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?
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.]]
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
扩展@Martin Thoma的答案
def one_hot_encode(y):
"""Convert an iterable of indices to one-hot encoded labels."""
y = y.flatten() # Sometimes not flattened vector is passed e.g (118,1) in these cases
# the function ends up creating a tensor e.g. (118, 2, 1). flatten removes this issue
nb_classes = len(np.unique(y)) # get the number of unique classes
standardised_labels = dict(zip(np.unique(y), np.arange(nb_classes))) # get the class labels as a dictionary
# which then is standardised. E.g imagine class labels are (4,7,9) if a vector of y containing 4,7 and 9 is
# directly passed then np.eye(nb_classes)[4] or 7,9 throws an out of index error.
# standardised labels fixes this issue by returning a dictionary;
# standardised_labels = {4:0, 7:1, 9:2}. The values of the dictionary are mapped to keys in y array.
# standardised_labels also removes the error that is raised if the labels are floats. E.g. 1.0; element
# cannot be called by an integer index e.g y[1.0] - throws an index error.
targets = np.vectorize(standardised_labels.get)(y) # map the dictionary values to array.
return np.eye(nb_classes)[targets]
你可以用numpy来做。眼和一个使用数组元素的选择机制:
import numpy as np
nb_classes = 6
data = [[2, 3, 4, 0]]
def indices_to_one_hot(data, nb_classes):
"""Convert an iterable of indices to one-hot encoded labels."""
targets = np.array(data).reshape(-1)
return np.eye(nb_classes)[targets]
indices_to_one_hot(nb_classes, data)的返回值现在是
array([[[ 0., 0., 1., 0., 0., 0.],
[ 0., 0., 0., 1., 0., 0.],
[ 0., 0., 0., 0., 1., 0.],
[ 1., 0., 0., 0., 0., 0.]]])
. remodeling(-1)的作用是确保标签格式正确(也可能有[[2]，[3]，[4]，[0]])。
