1

我有一个包含文件名的字符串列表，例如，

file_names = ['filei.txt','filej.txt','filek.txt','file2i.txt','file2j.txt','file2k.txt','file3i.txt','file3j.txt','file3k.txt']

然后我使用以下方法删除.txt扩展名：

extension = os.path.commonprefix([n[::-1] for n in file_names])[::-1]

file_names_strip = [n[:-len(extension)] for n in file_names]

然后返回列表中每个字符串的最后一个字符file_names_strip：

h = [n[-1:] for n in file_names_strip]

这使h = ['i', 'j', 'k', 'i', 'j', 'k', 'i', 'j', 'k']

我如何测试字符串的模式h？因此，如果i, j,k顺序发生，它将返回 True ，否则返回 False 。我需要知道这一点，因为并非所有文件名的格式都像file_names.

所以：

test_ijk_pattern(h) = True

no_pattern = ['1','2','3','1','2','3','1','2','3']

test_ijk_pattern(no_pattern) = False
4

2 回答 2

1

以下是我将如何攻击它：

def patternFinder(h):    #Takes a list and returns a list of the pattern if found, otherwise returns an empty list

    if h[0] in h[1:]:
        rptIndex = h[1:].index(h[0]) + 1 #Gets the index of the second instance of the first element in the list
    else:
        print "This list has no pattern"
        return []

    if len(h) % rptIndex != 0:
        h = h[:-(len(h) % rptIndex)]   #Takes off extra entries at the end which would break the next step

    subLists = [h[i:i+rptIndex] for i in range(0,len(h),rptIndex)]   #Divide h into sublists which should all have the same pattern

    hasPattern = True   #Assume the list has a pattern
    numReps = 0  #Number of times the pattern appears

    for subList in subLists:
        if subList != subLists[0]: 
            hasPattern = False
        else:
            numReps += 1

    if hasPattern and numReps != 1:
        pattern = subList[0]
        return pattern
    else:
        print "This list has no pattern"
        return []

假设这使得：

  • 模式显示在前几个字符中
  • 最后不完整的模式并不重要（[1,2,3,1,2,3,1,2]将提出 2 个实例[1,2,3]）
  • h至少有 2 个条目
  • 模式之间没有多余的字符

如果您对这些假设没意见，那么这对您有用，希望对您有所帮助！

于 2013-11-08T14:16:17.133 回答
0

你可以使用正则表达式。

import re
def test_pattern(pattern, mylist):
  print pattern
  print mylist
  print "".join(mylist)
  if re.match(r'(%s)+$' % pattern, "".join(mylist)) != None: # if the pattern matchtes at least one time, nothing else is allowed
    return True
  return False       

print test_pattern("ijk", ["i", "j", "k", "i", "j", "k"])

您可以这样做而无需删除最后一个字母和文件结尾。我更新了正则表达式以使其正常工作。一个问题是我使用了变量名，它寻找模式“mypattern”。使用 %s 将其替换为真实模式。我希望这个解决方案适合你。

myfiles = ["ai.txt", "aj.txt", "ak.txt", "bi.txt", "bj.txt", "bk.txt"]
mypattern = ["i", "j", "k"]

import re
# pattern as a list e.g. ["i", "j", "k"]
def test_pattern(pattern, filenames):
    mypattern = "["+"\.[a-zA-Z0-9]*".join(pattern) + "\.[a-zA-Z0-9]*]*"
    # this pattern matches any character, an "i", followed by a dot, any characters, followed by j., any characters, followd by k. (change it a bit if your file names contain numbers and/or uppercase)
    print mypattern
    print "".join(filenames)
    if re.search(r'%s' % mypattern, "".join(filenames)) != None: # if the pattern matchtes at least one time, nothing else is allowed
        return True
    return False



print test_pattern(mypattern, myfiles)

输出：

[i\.[a-zA-Z0-9]*j\.[a-zA-Z0-9]*k\.[a-zA-Z0-9]*]*
ai.txtaj.txtak.txtbi.txtbj.txtbk.txt
True
于 2013-11-08T14:02:20.157 回答