如何在列表中提取每个元组的'x'并以python方式连接？

Question

I have a value like this which I extract from NLTK tree.

[[('Happy', 'NNP'), ('Mother', 'NNP')], [('Day', 'NNP')], [('Joey', 'NNP'), ('M.', 'NNP'), ('Bing', 'NNP')], [('kind', 'NN')], [('happy', 'JJ'), ('wife', 'NN')], [('mother', 'NN')], [('friend', 'NN')]]

I would like the end result to be

['Happy Mother','Day','Joey M. Bing','kind','happy wife','mother','friend']

How do I do this in python way?

This is what I have done so far, which is very ugly I know. I'm a python virgin.

Y = []
for x in X:
    s = ""
    for z in x:
        s += z[0] + " "
    Y.append(s)

print Y

Answer 1

You can do it pretty easily with zip and str.join .

result = [' '.join(zip(*row)[0]) for row in data]

zip(*sequences)[i] is a common Python idiom for getting the ith value from each sequence (list, tuple, etc.)

It is similar to [seq[i] for seq in sequences] but it works even if the sequences are not subscriptable (for example iterators). In Cpython, it may be slightly faster due to using a builtin (though you should always profile if it's important). Also, it returns a tuple instead of a list.

For more information, see the documentation .

Answer 2

Y = [' '.join(t[0] for t in l) for l in X]

Answer 3

Use a list comprehension:

>>> X = [[('Happy', 'NNP'), ('Mother', 'NNP')], [('Day', 'NNP')], [('Joey', 'NNP'), ('M.', 'NNP'), ('Bing', 'NNP')], [('kind', 'NN')], [('happy', 'JJ'), ('wife', 'NN')], [('mother', 'NN')], [('friend', 'NN')]]
>>> Y = [' '.join(z[0] for z in x) for x in X]
>>> Y
['Happy Mother', 'Day', 'Joey M. Bing', 'kind', 'happy wife', 'mother', 'friend']