REGEXVAULTv2.0
Finance/Bank Identifiers
Verified Safe

US ACH/ABA Routing Number Regex for Python

/^(?:0[1-9]|1[0-2]|2[1-9]|3[0-2]|6[1-9]|7[0-2]|80)[0-9]{2}(?!0{5})[0-9]{5}$/

What this pattern does

This page provides a comprehensive, battle-tested regular expression for matching us ach/aba routing number, ported and verified for Python. Financial data validation has zero tolerance for false negatives — a missed invalid entry can corrupt downstream calculations. The snippet below is ready to drop into your Python project — whether you're validating in a Django view, a FastAPI endpoint, or a standalone data processing script.

Python Implementation

Python
# US ACH/ABA Routing Number
# ReDoS-safe | RegexVault — Finance > Bank Identifiers

import re

us_achaba_routing_number_pattern = re.compile(r'^(?:0[1-9]|1[0-2]|2[1-9]|3[0-2]|6[1-9]|7[0-2]|80)[0-9]{2}(?!0{5})[0-9]{5}$')

def validate_us_achaba_routing_number(value: str) -> bool:
    return bool(us_achaba_routing_number_pattern.fullmatch(value))

# Example
print(validate_us_achaba_routing_number("021000021"))  # True

Test Cases

Matches (Valid)
Rejects (Invalid)
02100002102100002
0114015330210000211
063100277021000000
121000248000000000
322271627abcdefghi
021-000-021

When to use this pattern

This pattern is drawn from the Finance > Bank Identifiers category and carries a ReDoS-safe certification. That matters for Python developers because particularly important in Python web servers where CPU-bound regex operations can stall concurrent request handling. RegexVault audits patterns against known backtracking attack vectors, ensuring you have the necessary context before using this regex in a high-stakes production environment.

Common Pitfalls

Always validate the ABA checksum — structurally valid 9-digit numbers that fail the checksum are not valid routing numbers. The Luhn-like checksum prevents most typos.

Technical Notes

The first two digits identify the Federal Reserve District (01-12), Federal Home Loan Bank (21-32), or special assignment (61-72, 80). Checksum: (3*(d1+d4+d7) + 7*(d2+d5+d8) + (d3+d6+d9)) mod 10 = 0.

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