计算两个字典之间的最小差值

Question

我试图计算两个字典之间的差异以返回一个特定的值。你知道吗

我输入了不同的值，这些值将返回不同的结果，但结果保持不变。你知道吗

diets = {"normal" : {'p': '32.50', 'c': '60', 'f': '40.86'},
         "oncology" : {'p': '35', 'c': '52.50', 'f': '37.63'},
         "cardiology" : {'p': '32.50', 'c': '30', 'f': '26.88'},
         "diabetes" : {'p': '20', 'c': '27.50', 'f': '27.95'},
         "kidney" : {'p': '15', 'c': '55', 'f': '23.65'}}

amounts = {'p': p, 'c': c, 'f': f}

value = { k : diets[k] for k in set(diets) - set(amounts) }

calculate_error = min(value)

print(calculate_error)

当我输入32、60和40时，返回的结果应该是正常的，但返回的是肿瘤学

Answer 1

看起来你完全搞不清楚value会是什么

>>> diets = {"normal" : {'p':'32.50', 'c':'60', 'f':'40.86'},
...          "oncology" : {'p':'35', 'c':'52.50', 'f':'37.63'},
...          "cardiology" : {'p':'32.50', 'c':'30', 'f':'26.88'},
...          "diabetes" : {'p':'20', 'c':'27.50', 'f':'27.95'},
...          "kidney" : {'p':'15', 'c':'55', 'f':'23.65'}}
>>> set(diets)
{'kidney', 'cardiology', 'oncology', 'normal', 'diabetes'}
>>> amounts = {'p':32, 'c':60, 'f':40}
>>> set(amounts)
{'c', 'f', 'p'}
>>> set(diets) - set(amounts)
{'cardiology', 'diabetes', 'kidney', 'oncology', 'normal'}
>>> value = { k : diets[k] for k in set(diets) - set(amounts) }
>>> value
{'cardiology': {'p': '32.50', 'c': '30', 'f': '26.88'},
 'diabetes': {'p': '20', 'c': '27.50', 'f': '27.95'},
 'kidney': {'p': '15', 'c': '55', 'f': '23.65'},
 'oncology': {'p': '35', 'c': '52.50', 'f': '37.63'},
 'normal': {'p': '32.50', 'c': '60', 'f': '40.86'}}
>>> min(value)
'cardiology'

也就是说，我希望你得到cardiology，也就是diets.keys()的最小值 也就是说，请注意饮食中的值是str，例如“32.50”，您需要在任何计算之前转换这些值。你知道吗

Answer 2

执行此操作时，应查看正在创建的值：

set(diets)

这只是一个钥匙清单。你知道吗

{'cardiology', 'diabetes', 'kidney', 'normal', 'oncology'}

当您减去其他键列表时，您只得到原始列表，因为没有共同的值。你知道吗

你需要一步一步地完成这些项目，然后做减法运算以得到差异。然后你就可以找到微分的和，和的最小值。你知道吗

一种方法是：

diets = {"normal" : {'p': '32.50', 'c': '60', 'f': '40.86'},
         "oncology" : {'p': '35', 'c': '52.50', 'f': '37.63'},
         "cardiology" : {'p': '32.50', 'c': '30', 'f': '26.88'},
         "diabetes" : {'p': '20', 'c': '27.50', 'f': '27.95'},
         "kidney" : {'p': '15', 'c': '55', 'f': '23.65'}}
amounts = {'p': 32., 'c': 60., 'f': 40.}

mins = [(diet, sum([abs(amounts[k] - float(d[k])) for k in amounts])) for diet, d in diets.items()]
the_min = min(mins, key = lambda x: x[1])

mins将是：

[('normal', 1.3599999999999994),
 ('oncology', 12.869999999999997),
 ('cardiology', 43.620000000000005),
 ('diabetes', 56.55),
 ('kidney', 38.35)]

the_min将是：

('normal', 1.3599999999999994)