Question

假设以下Python函数从字符串中删除破折号（“gap”），同时在此字符串上保留正确的注释。输入变量 instring 和 annotations 分别构成字符串和字典。

def DegapButMaintainAnno(instring, annotations):
    degapped_instring = ''
    degapped_annotations = {}
    gaps_cumulative = 0
    for range_name, index_list in annotations.items():
        gaps_within_range = 0
        for pos, char in enumerate(instring):
            if pos in index_list and char == '-':
                index_list.remove(pos)
                gaps_within_range += 1
            if pos in index_list and char != '-':
                degapped_instring += char
                index_list[index_list.index(pos)] = pos - gaps_within_range
        index_list = [i-gaps_cumulative for i in index_list]
        degapped_annotations[range_name] = index_list
        gaps_cumulative += gaps_within_range
    return (degapped_instring, degapped_annotations)

如果输入字典指定的范围都没有重叠，则所述函数按预期工作：

>>> instr = "A--AT--T"
>>> annot = {"range1":[0,1,2,3,4], "range2":[5,6,7]}
>>> DegapButMaintainAnno(instr, annot)
Out: ('AATT', {'range1': [0, 1, 2], 'range2': [3]})

但是，只要一个或多个范围重叠，代码就会失败：

>>> annot = {"range1":[0,1,2,3,4], "range2":[4,5,6,7]}
>>> DegapButMaintainAnno(instr, annot)
Out: ('AATTT', {'range1': [0, 1, 2], 'range2': [2, 3]}) # See additional 'T' in string

是否有人建议如何更正重叠范围的代码？

Answer 1

我认为你可能会过度思考问题。这是我的尝试：

from copy import copy

def rewriteGene(instr, annos):
    annotations = copy(annos)
    index = instr.find('-')
    while index > -1:
        for key, ls in annotations.items():
            if index in ls:
                ls.remove(index)
            annotations[key] = [e-1 if e > index else e for e in ls]
        instr = instr[:index] + instr[index+1:]
        index = instr.find('-')
    return instr, annotations

instr = "A--AT--T"
annos = {"range1":[0,1,2,3,4], "range2":[4,5,6,7]}

print rewriteGene(instr, annos)
# ('AATT', {'range2': [2, 3], 'range1': [0, 1, 2]})

它应该是非常易读的，但是如果你想澄清任何事情，请告诉我。

在从字符串中删除短划线时保持重叠注释

1 个答案: