Python 避免字典和元组的多重嵌套问题

Python 避免字典和元组的多重嵌套问题

目录

一、字典、元组的多重嵌套

二、嵌套结构重构为类

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

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:

① Gradebook()类,学生名是字典self._students的键;科目、成绩、权重是self._grades的值。

② Student()类,科目是字典self._subjects的键;成绩、权重是self._subjects的值。

③ 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。

''' 学习中遇到问题没人解答?小编创建了一个Python学习交流群:711312441 寻找有志同道合的小伙伴,互帮互助,群里还有不错的视频学习教程和PDF电子书! ''' from collections import namedtuple 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。

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