pandas 如何关联熊猫中的序数分类列？
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How to correlate an Ordinal Categorical column in pandas?
提问by yousraHazem
I have a DataFrame dfwith a non-numerical column CatColumn.
我有一个df带有非数字列的 DataFrame CatColumn。
A B CatColumn
0 381.1396 7.343921 Medium
1 481.3268 6.786945 Medium
2 263.3766 7.628746 High
3 177.2400 5.225647 Medium-High
I want to include CatColumnin the correlation analysis with other columns in the Dataframe. I tried DataFrame.corrbut it does not include columns with nominal values in the correlation analysis.
我想CatColumn与 Dataframe 中的其他列进行相关性分析。我试过，DataFrame.corr但它不包括相关分析中具有标称值的列。
回答by FatihAkici
I am going to stronglydisagree with the other comments.
我将强烈反对其他评论。
They miss the main point of correlation: How much does variable 1 increase or decrease as variable 2 increases or decreases. So in the very first place, order of the ordinal variable must be preserved during factorization/encoding. If you alter the order of variables, correlation will change completely. If you are building a tree-based method, this is a non-issue but for a correlation analysis, special attention must be paid to preservation of order in an ordinal variable.
他们忽略了相关性的要点：随着变量 2 的增加或减少，变量 1 增加或减少了多少。因此，首先，在分解/编码期间必须保留序数变量的顺序。如果您改变变量的顺序，相关性将完全改变。如果您正在构建基于树的方法，这不是问题，但对于相关分析，必须特别注意顺序变量中的顺序保持。
Let me make my argument reproducible. A and B are numeric, C is ordinal categorical in the following table, which is intentionally slightly altered from the one in the question.
让我使我的论点可重复。A 和 B 是数字，C 是下表中的序数分类，有意与问题中的那个略有不同。
rawText = StringIO("""
A B C
0 100.1396 1.343921 Medium
1 105.3268 1.786945 Medium
2 200.3766 9.628746 High
3 150.2400 4.225647 Medium-High
""")
myData = pd.read_csv(rawText, sep = "\s+")
Notice: As C moves from Medium to Medium-High to High, both A and B increase monotonically. Hence we should see strong correlations between tuples (C,A) and (C,B). Let's reproduce the two proposed answers:
注意：随着 C 从 Medium 移动到 Medium-High 再到 High，A 和 B 都单调增加。因此我们应该看到元组 (C,A) 和 (C,B) 之间的强相关性。让我们重现两个建议的答案：
In[226]: myData.assign(C=myData.C.astype('category').cat.codes).corr()
Out[226]:
A B C
A 1.000000 0.986493 -0.438466
B 0.986493 1.000000 -0.579650
C -0.438466 -0.579650 1.000000
Wait... What? Negative correlations? How come? Something is definitely not right. So what is going on?
等等……什么？负相关？怎么来的？有些事情肯定是不对的。那么发生了什么？
What is going on is that C is factorized according to the alphanumerical sorting of its values. [High, Medium, Medium-High] are assigned [0, 1, 2], therefore the ordering is altered: 0 < 1 < 2 implies High < Medium < Medium-High, which is not true. Hence we accidentally calculated the response of A and B as C goes from High to Medium to Medium-High. The correct answer must preserve ordering, and assign [2, 0, 1] to [High, Medium, Medium-High]. Here is how:
发生的事情是 C 根据其值的字母数字排序被分解。[High, Medium, Medium-High] 被分配为 [0, 1, 2]，因此排序发生了变化：0 < 1 < 2 意味着 High < Medium < Medium-High，这是不正确的。因此，当 C 从高到中再到中高时，我们意外地计算了 A 和 B 的响应。正确答案必须保持排序，并将 [2, 0, 1] 分配给 [高、中、中-高]。方法如下：
In[227]: myData['C'] = myData['C'].astype('category')
myData['C'].cat.categories = [2,0,1]
myData['C'] = myData['C'].astype('float')
myData.corr()
Out[227]:
A B C
A 1.000000 0.986493 0.998874
B 0.986493 1.000000 0.982982
C 0.998874 0.982982 1.000000
Much better!
好多了！
Note1: If you want to treat your variable as a nominal variable, you can look at things like contingency tables, Cramer's V and the like; or group the continuous variable by the nominal categories etc. I don't think it would be right, though.
注1：如果你想把你的变量当作名义变量，你可以看看列联表、Cramer's V等；或按名义类别等对连续变量进行分组。不过，我认为这是不对的。
Note2: If you had another category called Low, my answer could be criticized due to the fact that I assigned equally spaced numbers to unequally spaced categories. You could make the argument that one should assign [2, 1, 1.5, 0] to [High, Medium, Medium-High, Small], which would be valid. I believe this is what people call the art part of data science.
注2：如果您有另一个名为“低”的类别，我的回答可能会受到批评，因为我将等间距的数字分配给不等间距的类别。您可以论证应该将 [2, 1, 1.5, 0] 分配给 [High, Medium, Medium-High, Small]，这是有效的。我相信这就是人们所说的数据科学的艺术部分。
回答by cy-press
Basically, there is no a good scientifical way to do it. I would use the following approach: 1. Split the numeric field into n groups, where n = number of groups of the categorical field. 2. Calculate Cramer correlation between the 2 categorical fields.
基本上，没有一个好的科学方法来做到这一点。我将使用以下方法： 1. 将数字字段拆分为 n 组，其中 n = 分类字段的组数。2. 计算 2 个分类字段之间的 Cramer 相关性。
回答by ei-grad
The right way to correlate a categorical column with N values is to split this column into N separate boolean columns.
将分类列与 N 个值相关联的正确方法是将此列拆分为 N 个单独的布尔列。
Lets take the original question dataframe. Make the category columns:
让我们采用原始问题数据框。制作类别列：
for i in df.CatColumn.astype('category'):
df[i] = df.CatColumn == i
Then it is possible to calculate the correlation between every category and other columns:
然后可以计算每个类别与其他列之间的相关性：
df.corr()
Output:
输出：
A B Medium High Medium-High
A 1.000000 0.490608 0.914322 -0.312309 -0.743459
B 0.490608 1.000000 0.343620 0.548589 -0.945367
Medium 0.914322 0.343620 1.000000 -0.577350 -0.577350
High -0.312309 0.548589 -0.577350 1.000000 -0.333333
Medium-High -0.743459 -0.945367 -0.577350 -0.333333 1.000000
