编写循环级联和 if 语句的 Pythonic 方式？

Question

Maybe this is a super dumb question but I was wondering what is the pythonic way of write this conditions:

custom_field_labels = ['value1', 'value2']

def whatever(label):
    if label not in custom_field_labels:
        custom_field_labels.append(label)
    else:
        invalid_name = True

        while invalid_name:
            label += "_"
            if label not in custom_field_labels:
                custom_field_labels.append(label)
                invalid_name = False

whatever('value1')
whatever('value3')
print(custom_field_labels) # ['value1', 'value2', 'value1_', 'value3']

I've read about that recursion is a bad idea in python. Is that true? If yes, what are the other options?

Answer 1

You just need one while loop.

while label in custom_field_labels:
    label += "_"
custom_field_labels.append(label)

If label isn't in the list, the loop will never be entered so you'll append the original label; this is essentially the same as the first if .

I also recommend against the pattern

while boolean_variable:
    # do stuff
    if <some condition>:
        boolean_variable = False

Either test the condition itself in the while condition, or use:

while True:
    # do stuff
    if <some condition>:
        break;

Answer 2

I would like to append "_" into a string while it exists inside custom_field_labels.

Then do exactly that?

def whatever(label):
    while label in custom_field_labels:
        label += "_"
    custom_field_labels.append(label)

No idea why you're asking about recursion.

Answer 3

Rather than store the almost-the-same labels directly, you can use a Counter object, and reconstruct the labels as needed.

>>> from collections import Counter
>>> c = Counter(["value1", "value2"])
>>> c['value1'] += 1
>>> c['value3'] += 1
>>> c
Counter({'value1': 2, 'value2': 1, 'value3': 1})
>>> [x+"_"*i for x, n in c.items() for i in range(n)]
['value1', 'value1_', 'value2', 'value3']

(It would be nice if there were an add method so that c.add(x) was equivalent to c[x] += 1 . Oh, well.)