Question

我能够实现这一目标，但我认为必须有更好的方法。我们的想法是为这两种情况提供共同的解决方案。

问题：

a＆amp; b是两本词典。这些字典的值可以是简单的字符串（用例1）或相同长度的列表（用例2）。请注意，这些列表的长度可能会因a＆amp;而异。 b（例如： - 用例2中的键＆＃39; 2＆＃39;）但同一字典中的长度相同。

预期：

我们需要创建一个最终字典，其中包含来自字典a＆amp; b并将这些值（无论是字符串还是列表）合并为长度相当于任一字典中最长的列表。例如： - 在用例2中，键2在字典a中有[＆＃34; abc＆＃34;，＆＃34; def＆＃34;]但在字典b中只有[＆＃34; abc＆＃34;]。在这里，我们期待最终结果为2：[＆＃34; abc＆＃34;，＆＃34; def＆＃34;，＆＃34; abc＆＃34;，None]。

示例：

用例1：

a = {1:"xyz",2:999,3:1234,4:"boy"}
b = {1:"xyz",2:99,3:19,6:"toy"}

Result expected : {1: ['xyz', 'xyz'], 2: [999, 99], 3: [1234, 19], 4: ['boy', None],
                   6: ['None', 'toy']

用例2：

a = {1:["xyz","123"],2:["abc","def"],3:["zzz",""],4:["boy","abc"]}
b = {1:["xyz","1234"],2:["abc"],3:["zz",""], 6: ["toy"]}

result expected: {1: [('xyz', 'xyz'), ('123', '1234')], 2: [('abc', 'abc'), ('def', None)], 
                  3: [('zzz', 'zz'), ('', '')], 4: [('boy', None), ('abc', None)], 
                  6: [(None, 'toy')]})

Answer 1

算法：

from itertools import izip_longest

def merge(a, b):
    res = {}
    keys = set(a.keys()) | set(b.keys())
    if isinstance(a.values()[0], list):
        res = {k: list(izip_longest(a.get(k, [None]), b.get(k, [None]))) for k in keys}
    else:
        res = {k: [a.get(k, None), b.get(k, None)] for k in keys}
    return res

用例1

a = {1:"xyz",2:999,3:1234,4:"boy"}
b = {1:"xyz",2:99, 3:19,  6:"toy"}

exp = {1: ['xyz', 'xyz'], 
       2: [999, 99], 
       3: [1234, 19], 
       4: ['boy', None],
       6: [None, 'toy']}

print merge(a, b) == exp

输出：

True

用例2

a = {1:["xyz","123"], 2:["abc","def"],3:["zzz",""],4:["boy","abc"]}
b = {1:["xyz","1234"],2:["abc"],     3:["zz",""],  6: ["toy"]}
exp = {1: [('xyz', 'xyz'), ('123', '1234')], 
       2: [('abc', 'abc'), ('def', None)],
       3: [('zzz', 'zz'), ('', '')], 
       4: [('boy', None), ('abc', None)],
       6: [(None, 'toy')]}

print merge(a, b) == exp

输出：

True

注意 Python 2

用Python加入词典

1 个答案:

用例1

用例2