# Check if all elements of a list are of the same type

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### Question :

Check if all elements of a list are of the same type

How can I check if the elements of a list are of the same type, without checking individually every element if possible?

For example, I would like to have a function to check that every element of this list is an integer (which is clearly false):

``````x = [1, 2.5, 'a']

def checkIntegers(x):
# return True if all elements are integers, False otherwise
``````

Try using `all` in conjunction with `isinstance`:

``````all(isinstance(x, int) for x in lst)
``````

You can even check for multiple types with `isinstance` if that is desireable:

``````all(isinstance(x, (int, long)) for x in lst)
``````

Not that this will pick up inherited classes as well. e.g.:

``````class MyInt(int):
pass

print(isinstance(MyInt('3'),int)) #True
``````

If you need to restrict yourself to just integers, you could use `all(type(x) is int for x in lst)`. But that is a VERY rare scenario.

A fun function you could write with this is one which would return the type of the first element in a sequence if all the other elements are the same type:

``````def homogeneous_type(seq):
iseq = iter(seq)
first_type = type(next(iseq))
return first_type if all( (type(x) is first_type) for x in iseq ) else False
``````

This will work for any arbitrary iterable, but it will consume “iterators” in the process.

Another fun function in the same vein which returns the set of common bases:

``````import inspect
def common_bases(seq):
iseq = iter(seq)
bases = set(inspect.getmro(type(next(iseq))))
for item in iseq:
bases = bases.intersection(inspect.getmro(type(item)))
if not bases:
break
return bases
``````

Using `any()`, no need to traverse whole list. Just break as soon as object which is not `int` or `long` is found:

``````>>> not any(not isinstance(y,(int,long)) for y in [1,2,3])
True
>>> not any(not isinstance(y,(int,long)) for y in [1,'a',2,3])
False
``````

``````>>> def checkInt(l):
return all(isinstance(i, (int, long)) for i in l)

>>> checkInt([1,2,3])
True
>>> checkInt(['a',1,2,3])
False
>>> checkInt([1,2,3,238762384762364892364])
True
``````

The simplest way to check if a list is composed of omogeneous elements can be with the groupby function of the itertools module:

``````from itertools import groupby
len(list(groupby(yourlist,lambda i:type(i)))) == 1
``````

If th len is different from one it means that it found different kind of types in the list.
This has the problem of running trough the entire sequence.
If you want a lazy version you can write a function for that:

``````def same(iterable):
iterable = iter(iterable)
try:
first = type(next(iterable))
return all(isinstance(i,first) for i in iterable)
except StopIteration:
return True
``````

This function store the type of the first element and stop as soon as it find a different type in one of the elements in the list.

Both of this methods are strongly sensitive to the type, so it will see as different int and float, but this should be as close as you can get to your request

EDIT:

replaced the for cycle with a call to all as suggested by mgilson

in case of void iterator it returns True to be consistent with the behavior of the bulitin all function

Combining some of the answers already given, using a combination of map(), type() and set() provides a imho rather readable answer. Assuming the limitation of not checking for type polymorphisms is ok. Also not the most computationally efficient answer, but it allows to easily check whether all elements are of the same type.

``````# To check whether all elements in a list are integers
set(map(type, [1,2,3])) == {int}
# To check whether all elements are of the same type
len(set(map(type, [1,2,3]))) == 1
``````

You can also use `type()` if you want to exclude subclasses. See the difference between `isinstance()` and `type()`:

``````>>> not any(not type(y) is int for y in [1, 2, 3])
True
>>> not any(not type(y) == int for y in [1, 'a', 2.3])
False
``````

Although you may not want to, because this will be more fragile. If y changes its type to a subclass of int, this code will break, whereas `isinstance()` will still work.

It’s OK to use `is` because there is only one `<type 'int'>` in memory, so they should return the same identity if they are the same type.

I like the function by EnricoGiampieri (above) but there is a simpler version using `all_equal` from the Itertools recipes section of the `itertools` docs:

``````from itertools import groupby

def all_equal(iterable):
"Returns True if all the elements are equal to each other"
g = groupby(iterable)
return next(g, True) and not next(g, False)
``````

All of these recipes are packaged in `more_itertools`:

Substantially all of these recipes and many, many others can be installed from the more-itertools project found on the Python Package Index:

`pip install more-itertools`

The extended tools offer the same high performance as the underlying toolset. The superior memory performance is kept by processing elements one at a time rather than bringing the whole iterable into memory all at once. Code volume is kept small by linking the tools together in a functional style which helps eliminate temporary variables. High speed is retained by preferring “vectorized” building blocks over the use of for-loops and generators which incur interpreter overhead.

``````from more_itertools import all_equal

all_equal(map(type, iterable))
``````

or with `isinstance` and a known type `int` (as per the original question)

``````all_equal(map(lambda x: isinstance(x, int), iterable))
``````

These two ways are more concise than Enrico’s suggestion and handle ‘void iterators’ (e.g. `range(0)`) just as Enrico’s function does.

``````all_equal(map(type, range(0))) # True
all_equal(map(type, range(1))) # True
all_equal(map(type, range(2))) # True

all_equal(map(lambda x: isinstance(x, int), range(0))) # True
all_equal(map(lambda x: isinstance(x, int), range(1))) # True
all_equal(map(lambda x: isinstance(x, int), range(2))) # True
``````

``````    def c(x):