I have the following array in Python in the following format:
Array[('John', '123'), ('Alex','456'),('Nate', '789')]
Is there a way I can assign the array variables by field as below?
Name = ['john', 'Alex', 'Nate']
ID = ['123', '456', '789']
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1Possible duplicate of How to split a list of 2-tuples into two lists?joel– joel2019-03-11 13:54:48 +00:00Commented Mar 11, 2019 at 13:54
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2 Answers
In the spirit of "explicit is better than implicit":
data = [('John', '123'), ('Alex', '456'), ('Nate', '789')]
names = [x[0] for x in data]
ids = [x[1] for x in data]
print(names) # prints ['John', 'Alex', 'Nate']
print(ids) # prints ['123', '456', '789']
Or even, to be even more explicit:
data = [('John', '123'), ('Alex', '456'), ('Nate', '789')]
NAME_INDEX = 0
ID_INDEX = 1
names = [x[NAME_INDEX] for x in data]
ids = [x[ID_INDEX] for x in data]
Comments
this is a compact way to do this using zip:
lst = [('John', '123'), ('Alex','456'),('Nate', '789')]
name, userid = list(zip(*lst))
print(name) # ('John', 'Alex', 'Nate')
print(userid) # ('123', '456', '789')
note that the results are stored in (immutable) tuples; if you need (mutatble) lists you need to cast.
answered Mar 11, 2019 at 13:51
hiro protagonist
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4 Comments
Daniel Pryden
What do you mean by "if you need (mutable)
lists you need to cast"? There are no casts in Python.hiro protagonist
wouldn't you call
int('123') a cast? (here i mean: list(name).)Daniel Pryden
No, that's not a cast. You're just constructing a new instance, or, more pedantically, just calling a callable that happens to construct a new instance. When you call
int('123') you are simply calling int.__init__(..., '123'), and the int class can do whatever it wants with that. It can have the side effect of producing a new object with a new type that is otherwise equivalent, but that's not its primary purpose.hiro protagonist
hmmm. i agree... i'd just have called that a cast (maybe because there is no such thing in python and this is the closest to it). but maybe i really shouldn't.