根据条件将字典项拆分为更小的字典

Question

我有两个列表：一个包含部分事务，另一个包含其父事务：

partials = [1,2,3,4,5,6,7,8,9,10]
parents = ['a','b','c','d','a','d','f','c','c','a']

我把这些单子编入字典：

transactions = zip(partials, parents)

如您所见，某些部分事务具有相同的父事务。你知道吗

我需要把字典里的条目分成更小的组（更小的字典？）在每个组中，属于一个父级的事务不超过一个。例如，与父级“a”的所有事务都需要在不同的组中结束。你知道吗

我也需要尽可能少的组，因为在现实世界中，每个组将是一个文件手动上传。你知道吗

预期输出如下：

第1组将包含事务处理1a、2b、3c、4d、7f

第2组将包含交易5a、6d、8c

第3组将包含交易记录9c、10a

我已经为这件事绞尽脑汁好一阵子了，如果有任何建议我都会很感激的。到目前为止，我没有任何工作代码张贴。你知道吗

Answer 1

&barciewicz根据您提供的输入和预期的输出，我也尝试用我的方式解决这个问题。你知道吗

Note» I have used OrderedDict() from collections module to retain the order of keys in dictionary. I've also used json module to pretty print the dictionaries, list etc.

我展示了3种不同的方法来获得3个独立函数的结果，如下所示。你知道吗

{ "group1": "1a,2b,3c,4d,7f", "group2": "5a,8c,6d", "group3": "10a,9c" }

你知道吗

{ "group1": { "a": 1, "b": 2, "c": 3, "d": 4, "f": 7 }, "group2": { "a": 5, "c": 8, "d": 6 }, "group3": { "a": 10, "c": 9 } }

你知道吗

{ "a": [ 1, 5, 10 ], "b": [ 2 ], "c": [ 3, 8, 9 ], "d": [ 4, 6 ], "f": [ 7 ] }

你知道吗

»在http://rextester.com/OYFF74927上在线尝试下面的代码。你知道吗

from collections import OrderedDict
import json; 

def get_formatted_transactions(parents, partials):
    d = OrderedDict();

    for index, partial in enumerate(partials):
        if parents[index] in d:
            l = d[parents[index]]
            l.append(partial)
            d[parents[index]] = l;
        else:
            d[parents[index]] = [partial]

    return d;


def get_groups(transactions):
    i = 1;
    groups = OrderedDict();

    while transactions:
        group_name = "group" + str(i);
        groups[group_name] = {};
        keys = list(transactions.keys());

        for key in keys:
            if transactions[key]:
                groups[group_name][key] = transactions[key].pop(0);
                if not transactions[key]:
                    del transactions[key];
            else:
                del transactions[key]
        i += 1;

    return groups;

def get_comma_separated_data(groups):
    new_dict = OrderedDict();
    for group_name in groups:
        d = groups[group_name]
        new_dict[group_name] = ",".join([str(value) + key  for value, key in zip(d.values(), d.keys())])

    return new_dict;



# Starting point
if __name__ == "__main__":
    partials = [1,2,3,4,5,6,7,8,9,10];
    parents = ['a','b','c','d','a','d','f','c','c','a'];

    transactions = get_formatted_transactions(parents, partials);
    # Pretty pritining ordered dictionary
    print(json.dumps(transactions, indent=4));

    print("\n");

    # Creating groups to organize transactions
    groups = get_groups(transactions)
    # Pretty printing
    print(json.dumps(groups, indent=4))

    print("\n");

    # Get comma separated form 
    comma_separated_data = get_comma_separated_data(groups);
    # Pretty printing
    print(json.dumps(comma_separated_data, indent=4));

输出»

{
    "a": [
        1,
        5,
        10
    ],
    "b": [
        2
    ],
    "c": [
        3,
        8,
        9
    ],
    "d": [
        4,
        6
    ],
    "f": [
        7
    ]
}

{
    "group1": {
        "a": 1,
        "b": 2,
        "c": 3,
        "d": 4,
        "f": 7
    },
    "group2": {
        "a": 5,
        "c": 8,
        "d": 6
    },
    "group3": {
        "a": 10,
        "c": 9
    }
}

{
    "group1": "1a,2b,3c,4d,7f",
    "group2": "5a,8c,6d",
    "group3": "10a,9c"
}