Search for a pattern in a list of strings - Python

Question

I have a list of strings containing filenames such as,

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

I then remove the .txt extension using:

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

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

And then return the last character of each string in the list file_names_strip :

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

Which gives h = ['i', 'j', 'k', 'i', 'j', 'k', 'i', 'j', 'k']

How can i test for a pattern of strings in h ? So if i , j , k occur sequentially it would return True and False if not. I need to know this because not all file names are formatted like they are in file_names .

So:

test_ijk_pattern(h) = True

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

test_ijk_pattern(no_pattern) = False

Answer 1

Here's how I would attack this:

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 []

Assumptions that this makes:

• The pattern is shown in the first few characters
• Incomplete patterns at the end aren't important ( [1,2,3,1,2,3,1,2] will come up with having 2 instances of [1,2,3] )
• h has at least 2 entries
• There are no extra characters between patterns

If you're fine with these assumptions, then this will work for you, hope this helps!

Answer 2

You could use regex.

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"])

You could do it this way without stripping the last letters and the file endings. I updated the regular expression so that it works. One problem was that I used the variable name and it looked for the pattern "mypattern". Using %s replaces it with the real pattern. I hope this solution suits you.

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)

Output:

[i\.[a-zA-Z0-9]*j\.[a-zA-Z0-9]*k\.[a-zA-Z0-9]*]*
ai.txtaj.txtak.txtbi.txtbj.txtbk.txt
True