How to replace a template with regex in python?

Question

I have a dataset that looks like this:

Male    Name=Tony;  
Female  Name=Alice.1; 
Female  Name=Alice.2;
Male    Name=Ben; 
Male    Name=Shankar; 
Male    Name=Bala; 
Female  Name=Nina; 
###
Female  Name=Alex.1; 
Female  Name=Alex.2;
Male    Name=James; 
Male    Name=Graham; 
Female  Name=Smith;  
###
Female  Name=Xing;
Female  Name=Flora;
Male    Name=Steve.1;
Male    Name=Steve.2; 
Female  Name=Zac;  
###

I want to change the list so that it looks like this:

Male    Name=Class_1;
Female  Name=Class_1.1;
Female  Name=Class_1.2;
Male    Name=Class_1;
Male    Name=Class_1;
Male    Name=Class_1; 
Female  Name=Class_1;
###
Female  Name=Class_2.1; 
Female  Name=Class_2.2; 
Male    Name=Class_2; 
Male    Name=Class_2; 
Female  Name=Class_2;  
###
Female  Name=Class_3; 
Female  Name=Class_3; 
Male    Name=Class_3.1; 
Male    Name=Class_3.2; 
Female  Name=Class_3;
###

Each name must be changed to the class to which they belong. I noticed that in the dataset, each new class in the list is indicated by the symbol '###. Therefore, I can break the data set into blocks at ### and count the instances of ###. Then use regex to find the names and replace them with the account ###.

My code is as follows:

blocks = [b.strip() for b in open('/file', 'r').readlines()]
pattern = r'Name=(.*?)[;/]'
prefix = 'Class_'
triple_hash_count = 1

for line in blocks:
    match = re.findall(pattern, line)
    print match

for line in blocks:
    if line == '###':
        triple_hash_count += 1
        print line 
    else: 
        print(line.replace(match, prefix + str(triple_hash_count))) 

This does not seem to do the job - no changes are made.

Answer 1

When I ran the code you provided, I received the following trace output:

print(line.replace(match, prefix + str(triple_hash_count))) 
TypeError: Can't convert 'list' object to str implicitly

, type(match) . PDB, . , match , for-loops. :

for line in blocks:
    match = re.findall(pattern, line)
    print(match)

    if line == '###':
        triple_hash_count += 1
        print(line) 
    else: 
        print(line.replace(match, prefix + str(triple_hash_count)))

match, : re.findall - . str.replace(...) , .

print(line.replace(match[0], prefix + str(triple_hash_count))) - , , , ###. , , str.replace() .

:

for line in blocks:
    match = re.findall(pattern, line)
    print(match)

    if line == '###':
        triple_hash_count += 1
        print(line) 
    else:
        if match: 
            print(line.replace(match[0], prefix + str(triple_hash_count)))
        else:
            print(line)

---

:

• 11 . triple_hash_count, hash_count.
• , 1. line.replace(match, prefix + str(triple_hash_count)) , .

Answer 2

( ). .

import re

blocks = [b.strip() for b in open('/file', 'r').readlines()]
pattern = r'Name=([^\.\d;]*)'
prefix = 'Class_'
triple_hash_count = 1

for line in blocks:

    if line == '###':
        triple_hash_count += 1
        print line     
    else:
        match = re.findall(pattern, line)
        print line.replace(match[0], prefix + str(triple_hash_count)) 

Answer 3

, ( , ):

import re
hashrx = re.compile(r'^###$', re.MULTILINE)
namerx = re.compile(r'Name=\w+(\.\d+)?;')

new_string = '###'.join([namerx.sub(r"Name=Class_{}\1".format(idx + 1), part) 
                for idx,part in enumerate(hashrx.split(string))])
print(new_string)

:

• ### ^ $ MULTILINE.
• -, Name, 1 ( , ).
• -, ### enumerate(), .
• Finally, he rejoins the resulting list ###.

As a single line (although not recommended):

new_string = '###'.join(
                [re.sub(r'Name=\w+(\.\d+)?;', r"Name=Class_{}\1".format(idx + 1), part) 
                for idx, part in enumerate(re.split(r'^###$', string, flags=re.MULTILINE))])

Demo

The demo says more than a thousand words.