Python 从主列表和子列表创建新的词典列表

我有两个列表（主列表和子列表），其中包含不同数量的字典，如果它们的位置匹配，它们的值将是相同的

mainLst = [{"name":"M_AAA_X", "position":"1", "value":"8"}, 
           {"name":"M_AAA_X", "position":"2", "value":"10"}, 
           {"name":"M_AAA_X", "position":"4", "value":"14"}, 
           {"name":"M_AAA_X", "position":"5", "value":"16"},
           {"name":"M_AAA_X", "position":"7", "value":"20"}]

我想通过以下方式创建一个新列表：

从mainLst获取元素

如果mainLst缺少某个位置（例如位置：3,6），则从子列表中获取元素

按“位置”升序排序

输出应如下所示：

newLst = [{"name":"M_AAA_X", "position":"1", "value":"8"}, 
          {"name":"M_AAA_X", "position":"2", "value":"10"}, 
          {"name":"S_AAA_X", "position":"3", "value":"12"},
          {"name":"M_AAA_X", "position":"4", "value":"14"}, 
          {"name":"M_AAA_X", "position":"5", "value":"16"},
          {"name":"S_AAA_X", "position":"6", "value":"18"},
          {"name":"M_AAA_X", "position":"7", "value":"20"}]

谁能帮我完成这个任务？提前谢谢

试试这个

mainLst.extend([dictSub for dictSub in subLst if dictSub['position'] not in [pos['position'] for pos in mainLst]])
print(sorted(mainLst, key = lambda i: i['position']))

基本上，我们在

子列表

中收集位置，该子列表不在

主列表

中。然后我们根据这一点对目录列表进行排序

输出

[{'name': 'M_AAA_X', 'position': '1', 'value': '8'}, 
 {'name': 'M_AAA_X', 'position': '2', 'value': '10'}, 
 {'name': 'S_AAA_X', 'position': '3', 'value': '12'}, 
 {'name': 'M_AAA_X', 'position': '4', 'value': '14'}, 
 {'name': 'M_AAA_X', 'position': '5', 'value': '16'}, 
 {'name': 'S_AAA_X', 'position': '6', 'value': '18'}, 
 {'name': 'M_AAA_X', 'position': '7', 'value': '20'}]

试试这个

mainLst = [{"name":"M_AAA_X", "position":"1", "value":"8"}, 
           {"name":"M_AAA_X", "position":"2", "value":"10"}, 
           {"name":"M_AAA_X", "position":"4", "value":"14"}, 
           {"name":"M_AAA_X", "position":"5", "value":"16"},
           {"name":"M_AAA_X", "position":"7", "value":"20"}]
subLst = [{"name":"S_AAA_X", "position":"1", "value":"8"}, 
          {"name":"S_AAA_X", "position":"2", "value":"10"}, 
          {"name":"S_AAA_X", "position":"3", "value":"12"}, 
          {"name":"S_AAA_X", "position":"4", "value":"14"}, 
          {"name":"S_AAA_X", "position":"5", "value":"16"},
          {"name":"S_AAA_X", "position":"6", "value":"18"}]
          
test_list=[]
test_list2=[]
for j in mainLst:
    print(j.get("position"))
    test_list.append(j.get("position"))
for k in subLst:
    print(k.get("position"))
    test_list2.append(k.get("position"))
for l in test_list:
    if l in test_list2:
        print("True")
    else:
        test_list2.append(l)
newlst=[]
spe_list=[]
print(test_list2)
for i in test_list2:
    for j in mainLst:
        if j.get("position")==i:
            spe_list.append(j.get("position"))
            newlst.append(j)
            break
        else:
            pass
    for k in subLst:
        if k.get("position")==i:
            if k.get("position") in spe_list:
                pass
            else:
                newlst.append(k)
                break
        else:
            pass


for j in newlst:
    print(j)

---

另一种解决方案是使用

itertools.groupby

：

从itertools导入groupby
out=[]
对于groupby中的g(
已排序（mainLst+subLst，key=lambda k:int（k[“位置]）），
lambda k:int（k[“位置]），
):
out.append（下一个（g））
打印（输出）

印刷品：

[{'name'：'M_AAA_X'，'position'：'1'，'value'：'8'}，
{'name'：'M_AAA_X'，'position'：'2'，'value'：'10'}，
{'name'：'S_AAA_X'，'position'：'3'，'value'：'12'}，
{'name'：'M_AAA_X'，'position'：'4'，'value'：'14'}，
{'name'：'M_AAA_X'，'position'：'5'，'value'：'16'}，
{'name'：'S_AAA_X'，'position'：'6'，'value'：'18'}，
{'name'：'M_AAA_X'，'position'：'7'，'value'：'20'}]

你已经试过什么了吗？这是一个非常有趣的列表<代码>用于然后是

如果

然后是

用于

。从未见过这样的事情，但我真的很喜欢这个解决方案。也许对于初学英语的人来说可能很难理解，因为只有代码的答案很少有用。

mainLst = [{"name":"M_AAA_X", "position":"1", "value":"8"}, 
           {"name":"M_AAA_X", "position":"2", "value":"10"}, 
           {"name":"M_AAA_X", "position":"4", "value":"14"}, 
           {"name":"M_AAA_X", "position":"5", "value":"16"},
           {"name":"M_AAA_X", "position":"7", "value":"20"}]
subLst = [{"name":"S_AAA_X", "position":"1", "value":"8"}, 
          {"name":"S_AAA_X", "position":"2", "value":"10"}, 
          {"name":"S_AAA_X", "position":"3", "value":"12"}, 
          {"name":"S_AAA_X", "position":"4", "value":"14"}, 
          {"name":"S_AAA_X", "position":"5", "value":"16"},
          {"name":"S_AAA_X", "position":"6", "value":"18"}]
          
test_list=[]
test_list2=[]
for j in mainLst:
    print(j.get("position"))
    test_list.append(j.get("position"))
for k in subLst:
    print(k.get("position"))
    test_list2.append(k.get("position"))
for l in test_list:
    if l in test_list2:
        print("True")
    else:
        test_list2.append(l)
newlst=[]
spe_list=[]
print(test_list2)
for i in test_list2:
    for j in mainLst:
        if j.get("position")==i:
            spe_list.append(j.get("position"))
            newlst.append(j)
            break
        else:
            pass
    for k in subLst:
        if k.get("position")==i:
            if k.get("position") in spe_list:
                pass
            else:
                newlst.append(k)
                break
        else:
            pass


for j in newlst:
    print(j)