解起来比原来容易。