如何将字典的值（列表）与另一个列表进行比较？

Question

假设我有一本字典：

students = {"adrian":["a","b","c","d"], "jamie":["b","b","a","d"], "adam":["a","c","d","d"]}

并且这是一个列表：

answers=["a","b","c","d"]

我只想检查每个元素是否与给定索引处的答案匹配，并与字典中的值进行比较。简而言之，将每个列表与答案进行比较。然后我将打印出每个学生正确答题的次数。

例如，如果我将键“adrian”的值与答案进行比较，我将得到4。如果我将jamie与答案进行比较，我会得到2。如果我将adam与答案进行比较，我会得到2。

如何进行比较？

非常感谢您的帮助！

- Lander Carlson

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请编辑问题并添加您尝试过的代码。 - shad0w_wa1k3r

Answer 1

>>> students = {"adrian":["a","b","c","d"], "jamie":["b","b","a","d"], "adam":["a","c","d","d"]}
>>> answers = ["a","b","c","d"]
>>> {s:sum(t == a for t, a in zip(students[s], answers)) for s in students}
{'jamie': 2, 'adam': 2, 'adrian': 4}

对于每个学生，将他们的答案与标准答案进行zip压缩，然后进行比较，最终将结果布尔值相加。

- Dan D.

Answer 2

import pandas as pd
students = {"adrian":["a","b","c","d"], "jamie":["b","b","a","d"], "adam":["a","c","d","d"]}
answers=["a","b","c","d"]
df = pd.DataFrame(students)
df.apply(lambda x: x==answers).sum()

一些解释：

df.apply(lambda x: x==answers) 

将为每个学生比较答案，生成以下数组：

    adam adrian  jamie
0   True   True  False
1  False   True   True
2  False   True  False
3   True   True   True

.sum()函数将对每一列进行求和（将True转换为1，False转换为0），从而将数组降维为:

adam      2
adrian    4
jamie     2

- Allen Qin

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即使这种方法是可行的，我也不会说在不涉及问题标签时使用pandas是非常好的选择。至少要添加一些解释。 - Netwave

Answer 3

像这样的吗？

def get_correct_answers_count(given_answers,
                              correct_answers):
    return sum(1
               for student_answer, correct_answer in zip(given_answers,
                                                         correct_answers)
               if student_answer == correct_answer)


students_correct_answers_count = {
    student_name: get_correct_answers_count(given_answers=student_answers,
                                            correct_answers=answers)
    for student_name, student_answers in students.items()}

给我们

{'adrian': 4, 'jamie': 2, 'adam': 2}

- Azat Ibrakov

Answer 4

你可以使用zip()来获取值和答案的组合。

for k,v in students.items():
    ans = 0
    for i, j in zip(v,answers):
        if i==j:
            ans+=1
    print ans

- Akshay Apte

Answer 5

我认为这个可以工作

students = {"adrian":["a","b","c","d"],
          "jamie":["b","b","a","d"],
          "adam":["a","c","d","d"]}
answers = ["a","b","c","d"]

results = {}

for k,v in students.items():
    results [k] = 0
    for i,answer in enumerate(v):
        if answers[i] == v[i]:
            results[k]+=1

print (results)

{'adrian': 4, 'adam': 2, 'jamie': 2}

这段代码与编程有关。

- ragardner

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这取决于你对“复杂”的定义是什么。 - David Bern

Answer 6

基于 set() 的解决方案

示例

data = {'jamie': ['a', 'b', 'c', 'd']}
answers = ['b', 'd', 'c', 'd']
tuples = map(set, zip(data['jamie'], answers))
reduce(lambda y, coll: (len(coll) - 1) + y , tuples, 0)

Out[8]: 2

这个例子利用了数据类型集合只能包含唯一值的特点。因此，通过将两个列表jamie和answers压缩成元组，如果元组中包含两个相同的值，则set()会使其变为唯一值。

当我减少元组列表时，利用了集合中有多个元素的事实，这表明答案错误并计数。

- David Bern