CODEWITHSUNDEEP
pythonpython-3.xlambda
peer
peer

Reputation: 4719

Nested maps in Python 3

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?

Upvotes: 8

Views: 18162

Answers (2)

Khajan Singh
Khajan Singh

Reputation: 1

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 ']]

Upvotes: -1

vaultah
vaultah

Reputation: 46533

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]

Upvotes: 13

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