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[英]How do I return a new dictionary if the keys in one dictionary, match the keys in another dictionary?
[英]How do I delete nan-valued keys from one dictionary into another dictionary
我有這本字典:
d = {4: {0: [12], 1: [194]}, 8: {0: [222], 1: nan}, 18: {0: [60], 1: nan},
19: {0: [128], 1: nan}, 21: {0: [54], 1: nan}}
output 應如下所示:
d = {4: {0: [12], 1: [194]}, 8: {0: [222] },
18: {0: [60] }, 19: {0: [128]}, 21: {0: [54]}}
您可以遞歸地將字典傳遞給 function 過濾掉具有nan
值的鍵。 例如:
from math import nan
d = {4: {0: [12], 1: [194]}, 8: {0: [222], 1: nan}, 18: {0: [60], 1: nan},
19: {0: [128], 1: nan}, 21: {0: [54], 1: nan}}
def clean_nan(d):
if isinstance(d, dict):
return {k: clean_nan(v) for k, v in d.items() if v == v}
return d
clean_nan(d)
給出結果:
{4: {0: [12], 1: [194]},
8: {0: [222]},
18: {0: [60]},
19: {0: [128]},
21: {0: [54]}}
這假設列表不包含nan
,但如果它們包含,則很容易添加另一個子句來過濾它們。
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