Question

我试图学习Python的一些功能方面。我正在寻找一种可以转化的理解：

a = {'name': 'school a', 'grades': [3, 4, 5]}
b = {'name': 'school b', 'grades': [3, 4, 5]}
c = {'name': 'school c', 'grades': [6, 7, 8]}

为：

schools_by_grades = {3: [a, b], 4: [a, b], 5: [a, b], 6: [c], 7: [c], 8: [c]}

我能够为a和c创建此内容，但分两步：

schools_by_grade = {grade: [a] for grade in a['grades']}
schools_by_grade.update({grade: [c] for grade in c['grades']})

关于如何做到这一点的任何想法？

Answer 1

势在必行的是Pythonic：

d = defaultdict(lambda: [])
for school in a, b, c:
    for g in school['grades']:
        d[g].append(school)

这是＆＃34;功能性＆＃34;方法，但正如预测的那样，它并不漂亮：

fst = lambda (x,_): x
grade_to_school = ((g,x) for x in a,b,c for g in x['grades'])
d = { g : list(y) for g,y in groupby(sorted(grade_to_school, key=fst), key=fst) }

Answer 2

你可以这样做：

schools_by_grades = {i: [school['name'] for school in (a, b, c) 
                         if i in school['grades']] 
                     for i in range(20) if any(i in school['grades'] 
                                               for school in (a, b, c))}

但你可能不应该。

这给了我：

{3: ['school a', 'school b'], 
 4: ['school a', 'school b'], 
 5: ['school a', 'school b'], 
 6: ['school c'], 
 7: ['school c'], 
 8: ['school c']}

Answer 3

如果具体如此，我建议为该操作制作一个功能。

def AddSchoolsByGrades(schools_by_grades, school_dict):
    # if there is a space in the name, it will take the last word of the name
    name = school_dict["name"].split(" ")[-1]
    for g in school_dict["grades"]:
        # if there is no entry for the grade, then make one
        if not g in schools_by_grades.keys():
            schools_by_grades[g] = []
        # add the name of the school to the grade
        schools_by_grades[g].append(name)
        # simple trick to remove any duplicates
        schools_by_grades[g] = list(set(schools_by_grades[g]))

a = {'name': 'school a', 'grades': [3, 4, 5]}
b = {'name': 'school b', 'grades': [3, 4, 5]}
c = {'name': 'school c', 'grades': [6, 7, 8]}

schools_by_grades = {}
AddSchoolsByGrades(schools_by_grades, a)
AddSchoolsByGrades(schools_by_grades, b)
AddSchoolsByGrades(schools_by_grades, c)

功能Python - 一个词典的多个词典

3 个答案: