python的嵌套生成器或迭代器

Question

I have the next list:

texts = [['abcdD', 'asdfaD'], ['qerqD', 'asdfafdasD']]

I want to delete all character D from the right part of all strings.

For one list I can do it easily:

res1 = [el.strip('D') for el in texts[0]] # ['abcd', 'asdfa']

Now I'm trying the same for each text:

res2 = [el.strip('D') for text in texts for el in text]

But it returns ONE list (combine my two!):

['abcd', 'asdfa', 'qerq', 'asdfafdas']

When I need next:

[['abcd', 'asdfa'], ['qerq', 'asdfafdas']]

How to do it correct?

Answer 1

You can use this recursive approach for arbitrary deeply nested structures:

def f(s):
  return [q.strip('D') if isinstance(q, basestring) else f(q)
          for q in s]

(Use str instead of basestring in Python3.)

Example:

f([['abcdD', 'asdfaD'], ['qerqD', 'asdfafdasD']])

returns:

[['abcd', 'asdfa'], ['qerq', 'asdfafdas']]

Answer 2

You can try this:

texts = [['abcdD', 'asdfaD'], ['qerqD', 'asdfafdasD']]
new_texts = [[b[:-1] if b.endswith('D') else b for b in i] for i in texts]

Output:

[['abcd', 'asdfa'], ['qerq', 'asdfafdas']]

Answer 3

how about a ' map happy' approach - just for fun, not any advantages I can see

list(map(list, map(lambda txt: map(lambda x: x.rstrip('D'), txt), texts)))

Out[240]: [['abcd', 'asdfa'], ['qerq', 'asdfafdas']]

Answer 4

As mentioned in my comment, you'll need to nest one of the loops.

x = [['abcdD', 'asdfaD'], ['qerqD', 'asdfafdasD']]
y = [[j.rstrip('D') for j in i] for i in x]

y
[['abcd', 'asdfa'], ['qerq', 'asdfafdas']]

Which is easy to understand if you convert the above to a nested loop -

y = []
for i in x:
    y.append([])
    for j in i:
        y[-1].append(j.rstrip('D'))

y
[['abcd', 'asdfa'], ['qerq', 'asdfafdas']]

What you decide to use is really a matter of style. The list comprehension is concise, but suffers from readability issues. The loop is generally very fast and readable.