从多维列表python中删除不必要的层

Question

My plan is to turn the input of a multidimensional list with an unknown amount of layers ie [["a"], [["b"], ["c"]]] into ["a", ["b", "c"]] but currently my output is the same as the original with the below function:

def extract(tree):
    for x in range(len(tree)):
        if type(tree[x]) == list:
            extract(tree[x])
        else:
            tree = tree[x]

Basically, I want to remove any unnecessary layers in the array that have only one element,

Any ideas on what I'm doing wrong?

Thanks for any help

Answer 1

Maybe something like:

def tree(element):
  if isinstance(element, str):
    return element
  elif len(element) == 1:
    return element[0]

  unfolded = []
  for each in element:
    unfolded.append(tree(each))
  return unfolded

a = [["a"], [["b"], ["c"]]]
print(tree(a)) # => ['a', ['b', 'c']]

Answer 2

Just add a few returns and your code works just fine.

def extract(tree):
    for x in range(len(tree)):
        if type(tree[x]) == list:
            tree[x] = extract(tree[x])
        else:
            return tree[x]

    return tree

# test cases
test = [["a"], [["b"], ["c"]]]
print(extract(test))

test = [["a"], [["b"], ["c"], [["b"], ["c"]]]]
print(extract(test))

Answer 3

1. It's generally good to use isinstance() instead of type() for type checking in code.

Not exactly sure what you're trying to accomplish with this code. Axe319's answer will remove all the inner arrays/lists and return both [["a"], [["b"], ["c"]]] and ["a", ["b", "c"]] as ['a','b','c'] . If you wish to make a list of one, then two, then three etc. (ie. [['a'], ['b', 'c'], ['d', 'e', 'f']... ), you'll have to redesign your code.