将列表列表转换为元组列表

Question

I have got a list of lists:

[['AB', '132'], ['C D'], ['EFG'], ['HJ K', '2  1']]  

and i am trying to convert it into a list of tuples:

[('AB', '132'), ('C D', ''), ('EFG', ''), ('HJ K', '2  1')]

Answer 1

Exceedingly simple, just use a list comprehension to loop over the source list and convert each element to a tuple, by directly passing the items to the tuple builtin :

>>> example = [['AB', '132'], ['C D'], ['EFG'], ['HJ K', '2 1']]
>>> [tuple(i) for i in example]
[('AB', '132'), ('C D',), ('EFG',), ('HJ K', '2 1')]

Alternatively, if you like functional programming, use map instead:

>>> map(tuple, example)
[('AB', '132'), ('C D',), ('EFG',), ('HJ K', '2 1')]

Answer 2

list comprehension will convert each element of your list into a tuple:

data = [['AB', '132'], ['C D'], ['EFG'], ['HJ K', '2 1']]
[tuple(elem) for elem in data]

gives:

[('AB', '132'), ('C D',), ('EFG',), ('HJ K', '2 1')]

Answer 3

One possibility is to use map(tuple, l) :

In [1]: l=[['AB', '132'], ['C D'], ['EFG'], ['HJ K', '2  1']]  

In [2]: map(tuple, l)
Out[2]: [('AB', '132'), ('C D',), ('EFG',), ('HJ K', '2  1')]

The downside is that in Python 3, this would return an iterable instead of a list. If you have to have a list, this will need to be spelled out as list(map(tuple, l)) (this works in both Python 2 and 3).

Another approach that works in both Python 2 and 3 is to use a list comprehension:

[tuple(x) for x in l]