I have a list like this:
[['ok.txt', 'hello'], [10, 20], ['first_one', 'second_one'], ['done', 'pending']]
I want to convert this list into dictionary like this:
{'ok.txt' : ['10', 'first_one', 'done'], 'hello' : ['20', 'second_one', 'pending']}
How to do something like this?
Try this:
dict(zip(xs[0], zip(*xs[1:])))
For lists as values of the dict:
dict(zip(xs[0], map(list, zip(*xs[1:]))))
>>> lis = [['ok.txt', 'hello'], [10, 20], ['first_one', 'second_one'], ['done', 'pending']]
>>> keys, values = lis[0],lis[1:]
>>> {key:[val[i] for val in values]
for i,key in enumerate(keys) for val in values}
{'ok.txt': [10, 'first_one', 'done'], 'hello': [20, 'second_one', 'pending']}
You can make use of the built-in zip function to perform this easily like this:
list_of_list = [['ok.txt', 'hello'], [10, 20], ['first_one', 'second_one'], ['done', 'pending']]
dict_from_list = dict(zip(list_of_list[0], zip(*list_of_list[1:])))
In this, the interior zip(*list_of_list[1:]) will convert the list of lists from list_of_list(except the first element) to a list of tuples. The tuple is order preserved and is again zipped with the supposed keys to form a list of tuples, which is converted to a proper dictionary through the dict function.
Please note that this will have tuple as data type for the values in the dictionary. As per your example, the one-liner will give:
{'ok.txt': (10, 'first_one', 'done'), 'hello': (20, 'second_one', 'pending')}
In order to have a list, you have to map the interior zip with the list function. (i.e) change
zip(*list_of_list[1:]) ==> map(list, zip(*list_of_list[1:]))
For info about zip function, click here
Edit: I just noticed that the answer is same as that of the one given by Simon. Simon gave it lot quicker when I was trying the code in terminal and I didn't noticed his answer when I was posting.
