Question

让我说我有一本字典：

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

列表：

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

我只想检查每个元素匹配来自其给定索引的答案与字典中的值。总之，将每个列表与答案进行比较。然后我会打印学生正确的次数。

例如，如果我将关键字“adrian”的值与我得到的答案进行比较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.

我如何比较两者？

非常感谢您的帮助！

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}

这对每个学生zip的答案都有答案，然后对它们进行比较，然后总结得出的布尔人。

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()

有点解释：

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

将比较每个学生的答案，得到以下数组：

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

.sum（）将执行逐列求和（将True转换为1，将False转换为0），将数组缩减为：

adam      2
adrian    4
jamie     2

Answer 3

这样的事情？

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}

Answer 4

您可以使用zip（）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

Answer 5

我认为这有效

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)

{＆＃39; adrian＆＃39;：4，＆＃39; adam＆＃39;：2，＆＃39; jamie＆＃39;：2}

Answer 6

基于 set（）

的解决方案

实施例

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

该示例使用的是数据类型集只能包含唯一的事实。因此，通过压缩两个列表jamie和答案我得到元组。如果元组包含两个相同的元组set()将使其唯一。

当我减少元组列表时，我使用的事实是，如果集合中有多个元素，则表示错误的答案＆amp;算上它。

如何将字典的值（列表）与单独的列表进行比较？

6 个答案: