合并两个字典并保留第一个字典的值 [英] Merge two dictionaries and persist the values of first dictionaries

问题描述

我有两个字典:

first = {"phone": {
                "home": "[phone]"
            },
         "address":"xyz" 
            }


second = {"phone": {
                "home": "[phone]",
                "home1": "[phone]"
            },
         "address":{} 
            } 

我想先合并两个字典，这意味着第一个字典不会丢失其先前的值，而只有丢失的键值会被添加到第一个字典中.

I want merge two dictionaries first over second, meaning the first dictionary doesn't lose its previous values and only the missing key values are added into first.

最终的字典应如下所示:

The final dictionary should look like this:

final = {"phone": {
                "home": "[phone]",
                "home1": "[phone]"
            },
         "address":"xyz" 
            } 

如果我们使用字典列表，例如:-

Also what if when we use list of dictionaries like : -

    first = {"phone": {
                "home": "[phone]"
            },
         "address":[{"home":""},{"office":""}]
            }

推荐答案

Update()将起作用，只需复制第二个字典并应用第一个字典.因此，第一个的值将替换第二个的值，而不是相反的方法.

Update() will work, simply copy the second dict and apply the first one. So the value of the first will replace the second and not the reverse way.

first = {"phone": {"home": "[phone]"},"address":"xyz"}
second = {"phone": {"home": "[phone]","home1": "[phone]"},"address":{}}

# recursive deep update you can find here: http://stackoverflow.com/a/3233356/956660
import collections

def update(d, u):
    for k, v in u.iteritems():
        if isinstance(v, collections.Mapping):
            r = update(d.get(k, {}), v)
            d[k] = r
        else:
            d[k] = u[k]
    return d

third = second.copy()
update(third, first)

print(third)

{'phone':{'home':'[phone]'，'home1':'[phone]'}，'address':'xyz'}

{'phone': {'home': '[phone]', 'home1': '[phone]'}, 'address': 'xyz'}

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