I have a list as:

[id, 1234, name, Jack, gender, M, id, 2345,
 name, Gary, gender, M, id, 5678 ....]

my desired list will be like this:

[{id:1234, name: Jack, gender: M},
 {id:2345, name: Gary, gender: M},
 {...........}
]

Zipping them up only returned the last items, so how can I keep all of them?

CodePudding user response：

This works perfectly within the given output.

lst = ['id', 1234, 'name', 'Jack', 'gender', 'M', 'id', 2345, 
    'name', 'Gary', 'gender', 'M', 'id', 5678]
new_list = []
temp_dict ={} 

for a in range(len(lst))[::2]: # range(len(lst))[::2] is a list [0, 2, 4, 6, 8, 10, 12 ...]
    if lst[a] == 'id' and a:
        new_list.append(temp_dict)
        temp_dict = {}
    
    temp_dict[lst[a]]=lst[a 1]
else: # Else execute after the loop exit.
    new_list.append(temp_dict)
print(new_list)

Output

[{'id': 1234, 'name': 'Jack', 'gender': 'M'},
 {'id': 2345, 'name': 'Gary', 'gender': 'M'},
 {'id': 5678}]

CodePudding user response：

You could try:

values = [
    'id', 1234, 'name', 'Jack', 'gender', 'M',
    'id', 2345, 'name', 'Gary', 'gender', 'M',
    'id', 5678, 'name', 'Joe'
]

dicts = [
    dict(zip(values[i:i 6:2], values[i 1:i 6:2]))
    for i in range(0, len(values), 6)
]

Result:

[{'id': 1234, 'name': 'Jack', 'gender': 'M'},
 {'id': 2345, 'name': 'Gary', 'gender': 'M'},
 {'id': 5678, 'name': 'Joe'}]

Logic:

• Iterate over each sixth index of values (0, 5, 11, ...)
• Starting from these indices split the corresponding part of values (6 items) into 2 interlaced parts, zip them, and then build a dict out of the tuples.

The assumption is that values starts with an 'id' item.