create string pattern automatically

Question

Need to create a string based on a given pattern.

If the pattern is 222243243 string need to be created is "2{4,6}[43]+2{1,3}[43]+". Logic to create the above string is, check how many 2's sets in pattern and count them and add more two 2's .here contains two sets of 2's. The first one contains 4 2's and the seconds part contains 1 2's. So the first 2's can be 4 to 6(4+2) 2's and seconds 2's can be 1 to 3(1+2). when there are 3's or 4's, [43]+ need to add.

workings:

import re
data='222243243'
TwosStart=[]#contains twos start positions
TwosEnd=[]#contains twos end positions
TwoLength=[]#number of 2's sets

for match in re.finditer('2+', data):
    s = match.start()#2's start position
    e = match.end()#2's end position
    d=e-s
    print(s,e,d)
    TwosStart.append(s)
    TwosEnd.append(e)
    TwoLength.append(d)

So using the above code I know how many 2's sets are in a given pattern and their starting and ending positions. but I have no idea to automatically create a string using the above information.

Ex:

if pattern '222243243' string should be "2{4,6}[43]+2{1,3}[43]+"

if pattern '222432243' string should be "2{3,5}[43]+2{2,4}[43]+"

if pattern '22432432243' string should be "2{2,4}[43]+2{1,3}[43]+2{2,4}[43]+"

Dani Mesejo · Accepted Answer · 2021-10-04 12:53:16Z

2

One approach is to use itertools.groupby:

from itertools import groupby

s = "222243243"

result = []
for key, group in groupby(s, key=lambda c: c == "2"):
    if key:
        size = (sum(1 for _ in group))
        result.append(f"2{{{size},{size+2}}}")
    else:
        result.append("[43]+")

pattern = "".join(result)
print(pattern)

Output

2{4,6}[43]+2{1,3}[43]+

answered Oct 4, 2021 at 12:53

Dani Mesejo

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Comments

miranda1000 · Accepted Answer · 2021-10-04 13:14:31Z

1

Using your base code:

import re
data='222243243'
cpy=data
offset=0 # each 'cpy' modification offsets the addition

for match in re.finditer('2+', data):
    s = match.start() # 2's start position
    e = match.end() # 2's end position
    d = e-s
    regex = "]+2{" + str(d) + "," + str(d+2) + "}["
    cpy = cpy[:s+offset] + regex + cpy[e+offset:]
    offset+=len(regex)-d

# sometimes the borders can have wrong characters
if cpy[0]==']':
        cpy=cpy[2:] # remove "]+"
else:
        cpy='['+cpy

if cpy[len(cpy)-1]=='[':
        cpy=cpy[:-1]
else:
        cpy+="]+"

print(cpy)

Output

2{4,6}[43]+2{1,3}[43]+

answered Oct 4, 2021 at 13:14

miranda1000

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