我寫了一個小程序來執行以下操作。我想知道是否有明顯更優化的解決方案:將字符串列表拆分爲與字符串不同列表匹配的位置
1)取2個字符串列表。通常,第二個列表中的字符串將比第一個列表中的字符串長,但不保證
2)返回從第二個列表派生的字符串列表,該列表已從第一個列表中刪除任何匹配的字符串。因此,該列表將包含< =第二個列表中字符串的長度。
到目前爲止,我這就是我。它似乎工作正常,但我只是好奇,如果有一個更優雅的解決方案,我錯過了。順便說一下,我會跟蹤字符串每個開始和結束的「位置」,這對於本程序的後續部分很重要。
def split_sequence(sequence = "", split_seq = "", length = 8):
if len(sequence) < len(split_seq):
return [],[]
split_positions = [0]
for pos in range(len(sequence)-len(split_seq)):
if sequence[pos:pos+len(split_seq)] == split_seq and pos > split_positions[-1]:
split_positions += [pos, pos+len(split_seq)]
if split_positions[-1] == 0:
return [sequence], [(0,len(sequence)-1)]
split_positions.append(len(sequence))
assert len(split_positions) % 2 == 0
split_sequences = [sequence[split_positions[_]:split_positions[_+1]] for _ in range(0, len(split_positions),2)]
split_seq_positions = [(split_positions[_],split_positions[_+1]) for _ in range(0, len(split_positions),2)]
return_sequences = []
return_positions = []
for pos,seq in enumerate(split_sequences):
if len(seq) >= length:
return_sequences.append(split_sequences[pos])
return_positions.append(split_seq_positions[pos])
return return_sequences, return_positions
def create_sequences_from_targets(sequence_list = [] , positions_list = [],length=8, avoid = []):
if avoid:
for avoided_seq in avoid:
new_sequence_list = []
new_positions_list = []
for pos,sequence in enumerate(sequence_list):
start = positions_list[pos][0]
seqs, positions = split_sequence(sequence = sequence, split_seq = avoided_seq, length = length)
new_sequence_list += seqs
new_positions_list += [(positions[_][0]+start,positions[_][1]+start) for _ in range(len(positions))]
return new_sequence_list, new_positions_list
輸出示例:
In [60]: create_sequences_from_targets(sequence_list=['MPHSSLHPSIPCPRGHGAQKA', 'AEELRHIHSRYRGSYWRTVRA', 'KGLAPAEISAVCEKGNFNVA'],positions_list=[(0, 20), (66, 86), (136, 155)],avoid=['SRYRGSYW'],length=3)
Out[60]:
(['MPHSSLHPSIPCPRGHGAQKA', 'AEELRHIH', 'RTVRA', 'KGLAPAEISAVCEKGNFNVA'],
[(0, 20), (66, 74), (82, 87), (136, 155)])
'string.split()'接受一個子串分隔符。你的算法看起來像你可以迭代分隔字符串。 –