Clever Geek Handbook

Find an item in a list by type

I have a list that can contain several elements of different types. I need to check if this list has one or more elements of a certain type and get its index.

l = [1, 2, 3, myobj, 4, 5]

I can achieve this goal by simply traversing my list and checking the type of each element:

for i, v in enumerate(l):
  if type(v) == Mytype:
    return i

Is there a more pythonic way to achieve the same result?

+3


source to share


2 answers


You can use a generatornext

expression as well :

return next(i for i, v in enumerate(l) if isinstance(v, Mytype)):

The advantage of this solution is that it is as lazy as your current one: it will only check as many elements as needed.

Also, I used isinstance(v, Mytype)

instead type(v) == Mytype

because that is the preferred method of type checking in Python. See PEP 0008 .



Finally, it should be noted that this solution will raise an exception StopIteration

if the required element is not found. You can catch this with try / except, or you can provide a default return value:

return next((i for i, v in enumerate(l) if isinstance(v, Mytype)), None):

In this case, it None

will be returned if nothing was found.

+6


source


You can use a list comprehension.

Get all elements of a string like:

b = [x for x in a if isinstance(x,str)]


Get all indices of elements of a string of type:

b = [x[0] for x in enumerate(a) if isinstance(x[1],str)]
0


source






More articles:


All Articles