I want to transform my list list = [" a , 1 ", " b , 2 "] into a nested list [["a","1"],["b","2"]].
The following works:
f1_a = map(lambda x :[t.strip() for t in x.split(',',1)],list)
but this does not work with Python 3 (it does work with Python 2.7!):
f1_b = map(lambda x :map(lambda t:t.strip(),x.split(',',1)),list)
Why is that?
Is there a more concise way then f1_4 to achieve what I want?
Python 3's map returns a map object, which you need to convert to list explicitly:
f1_b = list(map(lambda x: list(map(lambda t: t.strip(), x.split(',', 1))), lst))
Though in most cases you should prefer list comprehensions to map calls:
f1_a = [[t.strip() for t in x.split(',', 1)] for x in lst]
[t.strip() for t in x.split(',', 1)] would become x.replace(' ', '').split(','). On the other hand, now I see that the suggestion is restricted to the data in the question and will fail on more complex examples. I removed it from my answer.Python3 map function returns map object and that object need to typecast in list.
num = [" a , 1 ", " b , 2 "]
list1 = list(map(lambda x:x.split(','), num))
print(list1)
Output:
[[' a ', ' 1 '], [' b ', ' 2 ']]
