57 lines
1.4 KiB
Python
57 lines
1.4 KiB
Python
# 假如又两个字典:
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a = {"x":1, "z":3}
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b = {"y":2, "z":4}
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# 怎么将这两个字典映射到一个新字典里呢?愚蠢的办法是使用字典合并
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# 这个方法会把右边的b刷写到a里
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c = a|b
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print(c)
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# 但是,你会发现,如果原字典被改了,那c不会有变化:
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a["x"] = 2
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print(c)
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# 这个时候使用映射方案的chainmap就会很好用
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from collections import ChainMap
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c = ChainMap(a,b)
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print(",".join(str(x) for x in [c['x'], c['y'], c['z']]))
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print(len(c))
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# 如果此时更改原字典的值:
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a["x"] = 1
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# 输出也跟着更改了
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print(",".join(str(x) for x in [c['x'], c['y'], c['z']]))
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print(len(c))
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# 注,chainmap中如果有重复的键,则会采用左值,比如上面的例子使用a["z"]作为重复键z的取值;
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# 如果修改字典值,也会作用在左值上
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del c["z"]
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print(",".join(str(x) for x in [c['x'], c['y'], c['z']]))
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print(a, b)
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# 如果尝试删除a里已经没有但b里有的键,那么会报错
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try:
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del c["z"]
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except:
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print("在a里没找到键")
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# 这东西的底层其实是一个链表，演示如下：
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dic = ChainMap()
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dic["x"] = 1
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dic = dic.new_child()
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print(dic)
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dic["x"] = 2
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dic = dic.new_child()
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print(dic)
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dic["x"] = 3
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print(dic)
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# 从这里我们开始回溯历史
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dic = dic.parents
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print(dic)
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dic = dic.parents
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print(dic)
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dic = dic.parents
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print(dic)
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# 可以看到我们用new_child()函数创建了一个新节点；用parents属性来回溯 |