Python在函数的递归调用中返回值

问题描述 投票:3回答:1

我有一些数据可以创建一个数字数组。有时可能会重复这些数字。我想为每个数字创建一个唯一的文件名。所以我想出了一个黑客,在每个数字的末尾包含2位数字,并在重复时增加它

该程序似乎在某些情况下工作正常,并且在某些时候无法正常工作。

numbers = [10, 20, 30, 30, 40, 50, 50, 50, 50, 60]
filenames = []


def check_name(checkFileName):
    if checkFileName in filenames:
        checkFileName += 1
        check_name(checkFileName)
        return checkFileName

    else:
        print("Def-Filename :", checkFileName)
        return checkFileName


for number in numbers:
    stringNumber = str(number)
    tempFileName = stringNumber + "00"
    tempFileInt = int(tempFileName)

    permFileName = check_name(tempFileInt)
    filenames.append(permFileName)
    print("Permanent File Name :", permFileName)
    print(filenames)

输出是

Def-Filename : 1000
Permanent File Name : 1000
[1000]
Def-Filename : 2000
Permanent File Name : 2000
[1000, 2000]
Def-Filename : 3000
Permanent File Name : 3000
[1000, 2000, 3000]
Def-Filename : 3001
Permanent File Name : 3001
[1000, 2000, 3000, 3001]
Def-Filename : 4000
Permanent File Name : 4000
[1000, 2000, 3000, 3001, 4000]
Def-Filename : 5000
Permanent File Name : 5000
[1000, 2000, 3000, 3001, 4000, 5000]
Def-Filename : 5001
Permanent File Name : 5001
[1000, 2000, 3000, 3001, 4000, 5000, 5001]
Def-Filename : 5002
Permanent File Name : 5001
[1000, 2000, 3000, 3001, 4000, 5000, 5001, 5001]
Def-Filename : 5002
Permanent File Name : 5001
[1000, 2000, 3000, 3001, 4000, 5000, 5001, 5001, 5001]
Def-Filename : 6000
Permanent File Name : 6000
[1000, 2000, 3000, 3001, 4000, 5000, 5001, 5001, 5001, 6000]   

我哪里错了?

python recursion return-value
1个回答
1
投票

问题是您返回的是顶级函数计算的文件名,而不是递归函数调用返回的文件名

更改

if checkFileName in filenames:
    checkFileName += 1
    check_name(checkFileName)
    return checkFileName

对此

if checkFileName in filenames:
    checkFileName += 1
    return check_name(checkFileName)

话虽这么说,一个更容易的解决方案是使用collections.Counter

>>> from collections import Counter
>>> numbers = [10, 20, 30, 30, 40, 50, 50, 50, 50, 60]
>>> 
>>> [n*100 + i for n,cnt in Counter(numbers).items() for i in range(cnt)]
[1000, 2000, 3000, 3001, 4000, 5000, 5001, 5002, 5003, 6000]
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