I would like to generate one hot encoding for a set of DNA sequences. For example the sequence ACGTCCA can be represented as below in a transpose manner. But the code below will generate the one hot encoding in horizontal way in which I would prefer it in vertical form. Can anyone help me?
我想为一组DNA序列生成一个热编码。例如,序列ACGTCCA可以以转置方式如下表示。但是下面的代码将以水平方式生成一个热编码,我希望它以垂直形式。谁能帮我?
ACGTCCA
1000001 - A
0100110 - C
0010000 - G
0001000 - T
Example code:
from sklearn.preprocessing import OneHotEncoder
import itertools
# two example sequences
seqs = ["ACGTCCA","CGGATTG"]
# split sequences to tokens
tokens_seqs = [seq.split("\\") for seq in seqs]
# convert list of of token-lists to one flat list of tokens
# and then create a dictionary that maps word to id of word,
# like {A: 1, B: 2} here
all_tokens = itertools.chain.from_iterable(tokens_seqs)
word_to_id = {token: idx for idx, token in enumerate(set(all_tokens))}
# convert token lists to token-id lists, e.g. [[1, 2], [2, 2]] here
token_ids = [[word_to_id[token] for token in tokens_seq] for tokens_seq in tokens_seqs]
# convert list of token-id lists to one-hot representation
vec = OneHotEncoder(n_values=len(word_to_id))
X = vec.fit_transform(token_ids)
print X.toarray()
However, the code gives me output:
但是,代码给了我输出:
[[ 0. 1.]
[ 1. 0.]]
Expected output:
[[1. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 1. 1. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0.]
[0. 0. 0. 1. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 1. 1. 0. 0. 0. 1. 0. 0. 0. 0. 1. 1. 0.]]
1 个解决方案
#1
4
I suggest doing it a slightly more manual way:
我建议采用稍微手动的方式:
import numpy as np
seqs = ["ACGTCCA","CGGATTG"]
CHARS = 'ACGT'
CHARS_COUNT = len(CHARS)
maxlen = max(map(len, seqs))
res = np.zeros((len(seqs), CHARS_COUNT * maxlen), dtype=np.uint8)
for si, seq in enumerate(seqs):
seqlen = len(seq)
arr = np.chararray((seqlen,), buffer=seq)
for ii, char in enumerate(CHARS):
res[si][ii*seqlen:(ii+1)*seqlen][arr == char] = 1
print res
This gives you your desired result:
这可以为您提供所需的结果:
[[1 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 1 0]]
#1
4
I suggest doing it a slightly more manual way:
我建议采用稍微手动的方式:
import numpy as np
seqs = ["ACGTCCA","CGGATTG"]
CHARS = 'ACGT'
CHARS_COUNT = len(CHARS)
maxlen = max(map(len, seqs))
res = np.zeros((len(seqs), CHARS_COUNT * maxlen), dtype=np.uint8)
for si, seq in enumerate(seqs):
seqlen = len(seq)
arr = np.chararray((seqlen,), buffer=seq)
for ii, char in enumerate(CHARS):
res[si][ii*seqlen:(ii+1)*seqlen][arr == char] = 1
print res
This gives you your desired result:
这可以为您提供所需的结果:
[[1 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 1 0]]