我正在尝试对以下数据应用一种热编码。但是我对输出感到困惑。在进行一次热编码之前，数据的形状为（5,10），在进行一次热编码之后，数据的形状为（5,20）。但是每个字母将被编码为4个元素。因此，在应用一种热编码后，形状应为（5，40）而不是（5,10）。我该如何解决？
X = [[‘A’, ‘G’, ‘T’, ‘G’, ‘T’, ‘C’, ‘T’, ‘A’, ‘A’, ‘C’],
[‘A’, ‘G’, ‘T’, ‘G’, ‘T’, ‘C’, ‘T’, ‘A’, ‘A’, ‘C’],
[‘G’, ‘C’, ‘C’, ‘A’, ‘C’, ‘T’, ‘C’, ‘G’, ‘G’, ‘T’],
[‘G’, ‘C’, ‘C’, ‘A’, ‘C’, ‘T’, ‘C’, ‘G’, ‘G’, ‘T’],
[‘G’, ‘C’, ‘C’, ‘A’, ‘C’, ‘T’, ‘C’, ‘G’, ‘G’, ‘T’]]
Y = np.array(X)
print('Shape of numpy array', Y.shape)
# one hot encoding
onehot_encoder = OneHotEncoder(sparse=False)
onehot_encoded = onehot_encoder.fit_transform(Y)
print(onehot_encoded)
print('Shape of one hot encoding', onehot_encoded.shape)
Output:
Shape of numpy array (5, 10)
[[1. 0. 0. 1. 0. 1. 0. 1. 0. 1. 1. 0. 0. 1. 1. 0. 1. 0. 1. 0.]
[1. 0. 0. 1. 0. 1. 0. 1. 0. 1. 1. 0. 0. 1. 1. 0. 1. 0. 1. 0.]
[0. 1. 1. 0. 1. 0. 1. 0. 1. 0. 0. 1. 1. 0. 0. 1. 0. 1. 0. 1.]
[0. 1. 1. 0. 1. 0. 1. 0. 1. 0. 0. 1. 1. 0. 0. 1. 0. 1. 0. 1.]
[0. 1. 1. 0. 1. 0. 1. 0. 1. 0. 0. 1. 1. 0. 0. 1. 0. 1. 0. 1.]]
Shape of one hot encoding (5, 20)
您需要对每个列分别进行一次热编码，以便为ndarray中的每个列获取4个新列：
X = np.array(X)
# Get unique classes.
classes = np.unique(X)
# Replace classes with itegers.
X = np.searchsorted(classes, X)
# Get an identity matrix.
eye = np.eye(classes.shape[0])
# Iterate over all columns
# and get one-hot encoding for each column.
X = np.concatenate([eye[i] for i in X.T], axis=1)
X.shape
# (5, 40)
考虑以下示例：
[['A', 'G'],
['C', 'C'],
['T', 'A']]
一键编码的ndarray中将获得8（2 x 4）列：
Column 0 Column 1
A C G T A C G T
1 0 0 0 0 0 1 0
0 1 0 0 0 1 0 0
0 0 0 1 1 0 0 0
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