如何按元素 os.path.join 不均匀列表

Question

说我有

folders = [['folder1/subfold'],['folder2/subfold/subsubfold']]
clas = [['class1'], ['class2', 'class3','class4']]

如何将类的路径加入到它们各自的文件夹中？期望：

['folder1/subfold/class1',
'folder2/subfold/subsubfold/class2',
'folder2/subfold/subsubfold/class3',
'folder2/subfold/subsubfold/class4']

以下内容似乎仅在列表具有相同维度时才有效：[os.path.join(*i, j) for i,j in zip(folders, clas)]

Answer 1

你想要的是嵌套for循环

[os.path.join(*i, j) for i, names in zip(folders,clas) for j in names]

how to os.path.join uneven lists element-wise