在列表推导中使用.rjust（）进行字符串对齐 - 格式化工作与预期不同

Question

所以，我有这个代码：

def pairwiseScore(seqA, seqB):

    score = 0
    length = len(seqA)
    similarity = []

    for x in xrange(length):

        if seqA[x] == seqB[x]:
            if (x >= 1) and (seqA[x - 1] == seqB[x - 1]):
                score += 3
                similarity.append(x)
            else:
                score += 1
                similarity.append(x)                
        else:
            score -= 1

    return ''.join((seqA, '\n', ''.join(['|'.rjust(x) for x in similarity]), '\n', seqB, '\n', 'Score: ', str(score)))

这是this exercise的解决方案。

它的效果差不多，但是，当我执行时：

print pairwiseScore("ATTCGT", "ATCTAT"), '\n', '\n', pairwiseScore("GATAAATCTGGTCT", "CATTCATCATGCAA")

我得到了这个输出：

ATTCGT
||    |
ATCTAT
Score: 2 

GATAAATCTGGTCT
| |    |     |      |         |
CATTCATCATGCAA
Score: 4

正如您所看到的那样，那些（或竖条）的管道格式不正确。

它应该是这样的：

>>> print pairwiseScore("ATTCGT", "ATCTAT")
ATTCGT
||   |
ATCTAT
Score: 2 
>>> print pairwiseScore("GATAAATCTGGTCT", "CATTCATCATGCAA")
GATAAATCTGGTCT
||  |||  |   
CATTCATCATGCAA
Score: 4 
>>>

我的问题是：

这有什么问题：

''.join(['|'.rjust(x) for x in similarity]

功能？我如何编辑它，使这些管道在输出上格式良好？ 干杯！

Answer 1

您的相似度值是条形的绝对位置，但是您使用rjust（）函数的方式，它应该相对于前一个位置传递位置。

例如，您可以在函数的底部执行此操作：

prev = -1
relative_similarity=[]
for x in similarity:
  relative_similarity.append(x-prev)
  prev=x

return ''.join((seqA, '\n', ''.join(['|'.rjust(x) for x in relative_similarity]), '\n', seqB, '\n', 'Score: ', str(score)))

Answer 2

所以，感谢Vaughn Cato，我找到了解决方案。使用几行代码，我的方法终于奏效了，但没有通过最后一个测试用例 - ＆gt;

这就是代码：

def pairwiseScore(seqA, seqB):

    prev = -1
    score = 0
    length = len(seqA)
    similarity = []
    relative_similarity = []

    for x in xrange(length):

        if seqA[x] == seqB[x]:
            if (x >= 1) and (seqA[x - 1] == seqB[x - 1]):
                score += 3
                similarity.append(x)
            else:
                score += 1
                similarity.append(x)                
        else:
            score -= 1

    for x in similarity:

        relative_similarity.append(x - prev)
        prev = x

    return ''.join((seqA, '\n', ''.join(['|'.rjust(x) for x in relative_similarity]), '\n', seqB, '\n', 'Score: ', str(score)))

所以我修改了你的例子，并做了这个：

def pairwiseScore(seqA, seqB):

    score = 0
    bars = [str(' ') for x in seqA] #create a list filled with number of spaces equal to length of seqA string. It could be also seqB, because both are meant to have same length
    length = len(seqA)
    similarity = []

    for x in xrange(length):

        if seqA[x] == seqB[x]: #check if for every index 'x', corresponding character is same in both seqA and seqB strings
            if (x >= 1) and (seqA[x - 1] == seqB[x - 1]): #if 'x' is greater than or equal to 1 and characters under the previous index, were same in both seqA and seqB strings, do..
                score += 3
                similarity.append(x)
            else:
                score += 1
                similarity.append(x)                
        else:
            score -= 1

    for x in similarity:
        bars[x] = '|' #for every index 'x' in 'bars' list, replace space with '|' (pipe/vertical bar) character 

    return ''.join((seqA, '\n', ''.join(bars), '\n', seqB, '\n', 'Score: ', str(score)))

此代码通过s3-q11练习中的所有测试用例。 Sooo，我想我得到了解决方案而且我已经完成了。

谢谢和欢呼：）