我有一个80%类别变量的机器学习分类问题。如果我想使用一些分类器进行分类,我必须使用一个热编码吗?我可以将数据传递给分类器而不进行编码吗?

我试图做以下的特征选择:

I read the train file: num_rows_to_read = 10000 train_small = pd.read_csv("../../dataset/train.csv", nrows=num_rows_to_read) I change the type of the categorical features to 'category': non_categorial_features = ['orig_destination_distance', 'srch_adults_cnt', 'srch_children_cnt', 'srch_rm_cnt', 'cnt'] for categorical_feature in list(train_small.columns): if categorical_feature not in non_categorial_features: train_small[categorical_feature] = train_small[categorical_feature].astype('category') I use one hot encoding: train_small_with_dummies = pd.get_dummies(train_small, sparse=True)

问题是,第三部分经常卡住,尽管我使用的是一个强大的机器。

因此,如果没有一个热编码,我就无法进行任何特征选择,以确定特征的重要性。

你有什么建议吗?


当前回答

一个在numpy中使用矢量化并在pandas中应用的简单示例:

import numpy as np

a = np.array(['male','female','female','male'])

#define function
onehot_function = lambda x: 1.0 if (x=='male') else 0.0

onehot_a = np.vectorize(onehot_function)(a)

print(onehot_a)
# [1., 0., 0., 1.]

# -----------------------------------------

import pandas as pd

s = pd.Series(['male','female','female','male'])
onehot_s = s.apply(onehot_function)

print(onehot_s)
# 0    1.0
# 1    0.0
# 2    0.0
# 3    1.0
# dtype: float64

其他回答

简短的回答

这里有一个函数,可以在不使用numpy、pandas或其他包的情况下进行一次性编码。它接受一个整数、布尔值或字符串(也可能是其他类型)的列表。

import typing


def one_hot_encode(items: list) -> typing.List[list]:
    results = []
    # find the unique items (we want to unique items b/c duplicate items will have the same encoding)
    unique_items = list(set(items))
    # sort the unique items
    sorted_items = sorted(unique_items)
    # find how long the list of each item should be
    max_index = len(unique_items)

    for item in items:
        # create a list of zeros the appropriate length
        one_hot_encoded_result = [0 for i in range(0, max_index)]
        # find the index of the item
        one_hot_index = sorted_items.index(item)
        # change the zero at the index from the previous line to a one
        one_hot_encoded_result[one_hot_index] = 1
        # add the result
        results.append(one_hot_encoded_result)

    return results

例子:

one_hot_encode([2, 1, 1, 2, 5, 3])

# [[0, 1, 0, 0],
#  [1, 0, 0, 0],
#  [1, 0, 0, 0],
#  [0, 1, 0, 0],
#  [0, 0, 0, 1],
#  [0, 0, 1, 0]]
one_hot_encode([True, False, True])

# [[0, 1], [1, 0], [0, 1]]
one_hot_encode(['a', 'b', 'c', 'a', 'e'])

# [[1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [1, 0, 0, 0], [0, 0, 0, 1]]

长(er)回答

I know there are already a lot of answers to this question, but I noticed two things. First, most of the answers use packages like numpy and/or pandas. And this is a good thing. If you are writing production code, you should probably be using robust, fast algorithms like those provided in the numpy/pandas packages. But, for the sake of education, I think someone should provide an answer which has a transparent algorithm and not just an implementation of someone else's algorithm. Second, I noticed that many of the answers do not provide a robust implementation of one-hot encoding because they do not meet one of the requirements below. Below are some of the requirements (as I see them) for a useful, accurate, and robust one-hot encoding function:

单热编码函数必须:

处理各种类型的列表(例如,整数,字符串,浮点数等)作为输入 处理带有重复项的输入列表 返回与输入相对应(与输入顺序相同)的列表的列表 返回一个列表的列表,其中每个列表都尽可能短

我测试了这个问题的许多答案,其中大多数都未能满足上述要求之一。

试试这个:

!pip install category_encoders
import category_encoders as ce

categorical_columns = [...the list of names of the columns you want to one-hot-encode ...]
encoder = ce.OneHotEncoder(cols=categorical_columns, use_cat_names=True)
df_train_encoded = encoder.fit_transform(df_train_small)

df_encoded.head ()

生成的数据框架df_train_encoded与原始数据框架相同,但是分类特征现在被它们的单热编码版本所取代。

更多关于category_encoders的信息请点击这里。

您可以使用numpy。眼睛的功能。

import numpy as np

def one_hot_encode(x, n_classes):
    """
    One hot encode a list of sample labels. Return a one-hot encoded vector for each label.
    : x: List of sample Labels
    : return: Numpy array of one-hot encoded labels
     """
    return np.eye(n_classes)[x]

def main():
    list = [0,1,2,3,4,3,2,1,0]
    n_classes = 5
    one_hot_list = one_hot_encode(list, n_classes)
    print(one_hot_list)

if __name__ == "__main__":
    main()

结果

D:\Desktop>python test.py
[[ 1.  0.  0.  0.  0.]
 [ 0.  1.  0.  0.  0.]
 [ 0.  0.  1.  0.  0.]
 [ 0.  0.  0.  1.  0.]
 [ 0.  0.  0.  0.  1.]
 [ 0.  0.  0.  1.  0.]
 [ 0.  0.  1.  0.  0.]
 [ 0.  1.  0.  0.  0.]
 [ 1.  0.  0.  0.  0.]]

假设在10个变量中,在数据帧中有3个分类变量,分别为cname1、cname2和cname3。 然后下面的代码将自动在新的数据框架中创建一个热编码变量。

import category_encoders as ce
encoder_var=ce.OneHotEncoder(cols=['cname1','cname2','cname3'],handle_unknown='return_nan',return_df=True,use_cat_names=True)
new_df = encoder_var.fit_transform(old_df)

熊猫的热编码非常简单:

def one_hot(df, cols):
    """
    @param df pandas DataFrame
    @param cols a list of columns to encode 
    @return a DataFrame with one-hot encoding
    """
    for each in cols:
        dummies = pd.get_dummies(df[each], prefix=each, drop_first=False)
        df = pd.concat([df, dummies], axis=1)
    return df

编辑:

使用sklearn的LabelBinarizer实现one_hot的另一种方法:

from sklearn.preprocessing import LabelBinarizer 
label_binarizer = LabelBinarizer()
label_binarizer.fit(all_your_labels_list) # need to be global or remembered to use it later

def one_hot_encode(x):
    """
    One hot encode a list of sample labels. Return a one-hot encoded vector for each label.
    : x: List of sample Labels
    : return: Numpy array of one-hot encoded labels
    """
    return label_binarizer.transform(x)