项目禁用while语句，Python标准库是否有等效循环函数？

Great question! Since you can't use the while keyword directly, there are a few clean ways to replicate your loop_while logic using Python's standard libraries or concise one-liners. Let's walk through them:

1. Standard Library Solution with itertools #

You can combine itertools.accumulate and itertools.takewhile to replicate your loop behavior without using the while keyword. This is the most readable standard-library-based approach:

from itertools import accumulate, takewhile

def loop_while(pred, g, f, *args, **kwargs):
    initial = f(*args, **kwargs)
    # Generate the sequence of values where the predicate holds true
    # accumulate creates initial, g(initial), g(g(initial)), ...
    valid_values = takewhile(pred, accumulate([initial], lambda current, _: g(current)))
    
    # Extract the last valid value (or stay with initial if no values passed the predicate)
    last_valid = None
    for last_valid in valid_values:
        pass
    
    # If there were valid values, apply g one final time to reach the terminating value
    # Otherwise, return initial (since pred(initial) was already False)
    return g(last_valid) if last_valid is not None else initial

Testing your example:

loop_while(lambda r: r != 0, lambda v: v - 1, lambda: 5)  # Returns 0

2. One-Liner (No Imports) #

If you want a concise one-liner without relying on external libraries, you can use a generator expression with next() and a mutable object to track the current value. It's a bit clever, but works:

# Your example parameters
pred = lambda r: r != 0
g = lambda v: v - 1
f = lambda: 5

result = next(
    current[0]
    for current in [[f()]]
    while pred(current.__setitem__(0, g(current[0])) or current[0])
)
# result equals 0

How this works:

• We use a list current to hold the mutable value (since integers are immutable in Python)
• current.__setitem__(0, g(current[0])) updates the value and returns None
• or current[0] gives us the new value to check against pred
• The generator runs until pred returns False, then next() pulls the final terminating value

3. Recursive Approach (Not Recommended for Large Iterations) #

You could also use recursion, but note this will hit Python's recursion depth limit for large sequences (e.g., if your initial value was 10,000):

def loop_while(pred, g, f, *args, **kwargs):
    res = f(*args, **kwargs)
    return loop_while(pred, g, lambda: g(res)) if pred(res) else res

Testing this with your example also returns 0, but avoid it for large loops to prevent a RecursionError.

内容的提问来源于stack exchange，提问作者uspectaculum