如何创建以列表为值的字典？

Question

How can I get this output:

{'Test 1': ['100', '88', '45', '59', '73', '89'], 'Test 2': ['90', '99', '56', '61', '79', '97'], 'Test 3': ['80', '111', '67', '67', '83', '101']}

with this data

grades =  [
  ['Students', 'Test 1', 'Test 2', 'Test 3'],
  ['Tomas', '100', '90', '80'],
  ['Marcos', '88', '99', '111'],
  ['Flavia', '45', '56', '67'],
  ['Ramon', '59', '61', '67'],
  ['Ursula', '73', '79', '83'],
  ['Federico', '89', '97', '101']
] 

Answer 1

The first element (row) of grades contains all the keys you want to use. Every other row contains the values of the keys in every column. You can use zip with unpacking like so: zip(grades[0], *grades[1:]) to obtain an iterator that gives a tuple containing the key as the first element, and the corresponding column from every other row in the remaining elements. Then, you can unpack this using k, *v in zip(...) to get the key in k , and a list of the values in the column in v .

Then, you can use a dictionary comprehension to create a dictionary out of the result of the zip operation. k:v is keys and values for the dictionary, for k in grades[0] so each iteration will give k one element from grades[0] , then *v is a full vertical column.

{k: v for k, *v in zip(grades[0], *grades[1:])}

# Result:
{'Students': ['Tomas', 'Marcos', 'Flavia', 'Ramon', 'Ursula', 'Federico'],
 'Test 1': ['100', '88', '45', '59', '73', '89'],
 'Test 2': ['90', '99', '56', '61', '79', '97'],
 'Test 3': ['80', '111', '67', '67', '83', '101']}

Answer 2

dictt ={}
for n,i in enumerate(grades[0][1:],start=1):
  dictt[i] = [x[n] for x in grades[1:]]

print(dictt)

>>> {'Test 1': ['100', '88', '45', '59', '73', '89'], 'Test 2': ['90', '99', '56', '61', '79', '97'], 'Test 3': ['80', '111', '67', '67', '83', '101']}

start = 1 omits from adding the labels to the dictionary values