我有一个dict of dict of dict,我想通过组合两个最外层的键来转换成dict of dict。有什么优雅的方法吗
这是最初的格言:
alldic={}
for a in range(3):
machinedic={}
for x in range(5):
machinetimedic={}
for i in range(4):
machinetimedic["variable0"]=0
machinetimedic["variable1"]=1
machinedic["time "+str(x)]=machinetimedic
alldic["machine "+str(a)]=machinedic
我想要的是这样一个口述:
{'machine 0-time 0': {'variable0': 0, 'variable1': 1},
'machine 0-time 1': {'variable0': 0, 'variable1': 1},
'machine 0-time 2': {'variable0': 0, 'variable1': 1},
'machine 0-time 3': {'variable0': 0, 'variable1': 1},
'machine 0-time 4': {'variable0': 0, 'variable1': 1},
'machine 1-time 0': {'variable0': 0, 'variable1': 1},
'machine 1-time 1': {'variable0': 0, 'variable1': 1},
'machine 1-time 2': {'variable0': 0, 'variable1': 1},
'machine 1-time 3': {'variable0': 0, 'variable1': 1},
'machine 1-time 4': {'variable0': 0, 'variable1': 1},
'machine 2-time 0': {'variable0': 0, 'variable1': 1},
'machine 2-time 1': {'variable0': 0, 'variable1': 1},
'machine 2-time 2': {'variable0': 0, 'variable1': 1},
'machine 2-time 3': {'variable0': 0, 'variable1': 1},
'machine 2-time 4': {'variable0': 0, 'variable1': 1}}
迭代效果非常好:
与其筑巢,不如像这样把它压扁
或者,你可以得到同样的结果,通过字典理解,就像这样
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