将列表列表的值与词典列表合并

Question

u= [['1', '2'], ['3'], ['4', '5', '6'], ['7', '8', '9', '10']]
v=[{'id': 'a', 'adj': ['blue', 'yellow']}, {'id': 'b', 'adj': ['purple', 'red']}, {'id': 'c', 'adj': ['green', 'orange']}, {'id': 'd', 'adj': ['black', 'purple']}]

我想要：

 result=[ {'id': 'a', 'adj': ['blue', 'yellow'], 'value': '1' },
        {'id': 'a', 'adj': ['blue', 'yellow'], 'value': '2' },
        {'id': 'a', 'adj': ['purple', 'red'], 'value': '3' },
        ...]

我已将u转换为字典：

m=[]
for i in u:
    s={}
    s['value']=i
    m.append(s)

#>>m= [{'value': ['1', '2']}, {'value': ['3']}, {'value': ['4', '5', '6']}, {'value': ['7', '8', '9', '10']}]

然后尝试应用zip函数

for i,j in enumerate(v):
    for s,t in enumerate(l):
        if i= =s:
            #zip 2 dictionary together. Stuck here

提前多谢了！这是我学习编程的第二周

Answer 1

您需要压缩、迭代u中的每个子列表、v中的每个dict并添加新的键/值对，最后将新dict附加到列表中：

from copy import deepcopy

u= [['1', '2'], ['3'], ['4', '5', '6'], ['7', '8', '9', '10']]
v=[{'id': 'a', 'adj': ['blue', 'yellow']}, {'id': 'b', 'adj': ['purple', 'red']}, {'id': 'c', 'adj': ['green', 'orange']}, {'id': 'd', 'adj': ['black', 'purple']}]

out = []
# match up corresponding elements fromm both lists
for dct, sub in zip(v, u):
    # iterate over each sublist
    for val in sub:
        # deepcopy the dict as it contains mutable elements (lists)
        dct_copy = deepcopy(dct)
        # set the new key/value pairing
        dct_copy["value"] = val
        # append the dict to our out list
        out.append(dct_copy)
from pprint import pprint as pp
pp(out)

这将给你：

[{'adj': ['blue', 'yellow'], 'id': 'a', 'value': '1'},
 {'adj': ['blue', 'yellow'], 'id': 'a', 'value': '2'},
 {'adj': ['purple', 'red'], 'id': 'b', 'value': '3'},
 {'adj': ['green', 'orange'], 'id': 'c', 'value': '4'},
 {'adj': ['green', 'orange'], 'id': 'c', 'value': '5'},
 {'adj': ['green', 'orange'], 'id': 'c', 'value': '6'},
 {'adj': ['black', 'purple'], 'id': 'd', 'value': '7'},
 {'adj': ['black', 'purple'], 'id': 'd', 'value': '8'},
 {'adj': ['black', 'purple'], 'id': 'd', 'value': '9'},
 {'adj': ['black', 'purple'], 'id': 'd', 'value': '10'}]

dict有一个.copy属性，或者你可以调用dict(dct)，但是因为你有可变的对象作为值，仅仅做一个浅层拷贝是不起作用的。下面的示例显示了实际的差异：

In [19]: d = {"foo":[1, 2, 4]}

In [20]: d1_copy = d.copy() # shallow copy, same as dict(d)

In [21]: from copy import  deepcopy

In [22]: d2_copy = deepcopy(d) # deep copy

In [23]: d["foo"].append("bar")

In [24]: d
Out[24]: {'foo': [1, 2, 4, 'bar']}

In [25]: d1_copy
Out[25]: {'foo': [1, 2, 4, 'bar']} # copy also changed

In [26]: d2_copy
Out[26]: {'foo': [1, 2, 4]} # deepcopy is still the same

what-exactly-is-the-difference-between-shallow-copy-deepcopy-and-normal-assignment

Answer 2

对两个列表应用zip，并创建一个新字典，其中旧的字典将来自相应列表的值添加为键值条目value：来自列表的数字：

>>> import pprint, copy
>>> result = [dict(copy.deepcopy(j), value = ind) for i, j in zip(u, v) for ind in i]
>>> pprint.pprint(result)
[{'adj': ['blue', 'yellow'], 'id': 'a', 'value': '1'},
 {'adj': ['blue', 'yellow'], 'id': 'a', 'value': '2'},
 {'adj': ['purple', 'red'], 'id': 'b', 'value': '3'},
 {'adj': ['green', 'orange'], 'id': 'c', 'value': '4'},
 {'adj': ['green', 'orange'], 'id': 'c', 'value': '5'},
 {'adj': ['green', 'orange'], 'id': 'c', 'value': '6'},
 {'adj': ['black', 'purple'], 'id': 'd', 'value': '7'},
 {'adj': ['black', 'purple'], 'id': 'd', 'value': '8'},
 {'adj': ['black', 'purple'], 'id': 'd', 'value': '9'},
 {'adj': ['black', 'purple'], 'id': 'd', 'value': '10'}]

Answer 3

您可以使用以下代码来获得所需的结果

result = []
for index, d in enumerate(u):
    for value in d:
        result.append(dict(v[index], value=value))

它在u的enumerate-离子上迭代，然后将正确的vdict和value的组合附加到result列表

您可以使用列表将其压缩为相对干净的一行

result = [dict(v[index], value=value) for index, d in enumerate(u) for value in d]