Initialize variable depending on another variables type

Question

In Python 2.7 I want to intialize a variables type depending on another variable.

For example I want to do something like:

var_1 = type(var_2)

Is there a simple/fast way to do that?

How many different types are we talking about? isinstance() is usually a better solution than type() (usually, not always). — cdarke
– cdarke, Commented Oct 13, 2016 at 7:12
Working on a function that takes an iterable object as argument and defines a variable of the same type — B. Kostas
– B. Kostas, Commented Oct 13, 2016 at 7:14

Jean-François Fabre · Accepted Answer · 2016-10-13 07:50:07Z

4

Just create another instance

var_1 = type(var_2)()

Note that if you're not sure whether the object has a non-default constructor, you cannot rely on the above, but you can use copy or deepcopy (you get a "non-empty" object.

import copy
var_1 = copy.copy(var_2)  # or copy.deepcopy

You could use both combined with the latter as a fallback mechanism

Note: deepcopy will ensure that your second object is completely independent from the first (If there are lists of lists, for instance)

edited Oct 13, 2016 at 7:50

answered Oct 13, 2016 at 7:09

Jean-François Fabre♦

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1 Comment

B. Kostas Over a year ago

Thank you! One more question regarding the copy library. By reading the docs it seems like copy.deepcopy() is more suitable for this purpose, or am I wrong?

Michel Touw · Accepted Answer · 2016-10-13 07:21:08Z

2

a = 1              # a is an int
a_type = type(a)   # a_type now contains the int-type
b = '1'            # '1' is a string
c = a_type(b)      # c is now an int with the value 1

So you can get the type of a variable using type(). You can then store this type in a variable and you can then use that variable just like you would use int(b), str(b), float(b) etc.

edited Oct 13, 2016 at 7:21

answered Oct 13, 2016 at 7:09

Michel Touw

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