處理離散型特徵和連續型特徵並存的情況，如何做歸一化。
參考博客進行了總結：
https://www.quora.com/What-are-good-ways-to-handle-discrete-and-continuous-inputs-together
總結如下：
1、拿到獲取的原始特徵，必須對每一特徵分別進行歸一化，比如，特徵A的取值範圍是[-1000,1000]，特徵B的取值範圍是[-1,1].
如果使用logistic迴歸，w1*x1+w2*x2，因爲x1的取值太大了，所以x2基本起不了作用。
所以，必須進行特徵的歸一化，每個特徵都單獨進行歸一化。
2、連續型特徵歸一化的常用方法：
2.1：Rescale bounded continuous features: All continuous input that are bounded, rescale them to [-1, 1] through x = (2x - max - min)/(max - min).線性放縮到[-1,1]
2.2:Standardize all continuous features: All continuous input should be standardized and by this I mean, for every continuous feature, compute its mean (u) and standard deviation (s) and do x = (x - u)/s.放縮到均值爲0，方差爲1
1、離散型特徵的處理方法：
a) Binarize categorical/discrete features: For all categorical features, represent them as multiple boolean features. For example, instead of having one feature called marriage_status, have 3 boolean features - married_status_single, married_status_married, married_status_divorced and appropriately set these features to 1 or -1. As you can see, for every categorical feature, you are adding k binary feature where k is the number of values that the categorical feature takes.對於離散的特徵基本就是按照one-hot編碼，該離散特徵有多少取值，就用多少維來表示該特徵。
爲什麼使用one-hot編碼來處理離散型特徵，這是有理由的，不是隨便拍腦袋想出來的！！！具體原因，分下面幾點來闡述：
1、Why do we binarize categorical features?
We binarize the categorical input so that they can be thought of as a vector from the Euclidean space (we call this as embedding the vector in the Euclidean space).使用one-hot編碼，將離散特徵的取值擴展到了歐式空間，離散特徵的某個取值就對應歐式空間的某個點。
2、Why do we embed the feature vectors in the Euclidean space?
Because many algorithms for classification/regression/clustering etc. requires computing distances between features or similarities between features. And many definitions of distances and similarities are defined over features in Euclidean space. So, we would like our features to lie in the Euclidean space as well.將離散特徵通過one-hot編碼映射到歐式空間，是因爲，在迴歸，分類，聚類等機器學習算法中，特徵之間距離的計算或相似度的計算是非常重要的，而我們常用的距離或相似度的計算都是在歐式空間的相似度計算，計算餘弦相似性，基於的就是歐式空間。
3、Why does embedding the feature vector in Euclidean space require us to binarize categorical features?
Let us take an example of a dataset with just one feature (say job_type as per your example) and let us say it takes three values 1,2,3.
Now, let us take three feature vectors x_1 = (1), x_2 = (2), x_3 = (3). What is the euclidean distance between x_1 and x_2, x_2 and x_3 & x_1 and x_3? d(x_1, x_2) = 1, d(x_2, x_3) = 1, d(x_1, x_3) = 2. This shows that distance between job type 1 and job type 2 is smaller than job type 1 and job type 3. Does this make sense? Can we even rationally define a proper distance between different job types? In many cases of categorical features, we can properly define distance between different values that the categorical feature takes. In such cases, isn't it fair to assume that all categorical features are equally far away from each other?
Now, let us see what happens when we binary the same feature vectors. Then, x_1 = (1, 0, 0), x_2 = (0, 1, 0), x_3 = (0, 0, 1). Now, what are the distances between them? They are sqrt(2). So, essentially, when we binarize the input, we implicitly state that all values of the categorical features are equally away from each other.
將離散型特徵使用one-hot編碼，確實會讓特徵之間的距離計算更加合理。比如，有一個離散型特徵，代表工作類型，該離散型特徵，共有三個取值，不使用one-hot編碼，其表示分別是x_1 = (1), x_2 = (2), x_3 = (3)。兩個工作之間的距離是，(x_1, x_2) = 1, d(x_2, x_3) = 1, d(x_1, x_3) = 2。那麼x_1和x_3工作之間就越不相似嗎？顯然這樣的表示，計算出來的特徵的距離是不合理。那如果使用one-hot編碼，則得到x_1 = (1, 0, 0), x_2 = (0, 1, 0), x_3 = (0, 0, 1)，那麼兩個工作之間的距離就都是sqrt(2).即每兩個工作之間的距離是一樣的，顯得更合理。
4、About the original question?
Note that our reason for why binarize the categorical features is independent of the number of the values the categorical features take, so yes, even if the categorical feature takes 1000 values, we still would prefer to do binarization.
對離散型特徵進行one-hot編碼是爲了讓距離的計算顯得更加合理。
5、Are there cases when we can avoid doing binarization?
Yes. As we figured out earlier, the reason we binarize is because we want some meaningful distance relationship between the different values. As long as there is some meaningful distance relationship, we can avoid binarizing the categorical feature. For example, if you are building a classifier to classify a webpage as important entity page (a page important to a particular entity) or not and let us say that you have the rank of the webpage in the search result for that entity as a feature, then 1] note that the rank feature is categorical, 2] rank 1 and rank 2 are clearly closer to each other than rank 1 and rank 3, so the rank feature defines a meaningful distance relationship and so, in this case, we don't have to binarize the categorical rank feature.
More generally, if you can cluster the categorical values into disjoint subsets such that the subsets have meaningful distance relationship amongst them, then you don't have binarize fully, instead you can split them only over these clusters. For example, if there is a categorical feature with 1000 values, but you can split these 1000 values into 2 groups of 400 and 600 (say) and within each group, the values have meaningful distance relationship, then instead of fully binarizing, you can just add 2 features, one for each cluster and that should be fine.
將離散型特徵進行one-hot編碼的作用，是爲了讓距離計算更合理，但如果特徵是離散的，並且不用one-hot編碼就可以很合理的計算出距離，那麼就沒必要進行one-hot編碼，比如，該離散特徵共有1000個取值，我們分成兩組，分別是400和600,兩個小組之間的距離有合適的定義，組內的距離也有合適的定義，那就沒必要用one-hot 編碼
離散特徵進行one-hot編碼後，編碼後的特徵，其實每一維度的特徵都可以看做是連續的特徵。就可以跟對連續型特徵的歸一化方法一樣，對每一維特徵進行歸一化。比如歸一化到[-1,1]或歸一化到均值爲0,方差爲1
有些情況不需要進行特徵的歸一化：
It depends on your ML algorithms, some methods requires almost no efforts to normalize features or handle both continuous and discrete features, like tree based methods: c4.5, Cart, random Forrest, bagging or boosting. But most of parametric models (generalized linear models, neural network, SVM,etc) or methods using distance metrics (KNN, kernels, etc) will require careful work to achieve good results. Standard approaches including binary all features, 0 mean unit variance all continuous features, etc。
基於樹的方法是不需要進行特徵的歸一化，例如隨機森林，bagging 和 boosting等。基於參數的模型或基於距離的模型，都是要進行特徵的歸一化。