如何转换。 按关键字转换为不同格式的示例字典？

Question

If I have a sample dictionary like this:

sample_field = {'XG102': '1/1:0,76:76:99:1|1:48306945_C_G:3353,229,0',
             'XG103': '1/1:0,52:52:99:.:.:1517,156,0',
             'XG104': '0/1:34,38:72:99:.:.:938,0,796'}

by giving the key word, I want to transform this sample dictionary like this:

    output = {'XG102': {'1: '0,76',
           '2': '76',
           '3': '99',
           '4': '1/1',
           '5': '1|1',
           '6': '48306945_C_G',
           '7': '3353,229,0'},
   'XG103': {'1': '0,52',
           '2': '52',
           '3': '99',
           '4': '1/1',
           '5': '.',
           '6': '.',
           '7': '1517,156,0'}
    ....

I try to map the sample dictionary to a new dictionary by using key like a index, but it doesn't work. Any suggestion ?

Answer 1

You may be looking for a dictionary comprehension using enumerate and split :

sample_field = {
    "XG102": "1/1:0,76:76:99:1|1:48306945_C_G:3353,229,0",
    "XG103": "1/1:0,52:52:99:.:.:1517,156,0",
    "XG104": "0/1:34,38:72:99:.:.:938,0,796",
}

out = {
    key: dict(enumerate(value.split(":"), 1))
    for (key, value) in sample_field.items()
}

-- out will be

{
    "XG102": {
        1: "1/1",
        2: "0,76",
        3: "76",
        4: "99",
        5: "1|1",
        6: "48306945_C_G",
        7: "3353,229,0",
    },
    "XG103": {
        1: "1/1",
        2: "0,52",
        3: "52",
        4: "99",
        5: ".",
        6: ".",
        7: "1517,156,0",
    },
    "XG104": {
        1: "0/1",
        2: "34,38",
        3: "72",
        4: "99",
        5: ".",
        6: ".",
        7: "938,0,796",
    },
}

Answer 2

This should work:

def create_split_dict(val):
  tokens = val.split(":")
  return {index+1:item for index, item in enumerate(tokens)}

output = {key:create_split_dict(val) for key, val in sample_field}