假设给了我一本字典：
students = {"adrian":["a","b","c","d"], "jamie":["b","b","a","d"], "adam":["a","c","d","d"]}
还有名单：
answers=["a","b","c","d"]
我只是想根据字典中的值检查给定索引处答案中的每个元素是否匹配。简而言之，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，我将得到 2。如果我将 adam 与答案进行比较，我将得到 2。
如何比较两者？
非常感谢您的帮助！
【问题讨论】：
标签： python list loops dictionary iteration
>>> 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}
这为每个学生zips 他们的答案与答案进行比较，然后将结果布尔值相加。
【讨论】：
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() 将按列求和（将 True 转换为 1，将 False 转换为 0），将数组减少到：
adam 2
adrian 4
jamie 2
【讨论】：
这样的？
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
{“阿德里安”：4，“杰米”：2，“亚当”：2}
【讨论】：
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
【讨论】：
一个基于set()
的解决方案例子
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
该示例使用了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 answers 我得到元组。如果元组包含两个相同的 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，则表示答案错误并计算它。
【讨论】：
我认为这可行
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)
{“阿德里安”：4，“亚当”：2，“杰米”：2}
【讨论】：