Python - 如何从另一个列表创建一个字典列表，该列表具有两个具有多个值的键？

Question

I am trying to create a dictionary list from another list which has two keys with multiple values as follows:

contact_items = [{'type': ['value1', 'value2', 'value3'],
                 'name': ['john', 'SCANIA', 'MARK']}]

list1 = ''
list2 = ''
list3 = []
test_dict = {}

for item in contact_items:
    list1 = item['type']
    list2 = item['name']

test_dict = {}

for k in list1:
    for l in list2:
        test_dict['type'] = k
        test_dict['name'] = l
        list3.append(test_dict)

print(list3)

The above code is not returning the dictionary list as expected.

Outputs:

[{'type': 'value3', 'name': 'MARK'}, {'type': 'value3', 'name': 'MARK'}, {'type': 'value3', 'name': 'MARK'}, {'type': 'value3', 'nam
e': 'MARK'}, {'type': 'value3', 'name': 'MARK'}, {'type': 'value3', 'name': 'MARK'}, {'type': 'value3', 'name': 'MARK'}, {'type': 'v
alue3', 'name': 'MARK'}, {'type': 'value3', 'name': 'MARK'}]

Expected:

result=[{'type':'value1','name':'john'},
{'type':'value2','name':'SCANIA'},
{'type':'value3','name':'MARK'}]

Answer 1

You're putting the same dict in the list multiple times. And you're assigning different values to the same variables in your first loop. And you're iterating through every combination of items in list1 and list2 in your nested loop.

Here is a way to get the outcome you specified from the input you specified:

types = contact_items[0]['type']
names = contact_items[0]['name']
list3 = [{'type': t, 'name': n} for (t,n) in zip(types, names)]

See list comprehensions , zip .

Answer 2

This is very trivial. Look up list comprehension and dictionary comprehension.

result = [{"type": contact_items[0]["type"][i], 
"name": contact_items[0]["name"][i]} 
for i in range(len(contact_items[0]["name"])) ]

Neater.

items = contact_items[0]

result = [{"type": items["type"][i], "name": items["name"][i]} for i in range(len(items["name"]))]
result

Answer 3

You can use list comprehension in combination with zip to achieve this.

result = [
    {"name": name, "type": type}
    for type, name in zip(*contact_items[0].values())
]