Python3：如何在for循环中增量匹配Scaffold(y)并按y存储对应行

Hey there! Let’s fix this so you don’t have to manually specify each Scaffold(y) number anymore. Instead of creating separate variables for each y value (which gets messy fast), we’ll use a dictionary to group lines by their scaffold number—it’s clean, scalable, and easy to maintain.

Step-by-Step Solution (Python Example) #

Here’s a script that will automatically detect all scaffold numbers, group their corresponding lines, and print each group:

import re

# Initialize a dictionary to hold lines grouped by scaffold number
scaffold_groups = {}

# Open and read your text file
with open("xfile", "r") as file:
    for line in file:
        line = line.strip()  # Remove extra whitespace/newlines
        if not line:
            continue  # Skip empty lines
        
        # Use regex to extract the number y from lines starting with Scaffold(y)_1
        match = re.match(r"^Scaffold(\d+)_1", line)
        if match:
            scaffold_num = match.group(1)  # Get the numeric part (y)
            
            # Add the line to the corresponding group in the dictionary
            if scaffold_num not in scaffold_groups:
                scaffold_groups[scaffold_num] = []
            scaffold_groups[scaffold_num].append(line)

# Print each group of lines
for num, lines in scaffold_groups.items():
    print(f"--- Scaffold{num} Lines ---")
    for line in lines:
        print(line)
    print("\n")  # Add a blank line between groups

How This Works: #

• Regex Matching: The pattern ^Scaffold(\d+)_1 targets lines starting with Scaffold, captures the numeric part y, and checks for the _1 suffix. The (\d+) is a capturing group that grabs the number.
• Dictionary Grouping: We use scaffold_groups to map each scaffold number to a list of its lines. This avoids the hassle of creating dynamic variables (which is generally not recommended in Python).
• Clean Output: Finally, we loop through the dictionary to print each group with a clear header, so you can easily see all lines for each scaffold number.

Optional: Sort Groups by Scaffold Number #

If you want the output to be ordered by the numeric value of y (instead of the order they appeared in the file), modify the final loop to sort the keys:

# Sort the scaffold numbers numerically before printing
for num in sorted(scaffold_groups.keys(), key=int):
    lines = scaffold_groups[num]
    print(f"--- Scaffold{num} Lines ---")
    for line in lines:
        print(line)
    print("\n")

This will ensure you see Scaffold1, Scaffold2, Scaffold3, etc., in order, regardless of how they’re arranged in the input file.

内容的提问来源于stack exchange，提问作者Cormac Tomás Ó Grífín