在python字典中单独合并公共键及其值

Question

我已经看到很多与我的问题相关的答案，但没有一个能解决我的问题

这是输入问题：

data = { "data": [{ key1:word},{ key1:hello},{key2:hey},{key2:hi}]}  

预期输出：

[{key1:word, key2:hey}, {key1:hello, key2:hi}]

如果我添加 key3 那么这是想要的输出：

data = { "data": [{ key1:word},{ key1:hello},{key2:hey},{key2:hi}, 
         {key3:somedata},{key3:somedata2}]}

# WANTED OUTPUT BELOW
 
[ {key1:word, key2:hey, key3:somedata}, {key1:hello, key2:hi, 
 key3:somedata2} ]

这是我尝试过的：

 for key,items in data.items():
    if key == "key1":
        do something
    if key == "key2":
        do something

但上面不能用于 1000 个密钥对值，所以代码应该是动态的。 将来可以添加key4，key5，key6....等，因此我无法在条件中定义每个关键项。

Answer 1

只要所有键的数量相同，我就找到了一个巧妙的解决方案！

from itertools import groupby

data = [{"key1": "value1"}, {"key1": "hello"}, {"key2": "value2"}, {"key2": "value3"}, {"key3": "test1"},
        {"key3": "test2"}]

data.sort(key=lambda x: list(x.keys())[0]) # you need to sort for groupby to work in every case
grouped = [list(g) for k, g in groupby(data, key=lambda x: list(x.keys())[0])]
# = [[{'key1': 'value1'}, {'key1': 'hello'}], [{'key2': 'value2'}, {'key2': 'value3'}], [{'key3': 'test1'}, {'key3': 'test2'}]]
output = [{k:v for dic in t for k,v in dic.items()} for t in zip(*grouped)]
# source https://stackoverflow.com/a/43529214/10875953
print(output) 
# prints [{'key1': 'value1', 'key2': 'value2', 'key3': 'test1'}, {'key1': 'hello', 'key2': 'value3', 'key3': 'test2'}]

Answer 2

有趣的请求.....我想这可能是你需要的？

data = [{ "key1":"value1"},{ "key1":"hello"},{"key2":"value2"},{"key2":"value3"}] 

odd_entries = data[::2]  # Iterate by twos, starting at 0
even_entries = data[1::2]  # Iterate by twos, starting at 1

print(list(zip(odd_entries, even_entries)))

[({'key1': 'value1'}, {'key1': 'hello'}), ({'key2': 'value2'}, {'key2': 'value3'})]

Answer 3

希望这能帮助您理解问题

data = { "data": [{"key1":"word"},{ "key1":"hello"},{"key2":"hey"},{"key2":"hi"}, 
     {"key3":"somedata"},{"key3":"somedata2"}]}  

dic1 = {}
dic2 = {}

[dic2.update(dic) if tuple(dic)[0] in dic1.keys() else dic1.update(dic) for dic in data["data"]]

上面的列表理解可以重写为blow（为了更好的理解）

    for dic in data["data"]:
       if tuple(dic)[0] in dic1.keys():
           dic2.update(dic)
       else:
           dic1.update(dic)

输出：

dic1

{'key1': 'word', 'key2': 'hey', 'key3': 'somedata'}

dic2

{'key1': 'hello', 'key2': 'hi', 'key3': 'somedata2'}