Question

如果我有2个列表，例如：

主题和功劳：

[['Math',6],['English',3],['Drama',2]]

学生和成绩：

[['Amy',,90,80],['Jack','60','30','50'],['Tom','80','','50']]

，并希望像这样计算它们：

对于艾米，数学没有分数。英语90分（3学分），戏剧80分（2学分）因此它将输出

[['Amy',,270,160],['Jack',360,90,100],['Tom'480,,100]]

编辑：

我自己解决了。我只需要使用类似的东西

a[i][j] = b[i][j] *c [i][1]

Answer 1

请执行以下操作：

void YourDSP::process(AUAudioFrameCount frameCount, AUAudioFrameCount bufferOffset) {

    for (int frameIndex = 0; frameIndex < frameCount; ++frameIndex) {
        int frameOffset = int(frameIndex + bufferOffset);

        float temp = 0;
        for (int channel = 0; channel < _nChannels; ++channel) {
            float *out = (float *)_outBufferListPtr->mBuffers[channel].mData + frameOffset;

            *out = ... whatever your magic is ...
        }
    }
}

也就是说，您可能应该实现这种更简单的方法，而无需列表理解。

Answer 2

如果要处理学生和分数，则可能要使用词典或对象而不是列表。例如：

from typing import List

class Student(object):
    def __init__(self, name: str):
        self.name = name
        self.subjects_ = []

    def __repr__(self):
        return self.name

    def add_subject(self, subject_, score_: int):
        subject = Subject(subject_, score_)
        self.subjects_.append(subject)
        print(f"added {subject_} to {self.name} with a score of {score_}")

    def remove_subject(self, subject_):
        self.subjects_.remove(subject_)

    def subjects(self):
        return [s.name for s in self.subjects_]`

class Subject(object):
    def __init__(self, name: str, credits: int):
        self.name = name
        self.credits = credits
    def __repr__(self):
        return self.name

students = ['Amy', 'Jack', 'Tom']
math = Subject('math', 4)
history = Subject('history', 8)

amy = Student(students[0])
amy.add_subject(math,10)
amy.add_subject(history,6)

print(amy.subjects())

给出2个嵌套列表，并计算最里面的一个元素

2 个答案: