在python中组合两个dict

我有一本python字典，如下所示

test={}
test['key1']={}
test['key1'] ['key2'] = {}
test['key1']['key2']['key3'] = 'val1'
test['key1']['key2']['key4'] = 'val2'

 check = {}
    check['key1']={}
    check['key1'] ['key2'] = {}
    check['key1']['key2']['key5'] = 'val3'
    check['key1']['key2']['key6'] = 'val4'

我有另一本字典如下

test={}
test['key1']={}
test['key1'] ['key2'] = {}
test['key1']['key2']['key3'] = 'val1'
test['key1']['key2']['key4'] = 'val2'

 check = {}
    check['key1']={}
    check['key1'] ['key2'] = {}
    check['key1']['key2']['key5'] = 'val3'
    check['key1']['key2']['key6'] = 'val4'

我想合并这本词典，所以我做了以下工作

test.update(check)

但如果我在打印测试字典时这样做，它就像打印一样

{'key1': {'key2': {'key5': 'val3', 'key6': 'val4'}}}

但预期产出是有限的

{'key1': {'key2': {'key3': 'val1', 'key4': 'val2','key5': 'val3', 'key6': 'val4'}}}

---

如果x和y是dict，那么

z={**x，**y}

是一个dict，将x和y合并在一起要深度合并dict，您需要使用：

def merge(source, destination):
    for key, value in source.items():
        if isinstance(value, dict):
            # get node or create one
            node = destination.setdefault(key, {})
            merge(value, node)
        else:
            destination[key] = value
    return destination

test = {'key1': {'key2': {'key3': 'val1', 'key4': 'val2'}}}
check = {'key1': {'key2': {'key6': 'val4', 'key5': 'val3'}}}

print(merge(test, check))
# {'key1': {'key2': {'key3': 'val1', 'key6': 'val4', 'key4': 'val2', 'key5': 'val3'}}}

---

这是来自vincent的代码这里有一种实现深度合并的方法：

test={}
test['key1']={}
test['key1'] ['key2'] = {}
test['key1']['key2']['key3'] = 'val1'
test['key1']['key2']['key4'] = 'val2'

check = {}
check['key1']={}
check['key1'] ['key2'] = {}
check['key1']['key2']['key5'] = 'val3'
check['key1']['key2']['key6'] = 'val4'

def deep_merge(a, b):
    for key, value in b.items():
        if isinstance(value, dict):
            # get node or create one
            node = a.setdefault(key, {})
            deep_merge(value, node)
        else:
            a[key] = value

    return a

deep_merge(test,check)
print(test)
# {'key1': {'key2': {'key3': 'val1', 'key6': 'val4', 'key5': 'val3', 'key4': 'val2'}}}

---

它变异了

test

，没有修改

check

。

显然你不会得到想要的输出

key2

更新了

key2

的

check

@Gahan我可以像测试中的key1和key2一样检查吗如果是的话，如果我给出这个表达式，它会工作吗测试['key1'['key2']['key5']='val3'用户的可能副本希望保留不同的键值，即在合并

x={'a'：1，'b'：{'y'：4}}y={'b'：{'x'：1}，'c'：4}

时，输出将是

{'a'：1，'b'：{'x'：1}，'c'：4}

但用户希望

{'a'：1，'b'：{'x'：1，'y'，'4}，'c''4}>output@Thaian我期望的op是{'key1'：{'key2'：{'key3'：{'val1'，'key4'：'val2'，'key5'：'val3'，'key6'：'val4'}}}}@wazza现在检查用户希望保留不同的键值，即在合并
x={'a'：1'，b'：{'y'：4}y={'b'：{'x'，'1'，'c'：4}
时，正如您所建议的，输出将是
{'a'：1'，'b'：{'x'：1'，'c4'>'c'，'code'
作为输出