我有三个列表和一个字符串变量:
var = "http:/domain.com"
directories = ['dir_A', 'dir_B', 'dir_C']
files = ['file_A', 'file_B', 'file_C']
extensions = ['ext_A', 'ext_B']
我想打印一个完全像这样的图案:
http:/domain.com/
http:/domain.com/dir_A/
http:/domain.com/dir_A/file_A.ext_A
http:/domain.com/dir_A/file_A.ext_B
http:/domain.com/dir_A/file_B.ext_A
http:/domain.com/dir_A/file_B.ext_B
http:/domain.com/dir_A/file_C.ext_A
http:/domain.com/dir_A/file_C.ext_B
http:/domain.com/dir_A/dir_B/
http:/domain.com/dir_A/dir_B/file_A.ext_A
http:/domain.com/dir_A/dir_B/file_A.ext_B
http:/domain.com/dir_A/dir_B/file_B.ext_A
http:/domain.com/dir_A/dir_B/file_B.ext_B
http:/domain.com/dir_A/dir_B/file_C.ext_A
http:/domain.com/dir_A/dir_B/file_C.ext_B
http:/domain.com/dir_A/dir_B/dir_C/
http:/domain.com/dir_A/dir_B/dir_C/file_A.ext_A
http:/domain.com/dir_A/dir_B/dir_C/file_A.ext_B
http:/domain.com/dir_A/dir_B/dir_C/file_B.ext_A
http:/domain.com/dir_A/dir_B/dir_C/file_B.ext_B
http:/domain.com/dir_A/dir_B/dir_C/file_C.ext_A
http:/domain.com/dir_A/dir_B/dir_C/file_C.ext_B
http:/domain.com/dir_A/dir_C/
http:/domain.com/dir_A/dir_C/file_A.ext_A
http:/domain.com/dir_A/dir_C/file_A.ext_B
http:/domain.com/dir_A/dir_C/file_B.ext_A
http:/domain.com/dir_A/dir_C/file_B.ext_B
http:/domain.com/dir_A/dir_C/file_C.ext_A
http:/domain.com/dir_A/dir_C/file_C.ext_B
http:/domain.com/dir_A/dir_C/dir_B/
http:/domain.com/dir_A/dir_C/dir_B/file_A.ext_A
http:/domain.com/dir_A/dir_C/dir_B/file_A.ext_B
http:/domain.com/dir_A/dir_C/dir_B/file_B.ext_A
http:/domain.com/dir_A/dir_C/dir_B/file_B.ext_B
http:/domain.com/dir_A/dir_C/dir_B/file_C.ext_A
http:/domain.com/dir_A/dir_C/dir_B/file_C.ext_B
这只适用于http:/domain.com/dir_A/...
案件{
我自己也试过,但没有成功达到要求的模式
任何帮助都将不胜感激
如果您想自己实现生成目录的各种排列,可以使用递归函数。一旦有了目录,文件和扩展名就很简单了
输出:
我猜你想要正斜杠,但我按照你的要求编码了。您忘记在提示符中引用一些字符串
您需要^{} 模块中的
combinations
、permutations
和product
来获取所有这些变体:输出(总共96个;请注意,我将所有
\
更改为/
并更改了域):相关问题 更多 >
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