使用相同键但值不同的Python字典

Question

I want to create a dictionary from key and value pairs. The problem is that I have same keys but different values. So my goal is to create

menu = [
    {"viewclass": "MDMenuItem",
     "text" : "option1"},
    {"viewclass": "MDMenuItem",
     "text" : "option2"}
]

I tried the creation of this variable by using a for loop

length = 2
menu = {}

view_class_keys = length * ["viewclass"]
view_class_values = length * ["MDMenuItem"]
text_keys = length * ["text"]
text_values = ["option1", "option2"]

for iterator in range(0, length):
    menu[view_class_keys[iterator]] = view_class_values[iterator]
    menu[text_keys[iterator]] = text_values[iterator]

print([menu])

# Output: [{'viewclass': 'MDMenuItem', 'text': 'option2'}]

I know the problem is that the keys are the same, but I do not know how to resolve this problem.

Answer 1

You are quite close. You should aggregate your dictionaries in the list while creating a new dictionary on each iteration appending it to the resulting list:

length = 2
menu_list = []
view_class_keys = length * ["viewclass"]
view_class_values = length * ["MDMenuItem"]
text_keys = length * ["text"]
text_values = ["option1", "option2"]

for iterator in range(0, length):
    menu = {}
    menu[view_class_keys[iterator]] = view_class_values[iterator]
    menu[text_keys[iterator]] = text_values[iterator]
    menu_list.append(menu)

print(menu_list)

Edit: Assuming the only variable part in your code is text_values list, your code can be simplified to

menu = [{"viewclass": "MDMenuItem", "text" : option} for option in text_values]

Answer 2

也许是这样的：

menu = [dict(zip(i[::2], i[1::2])) for i in zip(view_class_keys, view_class_values, text_keys, text_values)]

Answer 3

OR:

length = 2
text_values = ["option1", "option2"]
print([dict(viewclass="MDMenuItem", text=option) for option in text_values])

Answer 4

This should resolve your problem:

length = 2
# in your goals, "menu" type is a list
menu = []

view_class_keys = length * ["viewclass"]
view_class_values = length * ["MDMenuItem"]
text_keys = length * ["text"]
text_values = ["option1", "option2"]


for iterator in range(0, length):
    #first time you have to append to the list a new dictionary
    menu.append({view_class_keys[iterator]:view_class_values[iterator]})
    #than you can add a new key value to the dict 
    menu[iterator][text_keys[iterator]] = text_values[iterator]

#I leave the square brackets cause menu is already a list
print(menu)

# Output: [{'viewclass': 'MDMenuItem', 'text': 'option1'}, {'viewclass': 'MDMenuItem', 'text': 'option2'}]

However the dict may not be the best choise for you since you have same keys for different values. Maybe you can try with a single list like this:

length = 2
menu = []

view_class_keys = length * ["viewclass"]
view_class_values = length * ["MDMenuItem"]
text_keys = length * ["text"]
text_values = ["option1", "option2"]

for iterator in range(0, length):
    menu.append([view_class_keys[iterator],view_class_values[iterator]])
    menu.append([text_keys[iterator],text_values[iterator]])

print(menu)
# Output: [['viewclass', 'MDMenuItem'], ['text', 'option1'], ['viewclass', 'MDMenuItem'], ['text', 'option2']]