I have an original dataframe of sequences listed below and am trying to use one-hot encoding and then store these in a new dataframe, I am trying to do it with the following code but am not able to store because I get the following output afterwards:
Code:
onehot_encoder = OneHotEncoder()
sequence = np.array(list(x_train['sequence'])).reshape(-1, 1)
encoded_sequence = onehot_encoder.fit_transform(sequence).toarray()
encoded_sequence
but get error
ValueError: Wrong number of items passed 12755, placement implies 1
You get that strange array because it treats every sequence as an entry and tries to one-hot encode it, we can use an example:
import pandas as pd
from sklearn.preprocessing import OneHotEncoder
df = pd.DataFrame({'sequence':['AQAVPW','AMAVLT','LDTGIN']})
enc = OneHotEncoder()
seq = np.array(df['sequence']).reshape(-1,1)
encoded = enc.fit(seq)
encoded.transform(seq).toarray()
array([[0., 1., 0.],
[1., 0., 0.],
[0., 0., 1.]])
encoded.categories_
[array(['AMAVLT', 'AQAVPW', 'LDTGIN'], dtype=object)]
Since your entries are unique, you get this all zeros matrix. You can understand this better if you use pd.get_dummies
pd.get_dummies(df['sequence'])
AMAVLT AQAVPW LDTGIN
0 0 1 0
1 1 0 0
2 0 0 1
There's two ways to do this, one way is to simply count the amino acid occurrence and use that as a predictor, I hope I get the amino acids correct (from school long time ago):
from Bio import SeqIO
from Bio.SeqUtils.ProtParam import ProteinAnalysis
pd.DataFrame([ProteinAnalysis(i).count_amino_acids() for i in df['sequence']])
A C D E F G H I K L M N P Q R S T V W Y
0 2 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 1 0
1 2 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0
2 0 0 1 0 0 1 0 1 0 1 0 1 0 0 0 0 1 0 0 0
The other is to split the sequences, and do this encoding by position, and this requires the sequences to be equally long, and that you have enough memory:
byposition = df['sequence'].apply(lambda x:pd.Series(list(x)))
byposition
0 1 2 3 4 5
0 A Q A V P W
1 A M A V L T
2 L D T G I N
pd.get_dummies(byposition)
0_A 0_L 1_D 1_M 1_Q 2_A 2_T 3_G 3_V 4_I 4_L 4_P 5_N 5_T 5_W
0 1 0 0 0 1 1 0 0 1 0 0 1 0 0 1
1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0
2 0 1 1 0 0 0 1 1 0 1 0 0 1 0 0