Question

以下代码打印学生编号和学生分数。

def main():
    answers = [
        ['A', 'B', 'A', 'C', 'C', 'D', 'E', 'E', 'A', 'D'],
        ['D', 'B', 'A', 'B', 'C', 'A', 'E', 'E', 'A', 'D'],
        ['E', 'D', 'D', 'A', 'C', 'B', 'E', 'E', 'A', 'D'],
        ['C', 'B', 'A', 'E', 'D', 'C', 'E', 'E', 'A', 'D'],
        ['A', 'B', 'D', 'C', 'C', 'D', 'E', 'E', 'A', 'D'],
        ['B', 'B', 'E', 'C', 'C', 'D', 'E', 'E', 'A', 'D'],
        ['B', 'B', 'A', 'C', 'C', 'D', 'E', 'E', 'A', 'D'],
        ['E', 'B', 'E', 'C', 'C', 'D', 'E', 'E', 'A', 'D']]

    keys = ['D', 'B', 'D', 'C', 'C', 'D', 'A', 'E', 'A', 'D']

    for i in range(len(answers)):
        correctCount = 0
        for j in range(len(answers[i])):
            if answers[i][j] == keys[j]:
                correctCount += 1


        results = [i,correctCount]
        print(results, end = "") # this line used to be - print("Student", i, "'s correct count is", correctCount)


main()

我的目标是将学生分数从最低到最高排序，而不会混淆与该分数相关联的学号。我试图以这些形式使用sort函数：

results1 = sorted(results, key = lambda x: x[1])
results.sort(key = itemgetter(1))
results.sort()

使用所有这三个尝试我仍然无法使其工作。有任何想法吗？

预期产出：

[3, 4]
[2, 5]
[1, 6]
[0, 7]
[5, 7]
[6, 7]
[7, 7]
[4, 8]

第一列是学生＃，第二列是分数。

Answer 1

您没有构建results，results = [i, correctCount]的列表，只需覆盖结果，您需要：

results = []
for i in range(len(answers)):
    correctCount = 0
    for j in range(len(answers[i])):
        if answers[i][j] == keys[j]:
            correctCount += 1
    results.append([i, correctCount])
results = sorted(results, key=lambda x: x[1])

然后您的排序代码应该有效。

然而，python中更惯用的是迭代实际列表而不是range(len(list))，这通常可以简化代码，上面的内容可以重写：

results = []
for student, student_answers in enumerate(answers):
    correctCount = 0
    for answer, key in zip(student_answers, keys):
        if answer == key:
            correctCount += 1
    results.append([student, correctCount])
results = sorted(results, key=lambda x: x[1])

哪个比原始代码更具可读性您还可以使用理解和生成器来减少它：

results = []
for student, student_answers in enumerate(answers):
    results.append([student, sum(ans == key for ans, key in zip(student_answers, keys)])
results = sorted(results, key=lambda x: x[1])

如果你真的想走这么远，你可以一气呵成：

results = sorted(((student, sum(a == k for a, k in zip(ans, keys)))
                  for student, ans in enumerate(answers)), key=lambda x: x[1])
# [(3, 4), (2, 5), (1, 6), (0, 7), (5, 7), (6, 7), (7, 7), (4, 8)]

多维列表排序[多列]

1 个答案: