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?
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]