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[英]How to translate a Decision Tree in Orange to rules and print out the rules?
[英]How to print out a recursive decision tree?
我需要打印出一個遞歸樹,如下所示。 我希望這段代碼能夠通用,並適用於其他樹。 我只是在想出如何打印遞歸樹的麻煩。
使用此字典的PrintDt(Tennis),例如:
Tennis = ['skurple type', {'gloomp': 'Yes', 'murket': ['which murket', {'grabe': ['grabetime', {'sproon': 'No', 'midnort': 'Yes'}], 'carfer': 'Yes', 'glambit': 'No'}], 'snarzle': ['froovencity level', {'granky': 'No', 'slipely': 'Yes'}]}]
輸出應為:
彎頭型拆分
If skurple type == gloomp
Return: Yes
If skurple type == murket
Split on which murket
If which murket == grabe
Split on grabetime
If grabetime == sproon
Return: No
If grabetime == midnort
Return: Yes
If which murket == carfer
Return: Yes
If which murket == glambit
Return: No
If skurple type == snarzle
Split on froovencity level
If froovencity level == granky
Return: No
If froovencity level == slipely
Return: Yes
看起來像這樣可以工作:
def PrintDt(tree, depth=0):
if type(tree) == str:
print depth*'\t', 'Return:', tree
else:
print depth*'\t', 'Split on', tree[0]
for value in tree[1].keys():
print depth*'\t', 'If', tree[0], '==', value
PrintDt(tree[1][value], depth + 1)
例:
>>> def PrintDt(tree, depth=0):
... if type(tree) == str:
... print depth*'\t', 'Return:', tree
... else:
... print depth*'\t', 'Split on', tree[0]
... for value in tree[1].keys():
... print depth*'\t', 'If', tree[0], '==', value
... PrintDt(tree[1][value], depth + 1)
...
>>> Tennis = ['skurple type', {'gloomp': 'Yes', 'murket': ['which murket', {'grabe': ['grabetime', {'sproon': 'No', 'midnort': 'Yes'}], 'carfer': 'Yes', 'glambit': 'No'}], 'snarzle': ['froovencity level', {'granky': 'No', 'slipely': 'Yes'}]}]
>>> PrintDt(Tennis)
Split on skurple type
If skurple type == gloomp
Return: Yes
If skurple type == murket
Split on which murket
If which murket == grabe
Split on grabetime
If grabetime == sproon
Return: No
If grabetime == midnort
Return: Yes
If which murket == carfer
Return: Yes
If which murket == glambit
Return: No
If skurple type == snarzle
Split on froovencity level
If froovencity level == granky
Return: No
If froovencity level == slipely
Return: Yes
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