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How do I compare dictionary meanings (lists) to a single list?

Let's say I was given a dictionary:

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

And the list:

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

I just want to check that each item matches the responses from its pointer to the values ​​in the dictionary. Briefly compare each list with the answers. I would then print the number of times the student was correct.

For example, if I were to compare the value of the key "adrian" with the responses I would get 4. And if I were to compare jamie to the responses, I would get 2. And if I were to compare adam to the responses, I would get 2.

How can I compare the two?

Thanks a lot for any help!

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6 answers 


>>> 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}


This is for each student the zip

answers to their answers with the answers and then compares them and then sums up the resulting booleans.

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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()

Explanation of the bits:

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

will compare the answers for each student, resulting in the following array:



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

.sum () will do the column sum (cast True to 1 and False to 0), decreasing the array to:

adam      2
adrian    4
jamie     2
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something like that?

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()}

gives us

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

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You can use zip () to get a combination of values ​​and responses

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

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}

0


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Set () solution

Example

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

The example exploits the fact that a set of data types can only contain unique data. So by picking two lists of jamie and the responses, I get tuples. If a tuple contains two of them, it set()

will make it unique.

When I shorten the list of tuples, I am taking advantage of the fact that if the set contains more than one element, it is a sign of a wrong answer and counting it.

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