Python：将字典项除以取决于第一个键的值 [英] Python: DIvide items of a dictionary by a value that depends on the first key

问题描述

我相信这个问题与划分每个python字典非常相似，但是我无法处理多维密钥。

I believe that this questions is very similar to Divide each python dictionary by total value, however I am not able to deal with multidimentional key.

所以我有第一本字典：

x={'SBGL': 0.2631678018921199, 'SBAR': 0.00017329658914466182, 'SBKP': 0.05787906885949929, 'SBRJ': 0.01686853063415596, 'SBFZ': 0.002151268003175112, 'SBCH': 0.0, 'SBRF': 0.0010995369794006128, 'SBFI': 0.0, 'SBGR': 0.3555667545433087, 'SBCT': 0.12645173241811486, 'SBFL': 0.0020536641771051302, 'SBSP': 0.12206752790423671, 'SBSV': 0.002883296698699977, 'SBMG': 0.0, 'SBCA': 0.0, 'SBPA': 0.024485612897250063, 'SBPS': 0.0006125138064595806, 'SBCF': 0.005023609170377438, 'SBMO': 0.0020686035382382908, 'SBBR': 0.01744718188871371, 'SBJV': 0.0}

我有第二本字典：

   y= {('SBCF', 'TAM'): 294.0, ('SBCH', 'GLO'): 0.0, ('SBSP', 'ONE'): 0.0, ('SBGL', 'TAM'): 114094.0, ('SBCA', 'PTB'): 0.0, ('SBKP', 'GLO'): 0.0, ('SBMG', 'AZU'): 0.0, ('SBRF', 'GLO'): 1104.0, ('SBPA', 'AZU'): 23367.0, ('SBGR', 'AZU'): 313.0, ('SBGR', 'GLO'): 105170.0, ('SBCT', 'ONE'): 0.0, ('SBKP', 'TAM'): 160.0, ('SBGL', 'ONE'): 25330.0, ('SBGR', 'ONE'): 69043.0, ('SBCT', 'PTB'): 0.0, ('SBRJ', 'TAM'): 8118.0, ('SBPS', 'GLO'): 615.0, ('SBPA', 'TAM'): 1218.0, ('SBSV', 'GLO'): 1691.0, ('SBFI', 'AZU'): 0.0, ('SBSP', 'TAM'): 62158.0, ('SBKP', 'ONE'): 0.0, ('SBGR', 'PTB'): 0.0, ('SBGR', 'TAM'): 182484.0, ('SBCT', 'AZU'): 85180.0, ('SBBR', 'TAM'): 12685.0, ('SBBR', 'GLO'): 4833.0, ('SBGL', 'GLO'): 124812.0, ('SBJV', 'GLO'): 0.0, ('SBCT', 'TAM'): 11374.0, ('SBCT', 'GLO'): 30411.0, ('SBMO', 'GLO'): 1822.0, ('SBCF', 'AZU'): 4750.0, ('SBPA', 'GLO'): 0.0, ('SBCF', 'GLO'): 0.0, ('SBMO', 'TAM'): 255.0, ('SBFL', 'AZU'): 2062.0, ('SBFL', 'GLO'): 0.0, ('SBCA', 'AZU'): 0.0, ('SBRJ', 'GLO'): 7426.0, ('SBFI', 'ONE'): 0.0, ('SBKP', 'AZU'): 57954.0, ('SBFZ', 'TAM'): 97.0, ('SBSP', 'GLO'): 60405.0, ('SBRJ', 'AZU'): 1393.0, ('SBFZ', 'GLO'): 2063.0, ('SBSV', 'TAM'): 1204.0, ('SBAR', 'TAM'): 174.0, ('SBFI', 'GLO'): 0.0}

我想将y的所有项除以依赖于第一个键且来自x的值。更准确地说，所有具有第一个键 SBCF的y项都将除以x [ SBCF]。 y的所有具有第一个键等于 SBCH的项都应除以x ['SBCH']，依此类推...

I want to divide all the items of y by a value that depends on the first key and comes from x. To be more precise, all the items of y that has the first key equal to 'SBCF' will be divided by x['SBCF']. All the items of y that has the first key equal to has the first key equal to 'SBCH' should be divided by x['SBCH'] and so on...

我尝试过这样的事情：

y = {[i,j]: v / (x[i]) for i,j,v in y.items()} 

我遇到了错误

 not enough values to unpack (expected 3, got 2)

推荐答案

包装 i 和 j 带有附加括号，因为 y.items（）返回的形式为<（k1，k2），v> 。

Wrap i and j with an additional set of parentheses, since y.items() returns something of the form <(k1, k2), v>.

需要说明的一些事情：

1. 部门by 0


2. 如何处理 x

1. Division by 0
2. How to handle missing entries in x

{(i, j) : v / max(x.get(i, 1), 1) for (i, j), v in y.items()}

您可以选择处理除以0，或引发异常。另外，如果 x 中缺少 i ，则应考虑如何处理。我假设什么也没做（即除以1）。

You can either choose to handle the division by 0, or let an exception be thrown. Additionally, if i is missing in x, you should think how you'd want to handle that. I've assumed nothing is done (i.e., division by 1).

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